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45, 2nd Floor, Maharishi Dayanand Marg,, Corner Market, Malviya Nagar, New Delhi - 110017, Tel : 49842349 / 49842350, , No part of this publication may be reproduced in, any form without prior permission of the publisher., The author and the publisher do not take any legal, responsibility for any errors or misrepresentations that, might have crept in. We have tried and made our best, efforts to provide accurate up-to-date information in, this book., , All Right Reserved, , © Copyright, Disha, , Corporate, Office, , DISHA PUBLICATION, , By, Raghvendra Kumar Sinha, Brajendra Kumar Dubey, , www.dishapublication.com, , www.mylearninggraph.com, , Books &, ebooks for, School &, Competitive, Exams, , Etests, for, Competitive, Exams, , Write to us at feedback_disha@aiets.co.in

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Contents, • Latest Revised Syllabus for Academic Year (2021-2022), (Issued by CBSE on 28-07-2021), • CBSE Sample Paper 2021-22 (Basic) with solutions, , i–iii, , SQP 21-22-1–12, , (Issued by CBSE on 02-09-2021), • Objective Questions and Solutions , CBSE Sample Paper 2020-21 (Basic), , SQP 20-21-1–4, , • Objective Questions and Solutions , , SP 2020-1–2, , (Basic) CBSE Board 2020 Solved Paper All India, • Objective Questions and Solutions , , QB 1–14, , CBSE Questions Bank 2021, , 10 Sample Papers with OMR Answer Sheets, • Sample Paper-1, , SP-1–8, , • Sample Paper-2, , SP-9–16, , • Sample Paper-3, , SP-17–24, , • Sample Paper-4, , SP-25–32, , • Sample Paper-5, , SP-33–40, , • Sample Paper-6, , SP-41–48, , • Sample Paper-7, , SP-49–56, , • Sample Paper-8, , SP-57–64, , • Sample Paper-9, , SP-65–72, , • Sample Paper-10, , SP-73–78, , SOLUTIONS TO SAMPLE PAPERS 1-10, , S-1–42

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Note for Students, Dear Aspirants,, All sample papers of Disha’s “Super-10 Mock Test”, Class-10, Mathematics are as per latest CBSE, SAMPLE PAPER 2021-22 issued by CBSE on 02nd September, 2021, Each SAMPLE PAPER contains, Section-A has 20 MCQs , attempt any 16 out of 20, , (16×1=16 Marks), , Section-B has 20 MCQs , attempt any 16 out of 20, , (16×1=16 Marks), , Section-C has 10 MCQs based on two Case Studies, attempt any 8 out of 10, , (8×1=8 Marks), , Marking Scheme, •, , Each question carries 1 mark, , •, , There is no negative marking., , All SAMPLE PAPERS based on Revised Academic curriculum for the session 2021-22 issued by CBSE, on 28th July, 2021, For detailed revised CBSE Syllabus & Latest SAMPLE PAPERS, visit, http://www.cbseacademic.nic.in/web_material/CurriculumMain22/termwise/Secondary/Mathematics_, Sec_2021-22.pdf, http://www.cbseacademic.nic.in/web_material/SQP/ClassX_2021_22/MathsBasic-SQP.pdf, All the best, Disha Experts

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Latest Revised Syllabus Issued by CBSE on 28-07-2021, for Academic Year (2021-2022), Mathematics (Code no. 041), TERM-I, Unit No., I, Number Systems, , Unit Name, , Marks, 06, , II, III, IV, V, , Algebra, Coordinate Geometry, Geometry, Trigonometry, , 10, 06, 06, 05, , VI, VII, , Mensuration, Statistics & Probability, Total, , 04, 03, 40, 10, 50, , Internal Assessment, Total, Unit I: Number Systems, 1. Real Number, , Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating, through examples, Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals., , , , Euclid’s division lemma, Proofs of irrationality of, , 2, 3 5, , Unit II : Algebra, 2. Polynomials, Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials only., , , , Statement and simple problems on division algorithm for polynomials with real coefficients., , 3. Pair of Linear Equations in Two Variables, Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency., Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by, substitution, by elimination. Simple situational problems. Simple problems on equations reducible to linear equations., , , , cross multiplication method, , Unit III : Coordinate Geometry, 4. LINES (In two-dimensions), Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division)., , , , Area of a triangle., , Unit IV : Geometry, 5. Triangles, Definitions, examples, counter examples of similar triangles., 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides, are divided in the same ratio., (i)

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2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side., 3. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles, are similar., 4. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two, triangles are similar., 5. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are, proportional, the two triangles are similar., 6. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, the triangles on, each side of the perpendicular are similar to the whole triangle and to each other., 7. (Motivate) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides., 8. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides., 9. (Motivate) In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angles opposite to, the first side is a right angle., Unit V : Trigonometry, 6. Introduction to Trigonometry, Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); Values of the, trigonometric ratios of 30°, 45° and 60°. Relationships between the ratios., Trigonometric Identities, Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given., , , , Motive the ratios whichever are defined at 0° and 90°. Trigonometric ratios of complementary angles, , Unit VI : Mensuration, 7. Areas Related to Circles, Motivate the area of a circle; area of sectors and segments of a circle. Problems based on areas and perimeter / circumference, of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of, 60° and 90° only. Plane figures involving triangles, simple quadrilaterals and circle should be taken.), , , , Area of segment for central angle 120°, , Unit VII : Statistics and Probability, 8. Probability, Classical definition of probability. Simple problems on finding the probability of an event., , TERM-II, Unit No., I, Algebra (Cont.), II, Geometry (Cont.), III, Trigonometry (Cont.), IV, V, , Unit Name, , Mensuration (Cont.), Statistics & Probability (Cont.), Total, Internal Assessment, Total, , (ii), , Marks, 10, 09, 07, 06, 08, 40, 10, 50

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Unit : Algebra, 1. Quadratic Equations, Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization,, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic, equations related to day to day activities to be incorporated., , , , Problems on equations reducible to quadratic equations., , 2. Arithmetic Progressions, Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. application of, Arithmetic Progressions in solving daily life problems., , , , Applications based on sum to n terms of an A.P., , Unit : Geometry, 3. Circles, Tangent to a circle at, point of contact, 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact., 2. (Prove) The lengths of tangents drawn from an external point to a circle are equal., 4. Constructions, 1. Division of a line segment in a given ratio (internally)., 2. Tangents to a circle from a point outside it., , , , Construction of a triangle similar to a given triangle., , Unit : Trigonometry, 5. Some applications of trigonometry, Heights and Distances: Angle of elevation, Angle of Depression., Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation /, depression should be only 30°, 45°, 60°., Unit : Mensuration, 6. Surface Areas and Volumes, 1. Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right, circular cylinders/cones., , , , Frustum of a cone., , 2. Problems involving converting one type of metallic solid into another and other mixed problems. (Problems with combination of not more than two different solids be taken)., Unit VII : Statistics and Probability, 7. Statistics, Mean, median and mode of grouped data (bimodal situation and step deviation method for finding the mean to be avoided)., Mean by direct method and assumed mean method only., , , , Cumulative frequency graph., , (iii)

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CBSE Sample Paper 2021-2022, (Basic) with Solutions Term-1, Time Allowed: 90 minutes, , Sample Question Paper, Class- X Session- 2021-22, TERM 1, Subject- Mathematics (Basic), , Maximum Marks: 40, , General Instructions:, 1. The question paper contains three parts A, B and C., 2. Section A consists of 20 questions of 1 mark each. Attempt any 16 questions., 3. Section B consists of 20 questions of 1 mark each. Attempt any 16 questions., 4. Section C consists of 10 questions based on two Case Studies. Attempt any 8 questions., 5. There is no negative marking., , Q.NO., 1, , 2, , 3, , 4, , 5, , 6, , 7, , SECTION A, Section A consists of 20 questions. Any 16 questions are to be attempted, A box contains cards numbered 6 to 50. A card is drawn at random from the box. The, probability that the drawn card has a number which is a perfect square like 4,9….is, (a) 1/45, (b) 2/15, (c) 4/45, (d) 1/9, In a circle of diameter 42cm ,if an arc subtends an angle of 60 ˚ at the centre where, ∏=22/7,then the length of the arc is, (a) 22/7 cm, (b) 11cm, (c) 22 cm, (d) 44 cm, If sinƟ = x and secƟ = y , then tanƟ is, (a) xy, (b) x/y, (c) y/x, (d) 1/xy, The pair of linear equations y = 0 and y =-5 has, (a) One solution, (b) Two solutions, (c) Infinitely many solutions, (d) No solution, A fair die is thrown once. The probability of even composite number is, (a) 0, (b) 1/3, (c) 3/4, (d) 1, 8 chairs and 5 tables cost Rs.10500, while 5 chairs and 3 tables cost Rs.6450. The cost of, each chair will be, (a) Rs. 750, (b) Rs.600, (c) Rs. 850, (d) Rs. 900, If cosƟ+cos2Ɵ =1,the value of sin2Ɵ+sin4Ɵ is, (a) -1, (b) 0, (c) 1, (d) 2, , MARKS, 1, , 1, , 1, , 1, , 1, , 1, , 1

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Mathematics, , SQP 21-22-2, , 8, , 9, , 10, , 11, , The decimal representation of, , 23, , 23 × 52, , will be, , (a) Terminating, (b) Non-terminating, (c) Non-terminating and repeating, (d) Non-terminating and non-repeating, The LCM of 23X32 and 22X33 is, (a) 23, (b) 33, (c) 23X33, (d) 22X32, The HCF of two numbers is 18 and their product is 12960. Their LCM will be, (a) 420, (b) 600, (c) 720, (d) 800, In the given figure, DE II BC. Which of the following is true?, , (a) 𝑥𝑥 =, (b) 𝑦𝑦, (c), (d), , =, , 𝑥𝑥 =, , 𝑥𝑥 𝑎𝑎, =, 𝑦𝑦 𝑏𝑏, , 1, , 1, , 1, , 1, , 𝑎𝑎+𝑏𝑏, 𝑎𝑎𝑎𝑎, 𝑎𝑎𝑎𝑎, , 𝑎𝑎+𝑏𝑏, 𝑎𝑎𝑎𝑎, , 𝑎𝑎+𝑏𝑏, , 12, , The co-ordinates of the point P dividing the line segment joining the points A (1,3) and B (4,6), internally in the ratio 2:1 are, (a), (2,4), (b), (4,6), (c), (4,2), (d), (3,5), , 1, , 13, , The prime factorisation of 3825 is, (a) 3x52x21, (b) 32x52x35, (c) 32x52x17, (d) 32x25x17, , 1, , 14, , In the figure given below, AD=4cm,BD=3cm and CB=12 cm, then cotƟ equals, , 1, , (a), (b), (c), (d), , 3/4, 5/12, 4/3, 12/5

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CBSE Sample Paper 2021-2022, 15, , If ABCD is a rectangle , find the values of x and y, , (a), (b), (c), (d), 16, , 17, , 18, , 19, , 20, , 22, , 1, , X=10,y=2, X=12,y=8, X=2,y=10, X=20,y=0, , In an isosceles triangle ABC, if AC=BC and AB2=2AC2, then the measure of angle C will be, (a) 30˚, (b) 45˚, (c) 60˚, (d) 90˚, If -1 is a zero of the polynomial p(x)=x2-7x-8 , then the other zero is, (a) -8, (b) -7, (c) 1, (d) 8, In a throw of a pair of dice, the probability of the same number on each die is, (a) 1/6, (b) 1/3, (c) 1/2, (d) 5/6, , 1, , The mid-point of (3p,4) and (-2,2q) is (2,6) . Find the value of p+q, (a) 5, (b) 6, (c) 7, (d) 8, , 1, , The decimal expansion of, (a), (b), (c), (d), , 21, , SQP 21-22-3, , 1, 2, 3, 4, , 147, 120, , will terminate after how many places of decimals?, , SECTION B, Section B consists of 20 questions of 1 mark each. Any 16 questions are to be attempted, The perimeter of a semicircular protractor whose radius is ‘r’ is, (a), π + 2r, (b), π+r, (c), πr, (d), πr + 2r, If P (E) denotes the probability of an event E, then, (a), (b), (c), (d), , 0< P(E) ⩽1, 0 < P(E) < 1, 0 ≤ P(E) ≤1, 0 ⩽P(E) <1, , 1, , 1, , 1, , 1, , 1

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Mathematics, , SQP 21-22-4, , 23, , 24, , 25, , 26, , In ∆ABC, ˂B=90˚ and BD ꓕ AC. If AC = 9cm and AD = 3 cm then BD is equal to, (a) 2√2 cm, (b) 3√2 cm, (c) 2√3 cm, (d) 3√3 cm, The pair of linear equations 3x+5y=3 and 6x+ky=8 do not have a solution if, (a) K=5, (b) K=10, (c) k≠10, (d) k≠5, If the circumference of a circle increases from 2∏ to 4∏ then its area _____ the original, area, (a) Half, (b) Double, (c) Three times, (d) Four times, Given that sinƟ=a/b ,then tanƟ is equal to, 𝑏𝑏, (a) √𝑎𝑎 2 2, (b), (c), (d), , 27, , 28, , 29, , 30, , 31, , 32, , 1, , 1, , 1, , 1, , +𝑏𝑏, , 𝑏𝑏, , √𝑏𝑏2 −𝑎𝑎2, 𝑎𝑎, √𝑎𝑎2 −𝑏𝑏2, 𝑎𝑎, √𝑏𝑏2 −𝑎𝑎2, , If x = 2sin2Ɵ and y = 2cos2Ɵ+1 then x+y is, (a) 3, (b) 2, (c) 1, (d) 1/2, If the difference between the circumference and the radius of a circle is 37cm ,∏=22/7, the, circumference (in cm) of the circle is, (a) 154, (b) 44, (c) 14, (d) 7, The least number that is divisible by all the numbers from 1 to 10 (both inclusive), (a) 100, (b) 1000, (c) 2520, (d) 5040, Three bells ring at intervals of 4, 7 and 14 minutes. All three rang at 6 AM. When will they, ring together again?, (a) 6:07 AM, (b) 6:14 AM, (c) 6:28 AM, (d) 6:25 AM, What is the age of father, if the sum of the ages of a father and his son in years is 65 and, twice the difference of their ages in years is 50?, (a) 40 years, (b) 45 years, (c) 55 years, (d) 65 years, What is the value of (tanƟ cosecƟ)2-(sinƟ secƟ)2, (a) -1, (b) 0, (c) 1, (d) 2, , 1, , 1, , 1, , 1, , 1, , 1

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CBSE Sample Paper 2021-2022, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , The perimeters of two similar triangles are 26 cm and 39 cm.The ratio of their areas will be, (a) 2:3, (b) 6:9, (c) 4:6, (d) 4:9, There are 20 vehicles-cars and motorcycles in a parking area. If there are 56 wheels, together, how many cars are there?, (a) 8, (b) 10, (c) 12, (d) 20, A man goes 15m due west and then 8m due north. How far is he from the starting point?, (a) 7m, (b) 10m, (c) 17m, (d) 23m, What is the length of an altitude of an equilateral triangle of side 8cm?, (a) 2√3 cm, (b) 3√3 cm, (c) 4√3 cm, (d) 5√3 cm, If the letters of the word RAMANUJAN are put in a box and one letter is drawn at random., The probability that the letter is A is, (a), 3/5, (b), 1/2, (c), 3/7, (d), 1/3, Area of a sector of a circle is 1/6 to the area of circle. Find the degree measure of its minor, arc., (a) 90˚, (b) 60˚, (c) 45˚, (d) 30˚, A vertical stick 20m long casts a shadow 10m long on the ground. At the same time a tower, casts a shadow 50m long. What is the height of the tower?, (a) 30m, (b) 50m, (c) 80m, (d) 100m, What is the solution of the pair of linear equations 37x+43y=123, 43x+37y=117?, (a) x = 2,y = 1, (b) x = -1,y = 2, (c) x = -2,y = 1, (d) x = 1,y = 2, SECTION C, Case study based questions, Section C consists of 10 questions of 1 mark each. Any 8 questions are to be attempted., Case Study -1, , Pacific Ring of Fire, , SQP 21-22-5, , 1, , 1, , 1, , 1, , 1, , 1, , 1, , 1

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Mathematics, , SQP 21-22-6, , The Pacific Ring of Fire is a major area in the, basin of the Pacific Ocean where many, earthquakes and volcanic eruptions occur. In a, large horseshoe shape, it is associated with a, nearly continuous series of oceanic trenches,, volcanic arcs, and volcanic belts and plate, movements., , https://commons.wikimedia.org/wiki/File:Pacifick%C3%BD_ohniv%C3%BD_kruh.png, , Fault Lines, Large faults within the, Earth's crust result, from the action of, plate tectonic forces,, with the largest, forming the boundaries, between the plates., Energy release, associated with rapid, movement on active, faults is the cause of, most earthquakes., , https://commons.wikimedia.org/wiki/File:Faults6.png, Positions of some countries in the Pacific ring of fire is shown in the square grid below., , Based on the given information, answer the questions NO. 41-45

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CBSE Sample Paper 2021-2022, , 41, , 42, , 43, , SQP 21-22-7, , The distance between the point Country A and Country B is, (a) 4 units, (b) 5 units, (c) 6 units, (d) 7 units, Find a relation between x and y such that the point (x,y) is equidistant from the Country C and, Country D, (a) x-y = 2, (b) x+y = 2, (c) 2x-y = 0, (d) 2x+y = 2, , 1, , The fault line 3x + y – 9 = 0 divides the line joining the Country P(1, 3) and, Country Q(2, 7) internally in the ratio, , 1, , 1, , (a) 3 : 4, (b) 3 : 2, (c) 2 : 3, (d) 4 : 3, 44, , 45, , The distance of the Country M from the x-axis is, (a) 1 units, (b) 2 units, (c) 3 units, (d) 5 units, What are the co-ordinates of the Country lying on the mid-point of Country A and Country D?, (a) (1, 3), (b) (2, 9/2), (c) (4, 5/2), (d) (9/2, 2), , 1, , 1, , Case Study -2, ROLLER COASTER POLYNOMIALS, , Polynomials are everywhere. They play a key role in, the study of algebra, in analysis and on the whole, many mathematical problems involving them., Since, polynomials are used to describe curves of, various types engineers use polynomials to graph, the curves of roller coasters., https://images.app.goo.gl/WfcM1aRTHjjqtyT27, 46, , Based on the given information, answer the questions NO. 46-50., If the Roller Coaster is represented by the following graph y=p(x) , then name the type of the, polynomial it traces., , 1

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SQP 21-22-8, , 47, , (a) Linear, (b) Quadratic, (c) Cubic, (d) Bi-quadratic, The Roller Coasters are represented by the following graphs y=p(x). Which Roller Coaster has, more than three distinct zeroes?, (a), , (b), , (c ), , Mathematics, , 1

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CBSE Sample Paper 2021-2022, , SQP 21-22-9, , (d), , 48, , If the Roller Coaster is represented by the cubic polynomial t(x)= px3+qx2+rx+s ,then which of the, following is always true, (a) s≠0, (b) r≠0, (c) q≠0, (d) p≠0, , 49, , 50, , 1, , 1, , If the path traced by the Roller Coaster is represented by the above graph y=p(x), find the, number of zeroes?, (a) 0, (b) 1, (c) 2, (d) 3, , If the path traced by the Roller Coaster is represented by the above graph y=p(x), find its zeroes?, (a) -3, -6, -1, (b) 2, -6, -1, (c) -3, -1, 2, (d) 3, 1, -2, , 1

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Mathematics, , SQP 21-22-10, , Marking Scheme, Class- X Session- 2021-22, TERM 1, Subject- Mathematics (Basic), Q., N., 1, 2, 3, 4, 5, 6, , CORRECT, OPTION, (d), (c), (a), (d), (b), (a), , 7, , (c), , 8, 9, 10, , (a), (c ), (c), , 11, , (c), , 12, , (d), , 13, 14, , (c), (d), , 15, , (a), , 16, , (d), , 17, , (d), , 18, 19, , (a), (b), , 20, , (c ), , HINTS/SOLUTION, P(perfect Square)=5/45=1/9, length of the arc= Ɵ /360˚ (2πr)=(60˚/360˚)x2x(22/7)x21=22cm, TanƟ = sinƟ/cosƟ = sinƟxsecƟ = xy, The lines are parallel hence No solution, P(even composite no) =2/6=1/3, Let the cost of one chair=Rs. x, Let the cost of one table=Rs. y, 8x+5y=10500, 5x+3y=6450, Solving the above equations, Cost of each chair= x= Rs. 750, CosƟ=I-cos2Ɵ=sin2Ɵ, Therefore Sin2Ɵ+sin4Ɵ=cosƟ+cos2Ɵ=1, Terminating, 23x33, 1st No. x 2nd No. = HCF X LCM, 12960=18 X LCM, LCM=720, AE/AC=DE/BC=a/a+b=x/y, X=ay/(a+b), (2x4+1x1)/3 , (2x6+1x3)/3, =(3,5), 3825=32x52x17, AB2=AD2+BD2, AB=5cm, AC2=AB2+CB2, AC=13 cm, Cot 𝛉𝛉=CB/AB=12/5, x+y=12, X-y=8, Solving the above equations, X=10,y=2, AB2=AC2+AC2, =AC2+BC2, Hence, angle C=90°, Let the zeroes be a and b, Then, a=-1 , a+b=-(-7)/1, Hence, b=7+1=8, P(same no on each die)=6/36=1/6, (2,6)=((3p-2)/2, (4+2q)/2), 3p-2=4, 4+2q=12, P=2, q = 4 hence p+q = 6, 147/120= 49/40=49/23x5

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SQP 21-22-12, , 34, , (a ), , 35, , ( c), , 36, , ( c), , 37, 38, , (d), ( b), , 39, , ( d), , 40, , (d), , 41, , ( b), , 42, , (a), , 43, , (a ), , 44, 45, 46, 47, 48, 49, 50, , ( c), (b), (c), (d ), ( d), (d), ( c), , Mathematics, , A1/A2=(2/3)2=4/9, Let no of Cars=x, Let no of motorcycles=y, X+y=20, 4x+2y=56, Solving the above equations, No of cars=x=8, H2=P2+B2, H2=152+82, H=17m, (altitude)2=(side)2-(side/2)2, =82-42= 64-16 =48, Altitude=4√3 cm, P=3/9=1/3, Ɵ/360˚xπr2=1/6x πr2, Ɵ=60˚, Height of Vertical stick/Shadow of vertical stick=height of tower/shadow of tower, 20/10=Height of tower/50, Height of tower=100 m, 37x+43y=123 ____(1), 43x+37y=117 ____(2), Adding (1) and (2), X+y=3 ______(3), Subtracting (2) from (1), -x+y=1..............(4), Adding (3) and (4),, 2y=4, y=2, ⇒ x=1, ∴ solution is x=1 and y=2, AB=√{(4-1)2+(0-4)2}, =√(32+42), AB=5 units, 2, (x-7) +(y-1)2=(x-3)2+(y-5)2, X2+49-14x+y2+1-2y=x2+9-6x+y2+25-10y, Simplifying, x-y=2, 3x + y – 9 = 0, Let R divide the line in ratio k:1, R( 2k+1/k+1, 7k+3/k+1), 3(2k+1/k+1)+( 7k+3/k+1)-9=0, 4k-3=0, K=3/4, 3:4, Distance of M from X-axis=√(2-2)2+(0-3)2=√9=3units, ( (1+3)/2 , (4+5)/2) = (4/2, 9/2) = (2, 9/2), Cubic, Four Zeroes as the curve intersects the x-axis at 4 points, p≠0, 3 Zeroes as the curve intersects the x-axis at 3 points, -3,-1,2

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Objective Questions and Solutions, CBSE Sample Paper 2020-2021, SECTION-I, Section I has 12 questions of 1 mark each., 1., Express 156 as the product of primes., 2., Write a quadratic polynomial, sum of whose zeroes is 2, and product is –8., 3., Given that HCF (96,404) is 4, find the LCM (96,404)., OR, State the fundamental Theorem of Arithmetic., 4., On comparing the ratios of the coefficients, find out whether, the pair of equations x – 2y = 0 and 3x + 4y – 20 = 0 is, consistent or inconsistent., 5., If a and b are co-prime numbers, then find the HCF (a, b)., 6., Find the area of a sector of a circle with radius 6 cm if, , 22 ö, æ, angle of the sector is 60°. ç Take p =, 7 ÷ø, è, OR, A horse tied to a pole with 28m long rope. Find the, perimeter of the field where the horse can graze., , 7., , 22 ö, æ, ç Take p = 7 ÷, è, ø, In the given fig. DE || BC, ÐADE = 70° and ÐBAC = 50°,, then angle ÐBCA = ______., A, , 11., , OR, What type of straight lines will be represented by the, system of equations 2x + 3y = 5 and 4x + 6y = 7?, A bag contains 3 red balls and 5 black balls. A ball is, drawn at random from the bag. What is the probability, that the ball drawn is red?, OR, A die is thrown once. What is the probability of getting a, prime number?, , 12. Probability of an event E + Probability of the event E, (not E) is, ______., SECTION-II, The case study based questions are compulsory. Attempt any, 4 sub-parts of question. Each question carries 1 mark., 13. Class X students of a secondary school in Krishnagar, have been allotted a rectangular plot of a land for, gardening activity. Saplings of Gulmohar are planted on, the boundary at a distance of 1m from each other. There, is a triangular grassy lawn in the plot as shown in the fig., The students are to sow seeds of flowering plants on the, remaining area of the plot., , C, , B, P, , D, B, , C, OR, In the given figure, AD = 2cm, BD = 3 cm, AE = 3.5 cm and, AC = 7 cm. Is DE parallel to BC?, A, D, , 8., 9., 10., , R, , E, , E, , B, C, The cost of fencing a circular field at the rate of ` 24 per, metre is ` 5280. Find the radius of the field., If the perimeter and the area of a circle are numerically, equal, then find the radius of the circle., For what values of p does the pair of equations 4x + py +, 8 = 0 and 2x + 2y + 2 = 0 has unique solution?, , Q, A 1 2 3 4 5 6 7 8 910, D, Considering A as origin, answer question (i) to (v)., (i) Considering A as the origin, what are the coordinates, of A?, (a) (0, 1), (b) (1, 0), (c) (0, 0), (d) (–1, –1), (ii) What are the coordinates of P?, (a) (4, 6), (b) (6, 4), (c) (4, 5), (d) (5, 4), (iii) What are the coordinates of R?, (a) (6, 5), (b) (5, 6), (c) (6, 0), (d) (7, 4), (iv) What are the coordinates of D?, (a) (16, 0), (b) (0, 0), (c) (0, 16), (d) (16, 1)

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Mathematics, , SQP 20-21-2, , (v) What are the coordinate of P if D is taken as the, origin?, (a) (12, 2), (b) (–12, 6), (c) (12, 3), (d) (6, 10), 14., , (v) What is the area of the kite, formed by two, perpendicular sticks of length 6 cm and 8 cm?, (a) 48 cm2, (b) 14 cm2, 2, (c) 24 cm, (d) 96 cm2, 15. Due to heavy storm an electric wire got bent as shown in, the figure. It followed a mathematical shape. Answer the, following questions below., y, 6, 5, 4, 3, 2, 1, , Rahul is studying in X standard. He is making a kite to fly, it on a Sunday. Few questions came to his mind while, making the kite. Give answer to his questions by looking, at the figure., (i) Rahul tied the sticks at what angles to each other?, (a) 30°, (b) 60°, (c) 90°, (d) 45°, (ii) Which is the correct similarity criteria applicable for, smaller triangles at the upper part of this kite?, (a) RHS, (b) SAS, (c) SSA, (d) AAS, (iii) Sides of two similar triangles are in the ratio 4 : 9., Corresponding medians of these triangles are in the, ratio., (a) 2 : 3, (b) 4 : 9, (c) 81 : 16, (d) 16 : 81, (iv) In a triangle, if square of one side is equal to the sum, of the squares of the other two sides, then the angle, opposite the first side is a right angle. This theorem, is called as,, (a) Pythagoras theorem, (b) Thales theorem, (c) Converse of Thales theorem, (d) Converse of Pythagoras theorem, , –6 –5 –4 –3 –2 –1, –1, –2, –3, –4, –5, (i), (ii), , (iii), (iv), (v), , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , Name the shape in which the wire is bent, (a) spiral, (b) ellipse, (c) linear, (d) parabola, How many zeroes are there for the polynomial (shape, of the wire)?, (a) 2, (b) 3, (c) 1, (d) 0, The zeroes of the polynomial are, (a) –1, 5, (b) –1, 3, (c) 3, 5, (d) – 4, 2, What will be the expression of the polynomial?, (a) x2 + 2x – 3, (b) x2 – 2x + 3, 2, (c) x – 2x – 3, (d) x2 + 2x + 3, What is the value of the polynomial if x = –1?, (a) 6, (b) –18, (c) 18, (d) 0

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CBSE Sample Paper 2020-2021, , 1., 2., , 3., , 156 = 22 × 3 × 13, Quadratic polynomial is given by x2 – (a + b) x + ab, = x2 – 2x – 8, Here a + b = 2, ab = – 8, HCF × LCM = Product of two numbers, LCM (96, 404) =, , 96 ´ 404, 96 ´ 404, =, HCF(96, 404), 4, , SQP 20-21-3, , AD AE, ¹, BD EC, Hence, By converse of Thale's Theorem, DE is not parallel, to BC., , So,, , 8., , =, , LCM = 9696, , 4., , OR, Every composite number can be expressed (factorised), as a product of primes and this factorisation is unique,, apart from the order in which the factors occur., Given equations are x – 2y = 0, 3x + 4y – 20 = 0, , 1 -2, ¹, 3 4, a1 b1, As, a ¹ b is one condition for consistency.., 2, 2, 5., 6., , Therefore, the pair of equations is consistent., 1, HCF of co-prime numbers is always 1., , 220, 2p, , 220 × 7, = 35 m, 2 × 22, 9., Given Perimeter = Area (Numerically), 2pr = pr2, r = 2 units, 10. Let equations a, x + b, y + c1 = 0 and, a2x + b2y + c2 = 0, , So, r =, , then, , a1 b1, ¹, is the condition for the given pair of, a 2 b2, , equations to have unique solution., , 60° 22, ´ ´ (6)2, 360° 7, , 4 p, ¹, 2 2, p¹4, Therefore, for all real values of p except 4, the given pair, of equationsn will have a unique solution., OR, , a1 2 1, Here, a = 4 = 2, 2, , 22, ´ (28)cm = 176 cm, 7, By converse of Thale's theorem DE || BC, ÐADE = ÐABC = 70° (Corresponding angles), Given ÐBAC = 50°, ÐABC + ÐBAC + ÐBCA = 180° (Angle sum prop. of, triangles), 70° + 50° + ÐBCA = 180°, ÐBCA = 180° – 120° = 60°, OR, EC = AC – AE = (7 – 3.5) cm = 3.5 cm, , 5, b1 3 1, c, = = and 1 =, b2 6 2, c2 7, , = 2´, , AD 2, AE 3.5 1, = and, =, =, BD 3, EC 3.5 1, , ` 5280, = 220m, ` 24/metre, , r=, , 1 22, A = ´ ´ 36 = 18.86 cm2, 6 7, OR, Horse can graze in the field which is a circle of radius 28, cm., So, required perimeter = 2pr = 2.p(28) cm, , 7., , Total cost, Rate, , So, length of fence = Circumference of the field, 2pr = 220, , æ q ö 2, Given angle is q = 60°, Area of sector = ç, ÷ pr, è 360° ø, , A=, , Length of the fence of the circular field =, , 1 1 5, = ¹, 2 2 7, , a1 b1 c1, =, ¹, is the condition for which the given system, a 2 b 2 c2, , 11., , of equations will represent parallel lines., So, the given system of linear equations will represent a, pair of parallel lines., In the bag number of red balls = 3, Number of black balls, =5, Total number of balls = 5 + 3 = 8

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Mathematics, , SQP 20-21-4, , 3, 8, OR, Total number of possible outcomes (1, 2, 3, 4, 5, 6) = 6, There are 3 prime numbers = 2, 3, 5., , Probability of red balls =, , So, probability of getting a prime number is, 12., 13., , 1, (i), (ii), (iii), (iv), (v), , (c), (a), (a), (a), (b), , (0, 0) (Q A is origin), (4, 6), (6, 5), (16, 0), (–12, 6), , 3 1, =, 6 2, , 14. (i) (c) 90°, (ii) (b) SAS, (iii) (b) 4 : 9, Ratio of sides of similar triangles = Ratio of, corresponding medians, (iv) (d) Converse of Pythagoras theorem, (v) (a) 48 cm2, 15. (i) (d) parabola, (ii) (a) 2, Graph cuts at two distinct point on x-axis, (iii) (b) –1, 3, Points at where y-coordinate becomes zero, (iv) (c) x2 – 2x – 3, (v) (d) 0, As, x2 – 2x – 3 = (–1)2 – 2 (–1) – 3 = 1 + 2 – 3 = 0

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Objective Questions and Solutions, Basic Solved Paper 2020 (All India), SECTION-A, Question numbers 1 to 9 carry 1 mark each., Choice the correct option in question numbers 1 to 6., 1., The probability of an impossible event is, (a) 1, 2., , (c) not defined, , (d) 0, , (b) (6, –6), , (c) (6, –12), , (d), , æ3, ö, ç 2 , -3 ÷, è, ø, , (c) –8, , (d), , -, , 8 cot2A – 8 cosec2 A is equal to, (a) 8, , 4., , 1, 2, , If (3, –6) is the mid-point of the line segment joining (0, 0) and (x, y), then the point (x, y) is, (a) (–3, 6), , 3., , (b), , (b), , 1, 8, , 1, 8, , The point on x-axis which divides the line segment joining (2, 3) and (6, –9) in the ratio 1 : 3 is, (a) (4, –3), (b) (6, 0), (c) (3, 0), (d) (0, 3), 5., If a pair of linear equations is consistent, then the lines represented by them are, (a) parallel, (b) intersecting or coincident, (c) always coincident, (d) always intersecting, 6., 120 can be expressed as a product of its prime factors as, (a) 5 × 8 × 3, (b) 15 × 23, (c) 10 × 22 × 3, (d) 5 × 23 × 3, Fill in the blanks in question numbers 7 to 8., 7., If 2 is a zero of the polynomial ax2 – 2x, then the value of ‘a’ is ____________., 8., All squares are _____________ . (congruent/similar)., Answer the following question number 9., 9., A dice is thrown once. If getting a six, is a success, then find the probability of a failure.

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SP 2020-2, , 1., 2., , 3., 4., 5., 6., 7., 8., 9., , (d) 0, 0+ x, 0+ y, = 3,, = -6, 2, 2, x = 6, y = –12, (x, y) = (6, –12), (c) 8(cot2A – cosec2A) = 8 × (–1) = –8, , (c), , æ 1´ 6 + 3 ´ 2 ö æ 12 ö, , 0 ÷ = ç , 0 ÷ = (3, 0), (c) Required point = ç, è 1+ 3, ø è 4 ø, (b) Intersecting or coincident, (d) 5 × 23 × 3, 2 is a zero of ax2 – 2x then,, a(2)2 – 2 × 2 = 0, 4a = 4 Þ a = 1, similar, Probability of getting a success = P(getting a six), 1, 6, \ Probability of failure = 1 – P (getting a six), =, , =1-, , 1 5, = ., 6 6, , Mathematics

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Objective Questions and Solutions, CBSE Questions Bank-2021, Directions : Study the given case/study and answer the following questions., Case Study, To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library., There are two sections-section A and section B of grade X. There are 32 students in section A and 36 students in section B., , 1., 2., 3., 4., 5., , [From CBSE Question Bank-2021], What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among, students of Section A or Section B?, (a) 144, (b) 128, (c) 288, (d) 272, If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32, 36) is, (a) 2, (b) 4, (c) 6, (d) 8, 36 can be expressed as a product of its primes as, (a) 22 × 32, (b) 21 × 33, (c) 23 × 31, (d) 20 × 30, 7× 11 × 13 × 15 + 15 is a, (a) Prime number, (b) Composite number, (c) Neither prime nor composite (d) None of the above, If p and q are positive integers such that p = ab2 and q = a2b, where a, b areprime numbers, then the LCM (p, q) is, (a) ab, (b) a2b2, (c) a3b2, (d) a3b3, Case Study-II, , A seminar is being conducted by an Educational Organisation, where theparticipants will be educators of different subjects. The, number of participants in Hindi, English and Mathematics are 60, 84 and 108 respectively., , [From CBSE Question Bank-2021]

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Mathematics, , QB-2, , 6., 7., 8., 9., 10., , In each room the same number of participants are to be seated and all of them being in the same subject, hence maximum, number participants thatcan accommodated in each room are, (a) 14, (b) 12, (c) 16, (d) 18, What is the minimum number of rooms required during the event?, (a) 11, (b) 31, (c) 41, (d) 21, The LCM of 60, 84 and 108 is, (a) 3780, (b) 3680, (c) 4780, (d) 4680, The product of HCF and LCM of 60,84 and 108 is, (a) 55360, (b) 35360, (c) 45500, (d) 45360, 108 can be expressed as a product of its primes as, (a) 23 × 32, (b) 23 × 33, (c) 22 × 32, (d) 22 × 33, Case Study-III, , A Mathematics exhibition is being conducted in your school and one of your friendsis making a model of a factor tree. He has, some difficulty and asks for your help in completing a quiz for the audience., x, 5, , 2783, , y, , 253, , z, , 11, , [From CBSE Question Bank-2021], Observe the following factor tree and answer the following:, 11. What will be the value of x?, (a) 15005, (b) 13915, 12., 13., 14., , 15., , (c) 56920, , (d) 17429, , What will be the value of y?, (a) 23, (b) 22, , (c) 11, , (d) 19, , What will be the value of z?, (a) 22, (b) 23, , (c) 17, , (d) 19, , According to Fundamental Theorem of Arithmetic 13915 is a, (a) Composite number, (b) Prime number, (c) Neither prime nor composite, (d) Even number, The prime factorisation of 13915 is, (a) 5 × 113 × 132, (b) 5 × 113 × 232, (c) 5 × 112 × 23, , (d) 5 × 112 × 132, , Case Study-IV, The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is, an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges, and in architecture in a variety of forms.

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QB-3, , CBSE Questions Bank-2021, , [From CBSE Question Bank-2021], ax2, , 16. In the standard form of quadratic polynomial,, + bx + c, a, b and c are, (a) All are real numbers., (b) All are rational numbers., (c) ‘a’ is a non zero real number and b and c are any real numbers., (d) All are integers., 17. If the roots of the quadratic polynomial are equal, where the discriminant D = b2 – 4ac, then, (a) D > 0, (b) D < 0, (c) D ≥ 0, (d) D = 0, 18. If a and, (a) 4, , 1, α, , are the zeroes of the quadratic polynomial 2x2 – x + 8k, then k is, (b), , 1, 4, , (c), , 19. The graph of x2 + 1 = 0, (a) Intersects x-axis at two distinct points., (c) Neither touches nor intersects x-axis., , (d) 2, , (b) Touches x-axis at a point., (d) Either touches or intersects x-axis., , 20. If the sum of the roots is –p and product of the roots is –, , x , (a) k  – px 2 + + 1 (b), p , , , –1, 4, , , x , k  px 2 – –1, p , , , 1, , then the quadratic polynomial is, p, (c), , , 1, k  x 2 + px – , p, , , (d), , , 1, k  x 2 – px + , p, , , Case Study-V, An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and, modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In, the figure, one can observe that poses can be related to representation of quadratic polynomial., , TRIKONASANA, , ADHOMUKHA SAVASANA, , ADHO MUKHA SVANA, , [From CBSE Question Bank-2021], 21. The shape of the poses shown is, (a) Spiral, (b) Ellipse, , (c) Linear, , 22. The graph of parabola opens downwards, if__________., (a) a ≥ 0, (b) a = 0, (c) a < 0, , (d) Parabola, (d) a > 0

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Mathematics, , QB-4, , 23. In the graph, how many zeroes are there for the polynomial?, , 4, , –2, , –8, , (a) 0, (b) 1, 24. The two zeroes in the above shown graph are, (a) 2, 4, (b) –2, 4, , (c) 2, , (d) 3, , (c) –8, 4, , (d) 2, –8, , 2, , 25. The zeroes of the quadratic polynomial 4 3 x + 5 x – 2 3 are, (a), , 2, 3, ,, 3 4, , (b), , –, , 2, 3, ,, 3 4, , (c), , 2, 3, ,–, 4, 3, , (d), , –, , 2, 3, ,−, 4, 3, , Case Study-VI, Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player, uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor, on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing, quadratic polynomial., , v = 8.552 m/s, , 3, θ = 51.89°, , 1, , R = 7.239 m, , 2, , h = 3.048 m, , 4, , [From CBSE Question Bank-2021], , 26. The shape of the path traced shown is, (a) Spiral, (b) Ellipse, (c) Linear, 27. The graph of parabola opens upwards, if____________., (a) a = 0, (b) a < 0, (c) a > 0, 28. Observe the following graph and answer, , (d) Parabola, (d) a ≥ 0, , 6, , –4, , –3, , –2, , –1, , 2, , 1, , 2, , 3, , 4, , –2, –6, , In the above graph, how many zeroes are there for the polynomial?, (a) 0, (b) 1, (c) 2, , (d) 3

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QB-5, , CBSE Questions Bank-2021, 29. The three zeroes in the above shown graph are, (a) 2, 3, –1, (b) –2, 3, 1, 30. What will be the expression of the polynomial?, (a) x3 + 2x2 − 5x − 6 (b) x3 + 2x2 − 5x + 6, , (c) –3, –1, 2, (c) x3 + 2x2 + 5x − 6, Case Study-VII, , (d) –2, –3, –1, (d) x3 + 2x2 + 5x + 6, , A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while 1/4 mark is deducted for every, wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered, 120 questions and got 90 marks., Type of Question, True/False, , Marks given for correct answer, 1, , Marks deducted for wrong answer, 0.25, , [From CBSE Question Bank-2021], If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?, How many questions did he guess?, If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?, If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95, marks?, Case Study-VIII, Amit is planning to buy a house and the layout is given below. The design and the measurement has been made such that areas, of two bedrooms and kitchen together is 95 sq.m., x, 2, y, 31., 32., 33., 34., , 5m, , Bedroom 1, , Bath, room, , Kitchen, , 2m, Living Room, 5m, , Bedroom 2, 15 m, , [From CBSE Question Bank-2021], Based on the above information, answer the following questions:, 35. Form the pair of linear equations in two variables from this situation., 36. Find the length of the outer boundary of the layout., 37. Find the area of each bedroom and kitchen in the layout., 38. Find the area of living room in the layout., 39. Find the cost of laying tiles in kitchen at the rate of ` 50 per sq.m, Case Study-IX, It is common that Governments revise travel fares from time to time based on various factors such as inflation ( a general increase in, prices and fall in the purchasing value of money) on different types of vehicles like auto, rickshaws, taxis, radio cab etc. The auto charges, in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations., , Name of the city Distance travelled (km) Amount paid (`), City A, 10, 75, 15, 110, City B, 8, 91, 14, 145

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Mathematics, , QB-6, , Situation 1: In city A, for a journey of 10 km, the charge paid is ` 75 and for a journey of 15 km, the charge paid is ` 110., Situation 2: In a city B, for a journey of 8 km, the charge paid is ` 91 and for a journey of 14km, the charge paid is ` 145., [From CBSE Question Bank-2021], Refer situation 1, 40., , If the fixed charges of auto rickshaw be ` x and the running charges be ` y km/hr, the pair of linear equations representing, the situation is, (a) x + 10y = 110, x + 15y = 75, (b) x + 10y = 75, x + 15y = 110, (c) 10x + y = 110, 15x + y = 75, (d) 10x + y = 75, 15x + y = 110, , 41., , A person travels a distance of 50km. The amount he has to pay is, (a) ` 155, (b) ` 255, (c) ` 355, , (d) ` 455, , Refer situation 2, 42., 43., , What will a person have to pay for travelling a distance of 30km?, (a) ` 185, (b) ` 289, (c) ` 275, The graph of lines representing the conditions are: (situation 2), Y, 25, (20, 25), 20, 15, 10, (a), (b), (30, 5), 5 (0, 5), X¢ –5 0, –5, –10, , (c), , 50, 45, 40, 35, 30, 25, 20, 15, , Y¢, , (11, 10) (19, 9), , (47, 7), , (27, 8), , X¢ –5 0, –5, –10, , (20, 10), (12.5, 0), , 5 10 15 20 25 30 35 X, (25, –10), (5, –10), , Y, 25, 20, 15, 10, 5 (0, 10), , (d), , 10 (5, 10), 5, 0, , Y, 25, 20, 15, 10 (0, 10), 5, X¢ –5 0, –5, –10, Y¢, , 5 10 15 20 25 30 35 X, , (d) ` 305, , (15, 15), (35, 10), , 5 10 15 20 25 30 35 X, (15, –5), Y¢, , 5 10 15 20 25 30 35 40 45 50 55, , Case Study-X, In order to conduct Sports Day activities in your School, lines have been drawn with chalk powder at a distance of 1, m each, in a rectangular shaped ground ABCD, 100 flowerpots have been placed at a distance of 1 m from each other, along AD, as shown in given figure below. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag., Preet runs 1/5 th distance AD on the eighth line and posts a red flag., [From CBSE Question Bank 2021]

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QB-7, , CBSE Questions Bank-2021, C, , D, , G, R, , 2, 1, A, , 44., 45., 46., 47., , 48., , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , Find the position of green flag, (a) (2, 25), (b) (2, 0.25), (c) (25, 2), (d) (0, –25), Find the position of red flag, (a) (8, 0), (b) (20, 8), (c) (8, 20), (d) (8, 0.2), What is the distance between both the flags?, (a) √41, (b) √11, (c) √61, (d) √51, If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her, flag?, (a) (5, 22.5), (b) (10, 22), (c) (2, 8.5), (d) (2.5, 20), If Joy has to post a flag at one-fourth distance from green flag ,in the line segment joining the green and red flags, then where, should he post his flag?, (a) (3.5, 24), (b) (0.5, 12.5), (c) (2.25, 8.5), (d) (25, 20), Case Study-XI, , Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles.The height of, Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow, 50m long on the ground and the house of Ajay casts 20m shadow on the ground., [From CBSE Question Bank-2021], , Vijay's House, , 49., 50., 51., 52., 53., , Tower, , Ajay's House, , What is the height of the tower?, (a) 20m, (b) 50m, (c) 100m, (d) 200m, What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m?, (a) 75m, (b) 50m, (c) 45m, (d) 60m, What is the height of Ajay’s house?, (a) 30m, (b) 40m, (c) 50m, (d) 20m, When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house?, (a) 16m, (b) 32m, (c) 20m, (d) 8m, When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house?, (a) 15m, (b) 32m, (c) 16m, (d) 8m

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Mathematics, , QB-8, , Case Study-XII, Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond, as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with, another point C are a distance of 12m, connecting C to point D at a distance of 40m from point C and the connecting D to the point A, which is are a distance of 30m from D such the ∠ADC=90° ., [From CBSE Question Bank-2021], 12 m, , B, , A, , C, , 30, m, , 40, , m, , D, , 54., , 55., , Which property of geometry will be used to find the distance AC?, (a) Similarity of triangles , , (b) Thales Theorem, , (c) Pythagoras Theorem , , (d) Area of similar triangles, , What is the distance AC?, (a) 50m, , 56., , 58., , (c) 100m, , (d) 70m, , Which is the following does not form a Pythagoras triplet?, (a) (7, 24, 25), , 57., , (b) 12m, (b) (15, 8, 17), , Find the length AB?, (a) 12m, (b) 38m, Find the length of the rope used., (a) 120m, (b) 70m, , (c) (5, 12, 13), , (d) (21, 20, 28), , (c) 50m, , (d) 100m, , (c) 82m, , (d) 22m, , Case Study-XIII, , In ∆ABC, right angled at B, , C, 3cm, B, , A, , AB + AC = 9 cm and BC = 3cm., 59. The value of cot C is, , (a), , 3, 4, , [From CBSE Question Bank-2021], (b), , 1, 4, , (c), , 5, 4, , (d) None of these, , (b), , 5, 3, , (c), , 1, 3, , (d) None of these, , (c) –1, , (d) None of these, , 60. The value of sec C is, (a), , 4, 3, , 61. sin2C + cos2C =, (a) 0, , (b) 1

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QB-9, , CBSE Questions Bank-2021, Case Study-XIV, , Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam, in Tamil Nadu., During the festival of Onam, your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition, , suggests two designs given below., [From CBSE Question Bank-2021], , Observe these carefully., A, A, , B, , D, , C, , B, C, I, , II, , Design I: This design is made with a circle of radius 32cm leaving equilateral triangle ABC in the middle as shown in the, given figure., Design II: This Pookalam is made with 9 circular design each of radius 7cm., Refer Design I:, 62. The side of equilateral triangle is, (a) 12√3 cm, (b) 32√3 cm, 63., , The altitude of the equilateral triangle is, (a) 8 cm, (b) 12 cm, , (c) 48 cm, , (d) 64 cm, , (c) 48 cm, , (d) 52 cm, , Refer Design II:, 64., , The area of square is, (a) 1264 cm2, (b) 1764 cm2, , (c) 1830 cm2, , (d) 1944 cm2, , 65., , Area of each circular design is, (a) 124 cm2, (b) 132 cm2, , (c) 144 cm2, , (d) 154 cm2, , 66., , Area of the remaining portion of the square ABCD is, (a) 378 cm2, (b) 260 cm2, , (c) 340 cm2, , (d) 278 cm2, , Case Study-XV, A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat., Designs of some brooch are shown below. Observe them carefully., , A, , [From CBSE Question Bank-2021], , B, , Design A: Brooch A is made with silver wire in the form of a circle with diameter 28mm. The wire used for making 4 diameters, which divide the circle into 8 equal parts., Design B: Brooch b is made two colours-Gold and silver. Outer part is made with Gold. The circumference of silver part is 44mm, and the gold part is 3mm wide everywhere.

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Mathematics, , QB-10, , Refer to Design A, 67., 68., , The total length of silver wire required is, (a) 180 mm, (b) 200 mm, , (c) 250 mm, , (d) 280 mm, , The area of each sector of the brooch is, (a) 44 mm2, (b) 52 mm2, , (c) 77 mm2, , (d) 68 mm2, , The circumference of outer part (golden) is, (a) 48.49 mm, (b) 82.2 mm, , (c) 72.50 mm, , (d) 62.86 mm, , The difference of areas of golden and silver parts is, (a) 18 p, (b) 44 p, , (c) 51 p, , (d) 64 p, , Refer to Design B, 69., 70., 71., , A boy is playing with brooch B. He makes revolution with it along its edge.How many complete revolutions must it, take to cover 80 p mm ?, (a) 2, (b) 3, (c) 4, (d) 5, , Case Study-XVI, On a weekend Rani was playing cards with her family. The deck has 52 cards. If her brother drew one card ., , [From CBSE Question Bank-2021], 72., , Find the probability of getting a king of red colour., , 73., , 1, 1, (b), 13, 26, Find the probability of getting a face card., , 74., , 1, 1, (b), 26, 13, Find the probability of getting a jack of hearts., , 75., , 1, 1, (b), 26, 52, Find the probability of getting a red face card., , 76., , 1, 13, Find the probability of getting a spade., , (a), , (a), , (a), , (a), , 3, 26, , (b), , (a), , 1, 26, , (b), , 1, 13, , (c), , 1, 52, , (d), , 1, 4, , (c), , 2, 13, , (d), , 3, 13, , (c), , 3, 52, , (d), , 3, 26, , (c), , 1, 52, , (d), , 1, 4, , (c), , 1, 26, , (d), , 1, 4

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QB-11, , CBSE Questions Bank-2021, Case Study-XVII, Rahul and Ravi planned to play Business (board game) in which they were supposed to use two dice., , 77., , [From CBSE Question Bank-2021], Ravi got first chance to roll the dice. What is the probability that he got thesum of the two numbers appearing on the, top face of the dice is 8?, 1, 5, 1, (b), (c), (d) 0, 26, 36, 18, Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the, dice is 13?, (a), , 78., , 5, 1, (c), (d) 0, 36, 18, Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the, top face of the dice is less than or equal to 12 ?, (a) 1, , 79., , (b), , 5, 1, (c), (d) 0, 36, 18, Rahul got next chance. What is the probability that he got the sum of the two numbers appearing on the top face of the dice, is equal to 7 ?, (a) 1, , 80., , (b), , 5, 5, 1, (b), (c), (d) 0, 9, 36, 6, Now it was Ravi’s turn. He rolled the dice. What is the probability that he got the sum of the two numbers appearing on the, top face of the dice is greater than 8 ?, (a), , 81., , (a) 1, , (b), , 5, 36, , 1., (c) For getting least number of books,, taking LCM of 32, 36, 4, 8, 9, , 2., , 32, 36, 8, 9, 1, 9, 1, 1, ⇒ 4 × 8 × 9 = 288, (b) HCF of 32, 36 is, 4, , , =4, , 32, 36, 8, 9, , (c), , 1, 18, , (d), , 5, 18, , 3., (a) 36 is expressed as prime, 36 = 2 × 2 × 3 × 3 = 22 × 32, 4., (b) 7 × 11 × 13 × 15 + 15, ⇒ 15 (7 × 11 × 13 + 1), so given no. is a composite number., 5., (b) Given a, b are prime number. So, LCM of p, q, where p = ab2, q = a2b, p = a × b × b, q = a × b × a, a × b × b × a ⇒ a2b2, 6., (b) For maximum number of participants, taking HCF of, 60, 84 and 108 , 12 60, 84, 108, 5, 7, 9, , = 12

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Mathematics, , QB-12, , 7., (d), , 8., (a), , 9., (d), 10. (d), 11. (b), 12., 13., 14., 15., , Minimum number of rooms required are, −5 ± 25 + 4 × 4 3 × 2 3 −5 ± 11, 5 + 7 + 9 = 21, = =, 8 3, 8 3, , LCM of 60, 84, 108 is, 12 × 5 × 7 × 9 = 3780, −2 3, ⇒, ,, Product is = 12 × 3780 = 45360, 3 4, , 108 = 2 × 2 × 3 × 3 × 3 = 22 × 33, 26. (d) Parabola, x = 5 × 2783 = 13915, 27. (c) If a > 0, Graph of parabola looks like, (c) y = 253 ) 2783 ( = 11, 28. (d) Here graph cuts x-axis at 3 points, (b) z = 11) 253 ( = 23, so it has three zeros., (a) Composite number having more than 2 factors., (c) Prime factorisation of 13915 =, 5, 11, 11, 23, , 16., , 13915, 2783, 253, 23, 1, ⇒ 5 × 11 × 11 × 23, ⇒ 5 × 112 × 23, (c) a ≠ 0, a, b, c are real numbers, , 17. (d) For roots are equal, , b2 – 4ac = 0, or D = 0, 18. (b) For value of k,, 1 c, c, α. = (Product of roots = ), α a, a, 8k, 1 =, 2, 1, or k =, 4, 19. (c) For x2 + 1 = 0, roots are not real., So, graph of x2 + 1 = 0, neither touches nor intersects, x-axis., 20. (c) We know, for a quadratic polynomial, , k(x2 – (Sum of roots) x + Product of roots), , k(x2 – (–p) + (–1/p)), , k (x2 + p – 1/p), 21., , (d) Parabola., , 22. (c) a < 0, Graphs look like, , open downwards, 23. (c) According to graph, there are two zeros, one at (–2) and 2nd at 4, –2, 24., , (b) –2, 4, , 25., , (b) For zeros D =, , −b ± b 2 − 4ac, 2a, , Here, a = 4 3, b = 5, c = −2 3, , 29., , (c) Observing the graph we find –3, –1, 2 as zeros., , 30. (a) Given zeros are –3, –1, 2, then, Expression is (x – (–3)) (x – (–1)) (x – 2), = (x + 3)(x + 1)(x – 2), = x3 + 2x2 – 5x – 6, , x3 – (Sum of zeros)x2 + (Sum of zeros taking two at, a time)x – (Product of zeros), x3 – (– 3 – 1 + 2) x2 + ((–3)(–1) + (–1)(2) + (2)(–3)), x – (–3)(–1)(2), , x3 + 2x2 + (3 – 2 – 6)x – 6, , x3 + 2x2 – 5x – 6, Sol. (31-34):, Let x be number of known questions and y be number of, questions cheating by the student., Here, x + y = 120, 1, x− y=, 90, 4, On solving these two equations, We have, x = 96 and y = 24, 31., , No. of correct questions are 96, , 32., , He guessed 24 questions., , 33., , Marks = 80 –, , 34., , Here, x + y = 120, , 1, of 40 = 70, 4, , 1, x− y=, 95, , 4, On solving (i) & (ii) x = 100, 35., , ...(i), , ...(ii), , Given area of two bedrooms and a kitchen is 95 sq m., 2 × Area of bedroom + Area of kitchen = 95, 2 × 5 x + 5y = 95, or 2x + y = 19, , ...(i), , and x + 2 + y = 15, or x + y = 13, , ...(ii)

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QB-13, , CBSE Questions Bank-2021, 36., , Length of outer boundary = 12 + 15 + 12 +15 = 54 m, , 37., , On solving x + y = 13, , AB BC, 20, 12, =, ⇒, =, PQ QR, 100 QR, , , , , 2x + y = 19, , ⇒ QR = 60, , , , x = 6m, y = 7m, , 51. (b) Q DABC ~ DXYZ, , Area of a bedroom = 5x = 5 × 6 = 30 sq m, Area of kitchen = 5y = 5 × 7 = 35 sq m, , AB BC, 20 10, ∴, =, ⇒, =, XY 20, XY YZ, , 38., , Area of living room = 9 × 5 + 2 × 15 = 75 sq m, , ⇒ XY = 40, , 39., , Total cost of laying tiles in the kitchen = ` 50 × 35 = ` 1750, , 52. (a) Let QR = 40 m, PQ = 100 m and XY = 40 m, , 40., , (b) Given, fixed charges of auto rickshaw be ` x and, running charges be ` y km/hr, so representing situation, 1, , PQ QR, 100 40, ∴, =, ⇒, =, XY, YZ, 40 YZ, , , x + 10y = 75, , ⇒ YZ = 16 m., , x + 15y = 110, , 53. (d) Let QR = 40m, PQ =100m and AB = 20 m, , 41., , (c) On solving x + 10y = 75, , , , AB BC, 20 BC, ∵, =, ⇒, =, PQ QR, 100 40, , , x + 15y = 110, , , , we get x = 5 km,, , , , y = ` 7/km, , ⇒ BC = 8 m., 54. (c) Pythagoras theorem, , , , Charges to go 50 km., , , , x + 50y = 5 + 50 × 7 = ` 355, , 42., , 55. (a) AC2 = 302 + 402 = 2500 ⇒ AC = 50m, 56. (d) (21, 20, 28) [Q 282 ≠ (21)2 + (20)2], , (b) To cover 30 km distance,, , 57. (b) AB = 50 – 12 = 38m, , x + 30y = 19 + 30 × 9 = 289, , 58. (c) 82m, , 43., , (c), , 44., , (a) (2, 25), , 1, , , 2, y = × 100 =, 25, ∵ x =, 4, , , 45., , (c) (8, 20), , 1, , , 8, y = × 100 =, 20 , ∵ x =, 5, , , 46., , (c), , 47., ,  8 + 2 25 + 20 , ,, (a) ,  = (5, 22.5), 2, 2 , , 48., ,  2 + 5 25 + 22.5 , ,, (a) ,  = (3.5, 24), 2,  2, , , Sol. (59-61):, In ∆ABC, by Pythagoras theorem,, AC2 = AB2 + BC2 ⇒ AB = 4 cm., AC = 5 cm., BC 3, 59. (a) cot C =, =, AB 4, AC 5, 60. (b) sec C =, =, BC 3, 4, 3, 61. (b) sin C = , cos C =, 5, 5, 2, 2,  4 3, 2, 2, L.H.S = sin C + cos C =  +  , 5 5, 16 + 9, , =, = 1 = R.H.S, 25, Sol. (62-66), A, , (8 − 2)2 + (25 − 20)2 =, , 36 + 25 =, , 49. (c) Q DABC ~ DPQR, , 61, , AB BC, 20 10, ∴, =, ⇒, =, PQ QR, PQ 50, , ⇒ PQ = 100, , 32, , \ Height of the tower = 100 m, 50. (d) Let BC = 12 m and PQ = 100 m, , B, , cm, , 30°, , O, , D, , C

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Mathematics, , QB-14, , 62., , (b) cos 30° =, , 71., , BD, 32, , =, , BD = 16 3 cm., side BC = 32 3 cm, 63., , (c) AD =, , =, , =, , AB2 − BD 2, 2, , (32 3) − (16 3), , 2, , = 48 cm, 64., , (b) Side of square = 6 × 7 = 42 cm., Area of square = 42 × 42 = 1764 cm2, , 65., , 66., , (d) Area of each circular, 22, = p(7)2 =, × 49 = 154 cm2, 7, (a) Area of remaining, portion = 1764 – 9 × 154 = 378 cm2, , 67., , (b) Here r = 14 mm, Length of silverwire, , = 2pr + 8r, 22, =2×, × 14 + 8 × 14 = 200 mm, 7, 68. (c) Area of each sector, 1 22, = ×, × 14 × 14 = 77 mm2, 8 7, 69., , (d) Circumference of inner part = 44 mm, 22, ⇒2×, × r = 44, 7, ⇒ r = 7 mm, outer radius = 7 + 3 = 10 mm, , 70., , outer circumference, 22, =2×, × 10 = 62.86 mm, 7, (c) Difference of areas, 22, =, ( 102 – 72), 7, = 51 p mm2, , (c) Number of revolution, Distance, Outer circumference, 80π, = 4., 20π, , 72., , (a) P(king of red colour), =, , 73., , (d) P(getting a face card), =, , 74., 75., , 2, 1, =, 52 26, , 12 3, =, 52 13, 1, (b) P(getting a jack of hearts) =, 52, 3, (a) P(getting a red face card) =, 26, 13 1, =, 52 4, , 76., , =, (d) P(getting a spade), , 77., , (b) Sum of the two numbers appearing on the top face of, dice is 8., , , , (2, 6), (3, 5), (4, 4) (5, 3), (6, 2), 5, , \ Required probability =, 36, 78., , (d) Since, the sum of two numbers appearing on the top, face of dice cannot be 13., , , , So, required probability = 0., , 79., , (a) Since, the pair of number whose sum is less than 0 or, equal to 12 in a pair of dice is 36., 36, , \ Required probability, = = 1, 36, 80., , (c) Since, the pair of numbers on the top of dice whose, sum is 7 are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) , (6, 1), 6 1, , \ Required probability, = =, 36 6, 5, 81. (d), 18

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Sample Paper, , 1, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , 2., , 3., , Two numbers are in the ratio of 15 : 11. If their H.C.F. is 13, then numbers will be, (a) 195 and 143 , , (b), , 190 and 140, , (c) 185 and 163 , , (d), , 185 and 143, , 35, has ..................... decimal expansion., 50, (a) Terminating , , (b), , Non-terminating, , (c) Recurring , , (d), , Repeating, , Put suitable word in the sentence below:, , Which of the following is true?, , 22, 7, (b) The only real numbers are rational numbers, (a) π is equal to, , (c) Every non-terminating decimal can be written as a periodic decimal, (d) 0.21 lies between 0.2 and 0.3, 4., , A polynomial of degree 7 is divided by a polynomial of degree 4. Degree of the quotient is, (a) less than 3, , 5., , (b), , 3, , (c), , more than 3, , (d), , more than 5, , 2, , (c), , –2, , (d), , 3, , If 1 is zero of polynomial, p(x) = ax2 – 3(a – 1)x – 1, find a., (a) 1, , (b)

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Mathematics, , SP-2, , 6., , Two isosceles triangles have their corresponding angles equal and their areas are in the ratio 25 : 36. The ratio of their, corresponding height is, (a) 25 : 35, , 7., , 3, 16, , (c), , 5:6, , (d), , 6:5, , (b), , 5, 18, , (c), , 7, 36, , (d), , 7, 18, , (d), , (–3, 5), , (d), , 3.2 cm, , The coordinates of the point which is reflection of point (–3, 5) in x-axis are, (a) (3, 5), , 9., , 36 : 25, , Two dice are thrown at a time, then find the probability that the difference of the numbers shown on the dice is 1., (a), , 8., , (b), , (b), , (3, –5), , (c), , (–3, –5), , In the given figure, AD is the bisector of ∠A. If BD = 4 cm, DC = 3 cm and, AB = 6 cm, determine AC, A, , 6 cm, , B, , (a) 4.5 cm, 10., , (c), 11., , (b), , 3.5 cm, , a sin θ − b cos θ, If b tan θ = a, the value of, is, a sin θ + b cos θ, a−b, (a) 2, , a + b2, , 4 cm, , C, , D 3 cm, , (c), , (b), , a 2 + b2, , 4.8 cm, , a+b, a 2 + b2, a 2 − b2, , , (d), a 2 − b2, a 2 + b2, If the sum of the ages (in years) of a father and his son is 65 and twice the difference of their ages (in years) is 50, what is, the age of the father?, , (a) 45 years, , (b), , 40 years, , (c), , 50 years, , (d), , 55 years, , 12. If the point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3 : 1, then the value of y is, (a) 4, 13., , 14., 15., , 16., , (b), , 3, , If x = p sec q and y = q tan q, then, (a) x2 – y2 = p2q2z , 1, (c) x2q2 – y2p2 = 2 2 , p q, , (c), , 2, , (b), , x2q2 – y2p2 = pq, , (d), , x2q2 – y2p2 = p2q2, , (d), , 1, , If f (x) = 2x3 – 6x + 4x – 5 and g(x) = 3x2 – 9, then the value of f (1) + g(–2) is, (a) –3, (b) –2, (c) 3, (d) 2, A book containing 100 pages is opened at random. Find the probability that a doublet page is found., 7, 11, 8, 9, (a), (b), (c), (d), 100, 100, 25, 100, sin2q + cosec2q is always, (a) greater than 1 , , (b), , less than 1, , (c) greater than or equal to 2 , , (d), , equal to 2

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SP-3, , Sample Paper-1, 17., , 18., , Points A and B are 90 km. apart from each other on a highway. A car starts from A and another from B at the same time. If, they go in the same direction, they meet in 9 hrs and if they go in opposite directions, they meet in 9/7 hrs. Find their speeds., (a) 40 km/hr, 30 km/hr , , (b), , 10 km/hr, 20 km/hr, , (c) 20 km/hr, 30km/hr , , (d), , 50 km/hr, 40km/hr, , The two consecutive odd positive integers, the sum of whose squares is 290 are, (a) 9, 11, , 19., , (b), , 11, 13, , (c), , 13, 15, , (d), , 15, 17, , Determine the value of k for which the following system of equations becomes consistent :, 7x – y = 5, 21x – 3y = k., (a) k = 15, , 20., , (b), , k = 11, , (c), , k=4, , (d), , k, , 11, 2, , The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then find the greater number., (a) 111, , (b), , 137, , (c), , 37, , (d), , 311, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , ABCD is a square. F is the mid-point of A B, BE is one-third of BC. If the area of the ∆FBE is, 108 sq. cm find the length AC., (a), (c), , 22., , , , , , 36 2 cm , , 36 2 cm , , (b), , 37 2 cm, , (d), , (36)2 cm, , A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same, point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street., , E, D, 15m, , 15m, 9m, , A, (a) 21 m, 23., , 24., , (b), , 18 m, , 12 m, , C, (c), , B, 22 m, , (d), , 12 m, , The graphs of the equations x – y = 2 and kx + y = 3, where k is a constant, intersect at the point (x, y) in the first quadrant,, if and only if k is, (a) equal to – 1 , , (b), , greater than – 1, , (c) less than 3/2 , , (d), , lying between – 1 and 3/2, , (c), , 2, , If 0 < x ≤, (a) 0, , π, , then sin x + cosec x ≥, 2, (b) 1, , (d), , 3

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Mathematics, , SP-4, , 25., , If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2sin 3θ –, (a) sin2θ, , 26., , (b), , 1, 2, , (c), , 3 tan 3θ is equal to, , 1, 3, , (d), , 0, , Which among the following is correct?, (a) The ratios of the areas of two similar triangles is equal to the ratio of their corresponding sides., (b) The areas of two similar triangles are in the ratio of the corresponding altitudes., (c) The ratio of area of two similar triangles are in the ratio of the corresponding medians., (d) If the areas of two similar triangles are equal, then the triangles are congruent., , 27., , 28., , 29., , If the system of equations 2x + 3y = 7 and 2ax + (a + b)y = 28 represents coincident lines, which of the conditions holds, true?, (a) b = 2a , , (b), , a = 2b, , (c) 2a + b = 0 , , (d), , a + 2b = 0, , 2 (ax – by) + (a + 4b) = 0, 2 (bx + ay) + (b – 4a) = 0, (a) x = 0, y = 1 , , (b), , x = –1/2, y = 2, , (c) x = 1, y = 2 , , (d), , x = 1/2, y = –1/2, , Solve the following system of linear equations :, , 3, 2, Find a and b if x + 1 and x + 2 are factors of p (x) = x + 3 x − 2αx + β, , (a) 3, –1, 30., , 31., 32., , (c) 0, –3, (d) 5, 6, 5, If one zero of the quadratic polynomial 2x2 – 8x – m is , then the other zero is, 2, 2, 3, −15, 2, (a), (b) –, (c), (d), 3, 2, 2, 3, 3, 2, If x = 2 and x = 0 are roots of the polynomials f (x) = 2x – 5x + ax + b. Then values of a and b respectively are, (a) 2, 0, (b) 1, 2, (c) – 1, 1, (d) 0, 3, If cos A =, , 34., 35., , 36., , –1, 0, , , find the value of 9 cot2A – 1., , 16, 65, (c), (d) 0, 65, 16, Which of the following statement is false?, (a) All isosceles triangles are similar., (b) All equilateral triangles are similar., (c) All circles are similar. , (d) None of the above, If one root of the equation px2 – 14x + 8 = 0 is six times the other, then p is equal to, (a) 2, (b) 3, (c) 1, (d) none of these, Determine the values of a and b for which the following system of linear equations has infinitely many solutions:, 3x – (a + 1) y = 2b – 1, 5x + (1 – 2a) y = 3b, (a) a = 8, b = 5, (b) a = 4, b = 6, (c) a = 7, b = 1, (d) a = 5, b = 3, a 2 − b2, If sin θ = 2, , then find cosec θ + cot θ., a + b2, (a) 1, , 33., , (b), , (a), , a, a+b, , (b), , (b), , b+a, b−a, , (c), , a2, a+b, , (d), , a+b, a−b

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SP-5, , Sample Paper-1, 37., , Degree of polynomial y 3 − 2 y 2 − 3 y +, (a), , 38., , 1, 2, , (b), , 1, is, 2, , 2, , (c), , 3, , 3, 2, , (d), , Solve the following system of equations, ax + by = c;, bx – ay = c, , =, (a) x, , a, b, =, ,y, , 2, 2, 2, a +b, a + b2, , (b) =, x, , =, (c) x, , 2ab, 2ab, =, , y, , 2, (a + b), (a − b)2, , (d), , x=, , 1, 1, =, ,y, a, b, c(a + b), 2, , a +b, , 2, , , y= −, , c(a − b), a 2 + b2, , 21, is :, 45, , 39., , The decimal expansion of, , 40., , (a) terminating , (b) non-terminating and repeating, (c) non-terminating and non-repeating, (d) none of these, Find the value of a if (sin A + cosec A)2 + (cos A + sec A)2 = a + tan2A + cot2A, (a) 5, , (b), , 4, , (c), , 0, , (d), , 7, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, HCF of natural numbers is the largest factor which is common to all the number and LCM of natural numbers is the smallest, natural number which is multiple of all the numbers., 41. If p and q are two co-prime natural numbers, then their HCF is equal to, (a) p, , (b), , q, , (c), , 1, , (d), , pq, , (d), , equal, , 42. The LCM and HCF of two rational numbers are equal, then the numbers must be, (a) prime, , (b), , co-prime, , (c), , composite, , 43. If two positive integers a and b are expressible in the form a = pq2 and b = p3q; p, q being prime number, then LCM (a, b), is, (a) pq, , (b), , p3q3, , (c), , p3q2, , (d), , p2q2, , 44. The largest number which divides 285 and 1249 leaving remainders 9 and 7 respectively, is, (a) 46, , (b), , 6, , (c), , 12, , (d), , 138, , 45. The largest number which exactly divides 2011and 2623 leaving remainders 9 and 5 respectively is, (a) 11, , (b), , 22, , (c), , 154, , (d), , 13, , Q 46 - Q 50 are based on case study-II, Case Study-II, An honest person invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple, interest. He received yearly interest of 130, but if he had interchanged amounts invested, he would have received 4 more as interest., If x be the amount invested at the rate of 12% and y be the amount invested at the rate of 10%, then answer the following questions.

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Mathematics, , SP-6, , 46., , What is the yearly interest in terms of x and y ?, (a), , 47., , 12 x +10 y, 100, , (b) 12 x + 10 y = 13000, , (b) 6 x + 5 y = 6700, , (d), , 10 x +12 y, 100, , (c), , 6 x + 5 y = 6500 , , (d), , 5 x + 6 y = 6500, , (c), , 6 x + 5 y = 6300 , , (d), , 5 x + 6 y = 6300, , (c), , x – y = 100 , , (d), , x – y = 700, , (a) x = ` 500, y = ` 200 , , (b), , x = ` 500, y = ` 700, , (c) x = ` 100, y = ` 500 , , (d), , x = ` 400, y = ` 300, , Which of the following is true for x and y ?, (a) x + y = 120 , , 50., , 10 x + 12 y , , Find the equation corresponding to x and y when invested amount is interchanged., (a) 5 x + 6 y = 6700, , 49., , (c), , Find the equation corresponding to yearly received interest of `130., (a) 12 x + 10 y = 130, , 48., , (b) 12 x + 10 y , , (b), , x + y = 1200, , How much amount did he invest at different rates ?

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Page for Rough Work

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Sample Paper, , 2, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , Solve, , cos θ, cos θ, +, = 2, θ < 90º, cosec θ + 1 cosec θ − 1, , (a) 0° (b) 30° , (c), 2., , is equal to, , 2, , tan θ, 2, , 6., , (b) 2 , , (c), , 1, 2, , 2, , 2sin θ − cos θ, , (b), , 1, (c), 2, , 1, 3, , , , (d), , 1, 2, , sin θ − cos 2 θ, , 3 tan 3q is equal to, , (d), , 0, , Determine the value of k for which the following system of equations becomes consistent : 7x – y = 5, 21x – 3y = k., 11, 2, A railway half -ticket costs half the full fare but the reservation charges are the same on a half ticket as on full ticket. One, reserved first class ticket from station A to station B costs ` 2125. Also, one reserved first class ticket and one reserved half, first class ticket from A to B costs ` 3200. Find the full fare from station A to B and also the reservation charges for a ticket., (a) ` 1100, ` 15 , (b) ` 2100, ` 25 (c), ` 1000, ` 25 , (d), ` 2000, ` 40, Mrs. Vidya bought a piece of cloth as shown in the figure. The portion of the cloth that is not coloured consists of 6 identical, semi-circles., (a) k = 15 , , 5., , cosec θ, 2, , If 5q and 4q are acute angles satisfying sin 5q = cos 4q, then 2sin 3q –, (a) sin2q , , 4., , 60°, , tan θ−1 sec θ− cosec2 θ, , (a) 0 , 3., , +, , 45° (d), , 2, , (b) k = 11 , , (c), , k = 4 , , 42 cm, , (d), , k

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Mathematics, , SP-10, , Find the area of the coloured portion., (a) 144 cm2, (b) 126 cm2, 7., , 3, , 4, , (b), , 4, (c), 9, , 2, (d), 3, , 1, 2, , 14 , , None of these, , When 2256 is divided by 17, then remainder would be, (a) 1 , , 9., , (b) 16 , , (c), , (d), , In the given figure, P and Q are points on the sides AB and AC respectively of a triangle ABC. PQ is parallel to BC and, divides the triangle ABC into 2 parts, equal in area. The ratio of PA : AB =, A, , Q, , P, , C, , B, , (a) 1 : 1 , 10., , 243 cm2, , A factory has 120 workers in January, 90 of them are female workers. In February, another 15 male workers were employed., A worker is then picked at random. Calculate the probability of picking a female worker., (a), , 8., , 195 cm2 (d), , (c), , (b), , ( 2 − 1) : 2 , , (c), , 1: 2 (d), , ( 2 − 1) :1, , The figure given shows two identical semi-circles cut out from a piece of coloured paper., 22, Find the area of the remaining piece of paper (Use π =, ), 7, 15 cm, , 4 cm, , 20 cm, , 7 cm, , (a) 296.1 cm2 , 11., , 3, , 5, , (b) 1 : 6 , , (c), , (b), , 7, , 20, , (c), , (b) 18 cm , , (c), , 201.7 cm2, , 6 : 1 , , (d), , 2:1, , 3, (d), 4, , 1, 3, , 17 cm , , (d), , Data insufficient, , The sum of exponents of prime factors in the prime-factorisation of 196 is, (a) 3 (b) 4 (c), , 15., , (d), , The area of a right angled triangle is 40 sq. cm. and its perimeter is 40 cm. The length of its hypotenuse is, (a) 16 cm , , 14., , 221.5 cm2, , A box contains a number of marbles with serial number 18 to 38. A marble is picked at a random. Find the probability that, it is a multiple of 3., (a), , 13., , (c), , In what ratio does the point (–2, 3) divide the line-segment joining the points (–3, 5) and (4, –9) ?, (a) 2 : 3 , , 12., , (b) 265.4 cm2 , , 5 (d), , 2, , A drain cover is made from a square metal plate of side 40 cm having 441 holes of diameter 1 cm each drilled in it. Find the, area of the remaining square plate., (a) 1250.5 cm2 , , (b) 1253.5 cm2, , (c), , 1240.2 cm2 , , (d), , 1260.2 cm2

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Sample Paper-2, 16., , SP-11, , Which of the following statement is true?, (a) Every point on the number line represents a rational number., (b) Irrational numbers cannot be represented by points on the number line., 22, is a rational number., 7, (d) None of these., , (c), , 17., , Given ∆ABC ~ ∆DEF, if AB = 2DE and area of ∆ABC is 56 cm2, find the area of ∆DEF., (a) 14 sq.cm , , 18., , (b) 5 sq.cm , , (c), , (d), , 56 sq.cm, , Given that L.C.M. (91, 26) = 182, then H.C.F. (91, 26) is, (a) 13 (b) 26 , (c), , 19., , 18 sq.cm , , 17 (d), , 9, , III and IV , , None of these, , One card is drawn from a well shuffled deck of 52 cards., I., , The probability that the card will be diamond, is 1/2., , II. The probability of an ace of heart is 1/52., III. The probability of not a heart is 3/4., IV. The probability of king or queen is 1/26., Which of the statement(s) is/are true?, (a) I and II , 20., , (b) II and III , , (c), , (d), , In what ratio is the line segment joining the points (3, 5) & (–4, 2) divided by y–axis?, (a) 3 : 2 , , (b) 3 : 4 , , (c), , 2 : 3 , , (d), , 4:3, , SECTION-B, Section B consists of 20 quesions of 1 mark each. Any 16 quesions are to be attempted., 21., , Find an acute angle q, when, , cos θ − sin θ 1 − 3, =, cos θ + sin θ 1 + 3, , (a) 0° (b) 15° , (c), 22., , 24., , a 2 + b2, 2, , 2, , , , ab (d), , a, b, , (b) (0, 3) , , (–2, 2) , , (3, 0), , (c), , (d), , Find the point of trisectionof the line joining the points (–2, –19) and (5, 4)., (a) (2, –3) , , 26., , (b) a2 – b2 , (c), , b(1 − cos θ), , then xy =, sin θ, , a −b, Which of the following is not correct?, (a) If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium., (b) The line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram., (c) If corresponding sides of two similar triangles are in the ratio 4 : 5, then corresponding medians of the triangles must, be in the ratio 4 : 5., (d) None of the above, Find a point on the x-axis which is equidistant from the points (5, 4) and (–2, 3)., (a) (2, 0) , , 25., , 60°, , If x = a (cosec q + cot q) and y =, (a), , 23., , 30° (d), , (b) (1, 2) , , (c), , æ, ö, çç 1 , - 34 ÷÷ (d), çè 3, 3 ø÷, , æ 8 11ö÷, çç , ÷, èç 3 3 ø÷, , If the mid point of the line joining (3, 4) and (k, 7) is(x, y) and 2x + 2y + 1 = 0. Find the value of k., (a) 10 (b) –15 , (c), , 15 (d), , –10

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Mathematics, , SP-12, , 27., , For which value of p, will the lines represented by the following pair of linear equations be parallel, 3x – y – 5 = 0, 6x – 2y – p = 0, , 28., , 29., , (a) all real values except 10 , , (b), , 10, , (c) 5/2 , , (d), , 1/2, , If ABC and EBC are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC, and BDE is, (a) 2 : 1 , (b) 1 : 2 , (c), 1 : 4 , (d), 4:1, a , If  , 4  is the midpoint of the line segment joining A(–6, 5) and B(–2, 3), then what is the value of ‘a’?, 3 , (a) –4 (b) –12 , (c), , 30., , 1, , 3, , (b), , 1, (c), 6, , 3y, , 4, , (b) 6 + 6y, , (c), , 0, , 6+, , 4y, (d), 3, , 7x + 8y, 4, , If the zeroes of the polynomial f (x) = k2x2 – 17x + k + 2, (k > 0) are reciprocal of each other than value of k is, (a) 2 (b) –1 , (c), , 33., , 2, (d), 3, , ABC is an isosceles triangle in which AB = AC = 10 cm, BC = 12 cm. PQRS is a rectangle inside the isosceles triangle., Given PQ = SR = y cm and PS = QR = 2x cm, then x =, (a) 6 −, , 32., , –6, , A fair die is thrown once. The probability of getting a composite number less than 5 is, (a), , 31., , 12 (d), , –2 (d), , 1, , The figure shows two concentric circleswith centre O and radii 3.5 m and 7 m. If ∠BOA = 40°, find the area of the shaded, region., B, A, , D, C, O, , (a), , 34., , 77 2, cm , 6, , 73, , 6, , (d), , None of these, , (b), , 225, (c), 64, , 156, (d), 7, , –1, , If a letter is chosen at random from the letter of English alphabet, then the probability that it is a letter of the word ‘DELHI’ is, (a), , 36., , 76, (c), 5, ,  15 , (2 + 2sin θ) (1 − sin θ), If cot θ =   , then evaluate, 8, (1 + cos θ) (2 − 2 cos θ), (a) 1 , , 35., , (b), , 1, , 5, , (b), , 1, (c), 26, , 5, (d), 26, , 21, 26, , What is the largest number that divides 70 and 125, leaving remainders 5 and 8 respectively?, (a) 13 (b) 9 , (c), , 3 (d), , 585

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Sample Paper-2, 37., , SP-13, , The diagonal BD of a parallelogram ABCD intersects the segment AE at the point F, where E is any point on the side BC. Then, D, , C, , 4, E, , 1 F 2, 3, B, , A, , EF FB, , , FA AB, , (b), , DF × EF = FB × FA, , (c) DF × EF = (FB)2 , , (d), , None of these, , (a), , 38., , If P = (2, 5), Q = (x, –7) and PQ = 13, what is the value of ‘x’?, (a) 5 (b, , 39., , 3 , (c), , –5, , What is the largest number that divides 245 and 1029, leaving remainder 5 in each case?, (a) 15 (b) 16 , (c), , 40., , –3 (d), 9 (d), , 5, , If p, q are two consecutive natural numbers, then H.C.F. (p, q) is, (a) p , , (b) q , (c), , 1 (d), , pq, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Place a lighted bulb at a point O on the ceiling and directly below it a table in classroom. Place DABC shape cardboard parallel, to the ground between the lighted bulb and the table. Then a shadow of DA′B′C′ is cost on the table such that DABC ~ DA′B′C′, shown in figure., If AB = 5 cm, A′B′ = 15 cm; B′C′ = 12 cm,, AC = 3 cm, ∠B′ = 60° and ∠A = 80°., O, , A, , C, , B, A, , B, , C

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Mathematics, , SP-14, , Answer the following questions., 41., , Length of A′C′ is :, (a) 3 cm , , 42., , (c), , 9 cm , , (d), , 12 cm, , (b) 12 cm , , (c), , 3 cm , , (d), , 15 cm, , Length of BC is :, (a) 4 cm , , 43., , (b) 4 cm , , Measure of ∠A′ is :, (a) 60° (b) 80° (c), , 44., , 40°, , 80° (d), , 180°, , 80° (d), , 180°, , Find the measure of ∠B., (a) 60° (b) 40° (c), , 45., , 180° (d), , Find the measure of ∠C., (a) 60° (b) 40° (c), , Q 46 - Q 50 are based on case study-II, Case Study-II, A two digit number is obtained by either multiplying sum of the digits by 8 and adding 1 or by multiplying the difference of the, digits by 13 and adding 2., If x be the digit in ten’s place and y be the digit at unit place with x > y, then answer the following questions., 46. Find the equation corresponding to multiplying sum of the digits by 8 and adding 1., (a) 2x – 7y = 1 (b), 2x + 7y = 4, (c) 2x – 7y = 4 , (d), 2x + 7y = 1, 47. Find the equation corresponding to multiplying the difference of the digits by 13 and adding 2., (a) 14y – 3x = 2 (b), 3x – 14y = 4 (c) 14x – 3y = 2 , (d), 3y – 14x = 6, 48. What is the value of x ?, (a) 2 , (b), 3 , (c), 4 (d), 5, 49. What is the value of y ?, (a) 0 , (b), 1 , (c), 3 (d), 4, 50. What is the number ?, (a) 21 , (b), 31 , (c), 41 (d), 51

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Page for Rough Work

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Sample Paper, Time : 90 Minutes , , 3, Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., SECTION-A, , Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , Two numbers differ by 3 and their product is 54. Find the numbers., (a) 9 and 6 , (b) – 9 and – 6, (c) Both (a) and (b), , 2., , ∆ABC ~ ∆PQR and, , (d) 9 and – 4, area ∆ABC 16, = . If PQ = 18 cm and BC = 12 cm. then AB and QR are respectively , area ∆PQR 9, , (a) 9 cm, 24 cm , , (b) 24 cm, 9 cm, , (c) 32 cm, 6.75 cm, 3., , 4., , (d) 13.5 cm, 16 cm, 1, What is the maximum value of, ?, sec θ, , (a) 0 , , (b) 1, , (c) –1 , , (d) –2, , If the zeros of the polynomial, f(x) = k2x2 – 17x + k + 2, (k > 0 ) are reciprocal of each other, then the value of k is, (a) 2 , (b) – 1, (c) – 2 , , 5., , 6., 7., , (d) 1, , If the value of a quadratic polynomial p(x) is 0 only at x = – 1 and p(– 2) = 2, then the value of p(2) is, (a) 18 , (b) 9, (c) 6 , (d) 3, The probability of raining on day 1 is 0.2 and on day 2 is 0.3. The probability of raining on both the days is, (a) 0.2 , (b) 0.1, (c) 0.06 , (d), 0.25, Which of the following statement is false?, (a) All isosceles triangles are similar., (b) All quadrilateral triangles are similar., (c) All circles are similar. , (d) None of the above

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Mathematics, , SP-18, , 8., , A race track is in the form of a ring whose inner and outer circumference are 437m and 503m respectively. The area of the, track is, (a) 66 sq. cm , (b) 4935 sq. cm (c), 9870 sq. cm, (d), None of these, , 9., , Which of the following will have a terminating decimal expansion?, , 10., , (a), , 77, , 210, , I., , The L.C.M. of x and 18 is 36., , (b), , 23, 30, , 125, , 441, , (c), , (d), , 23, 8, , II. The H.C.F. of x and 18 is 2., What is the number x ?, 11., 12., , 13., , 14., , 15., , 16., , 17., , (a) 1 , (b) 2, (c) 3 , (d), 4, Which of the following cannot be the probability of an event?, (a) 2/3 , (b) – 1/5, (c) 15% , (d), 0.7, P, Q, R are three collinear points. The coordinates of P and R are (3, 4) and (11, 10) respectively and PQ is equal to 2.5, units. Coordinates of Q are, (a) (5, 11/2) , (b) (11, 5/2) (c) (5, –11/2) , (d), (–5, 11/2), A number lies between 300 and 400. If the number is added to the number formed by reversing the digits, the sum is 888, and if the unit’s digit and the ten’s digit change places, the new number exceeds the original number by 9. Then, the number, is, (a) 339 , (b) 341, (c) 378 , (d), 345, A fraction becomes 4 when 1 is added to both the numerator and denominator and it becomes 7 when 1 is subtracted from, both the numerator and denominator. The numerator of the given fraction is, (a) 2 , (b) 3, (c) 5 , (d), 15, The sum of the areas of two circles, which touch each other externally, is 153 π. If the sum of their radii is 15, then the ratio, of the larger to the smaller radius is, (a) 4 : 1 , (b) 2 : 1, (c) 3 : 1 , (d), None of these, The zeroes of the polynomial x2 – 3x – m(m + 3) are, (a) m, m + 3 , , (b) –m, m +3, , (c) m, –(m + 3) , , (d) –m, –(m + 3), , If a and b are zeroes of the polynomial, 2t2 – 4t + 3, then the value of a2b + ab2 is :, (a), , 18., , 3, , 4, , (b) 2, , (c) 3 , , (d), , 4, , In the given figure, DE || BC. The value of EC is, A, cm 1 cm, 5, ., D1, E, , 3, B, , 19., , cm, C, , (a) 1.5 cm , (b) 3 cm , (c), 2 cm , (d), 1 cm, At present ages of a father and his son are in the ratio 7 : 3, and they will be in the ratio 2 : 1 after 10 years. Then the present, age of father (in years) is, (a) 42 , (b) 56, (c) 70 , (d), 77

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Sample Paper-3, 20., , SP-19, , The probability that a two digit number selected at random will be a multiple of ‘3’ and not a multiple of ‘5’ is, (a), , 2, , 15, , (b), , 4, 15, , (c), , 1, , 15, , (d), , 4, 90, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , Solve 2x2 + 3y2 = 35;, , x 2 y2, 5, +, =, 2, 3, , ± 4, y =, ± 9 , (a) x =, 22., , 23., , 24., , 25., , (b) x = 3, y = 4, ± 2, y =, ±3, (c) x = 1, y = 1 , (d) x =, In ∆ABC, E divides AB in the ratio 3 : 1 and F divides BC in the ratio 3 : 2, then the ratio of areas of ∆BEF and ∆ABC is, (a) 3 : 5 , (b) 3 : 10, (c) 1 : 5 , Given that sin θ + 2 cos θ = 1, then, 2 sin θ – cos θ =, (a) 0 , , (b) 2, , (c) 1 , (d) None of these, If α and β are the zeroes of the polynomial, f(x) = x2 – 5x + k such that α – β = 1, the value of K is(a) 12 , (b) 6, (c) 4 , (d) 1, 3, 3, If x + y = 1, then x + y + 3xy = ..............., (a) 0 , (b) 1, (c) 2 , , 26., , (d) 3 : 20, , (d) None of these, , The zeroes of the polynomial are, p(x) = x2 –10x –75, (a) 5, – 15 , , 27., , (b) 5, 15, (c) 15, – 5 , 1, If cosec x – cot x = , where x ≠ 0, then the value of cos2x – sin2x is, 3, , (a), , 28., , 29., , 30., 31., , 16, , 25, , (b), , 9, 25, , The points (7, 2) and (–1, 0) lie on a line, (a) 7y = 3x – 7 , , (c), , 8, , 25, , (d), , – 5, – 15, , (d), , 7, 25, , (b) 4y = x + 1, , (c) y = 7x + 7 , (d) x = 4y + 1, X’s salary is half that of Y’s. If X got a 50% rise in his salary and Y got 25% rise in his salary, then the percentage increase, in combined salaries of both is, 1, 1, (a) 30 , (b) 33, (c) 37 , (d), 75, 3, 2, The perimeter of a sector of a circle with central angle 90° is 25 cm. Then the area of the minor segment of the circle is., (a) 14 cm2, (b) 16 cm2, (c), 18 cm2 (d), 24 cm2, The perimeters of two similar triangles ABC and PQR are respectively 36 cm and 24 cm. If PQ = 10 cm, then AB =, (a) 10 cm , (b) 20 cm, (c) 25 cm , (d), 15 cm

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Mathematics, , SP-20, , 32., , The least number which when divided by 15, leaves a remainder of 5, when divided by 25, leaves a remainder of 15 and, when divided by 35, leaves a remainder of 25, is, (a) 515 , , 33., , (b) 525, , (c) 1040 , (d) 1050, Out of one digit prime numbers, one number is selected at random. The probability of selecting an even number is, (a), , 1, , 2, , (b), , 1, 4, , (c), , 4, , 9, , (d), , 2, 5, , 34., , A can do a piece of work in 24 days. If B is 60% more efficient than A, then the number of days required by B to do the twice, as large as the earlier work is, (a) 24 , (b) 36, (c) 15 , (d), 30, , 35., , If n is an even natural number, then the largest natural number by which n (n + 1) (n + 2) is divisible is, (a) 6 , , 36., , 37., , (c) 12 , , (d), , 24, , The least number which is a perfect square and is divisible by each of 16, 20 and 24 is, (a) 240 , , (b) 1600, , (c) 2400 , , (d) 3600, , It is given that ∆ABC ~ ∆PQR with, , (a) 9 , 38., , (b) 8, , BC 1, ar(∆PQR), = . Then, is equal to, QR 3, ar(∆ABC ), , (b) 3, , (c), , 1, , 3, , (d), , 1, 9, , The figure given shows a rectangle with a semi-circle and 2 identical quadrants inside it., 28 cm, , 16 cm, , 23 cm, , 22, ), 7, (a) 363 cm2 , (b) 259 cm2 , (c), 305 cm2 , (d), The value of k for which the system of linear equations x + 2y = 3, 5x + ky + 7 = 0 is inconsistent is, What is the shaded area of the figure? (Use π =, , 39., , 14, , 3, (c) 5 , (a) −, , 40., , 2, 5, (d) 10, , (b), , The probability of getting a number greater than 2 in throwing a die is, (a) 2/3 , (b) 1/3, (c) 4/3 , , (d) 1/4, , 216 cm2

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Sample Paper-3, , SP-21, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, To enhance the reading skills of grade X students, the school nominates you and two of your friends to set up a class library. There, are two sections-section A and section B of grade X. There are 64 students in section A and 72 students in section B., , 41. What is the minimum number of books you will acquire for the class library, so that they can be distributed equally among, students of Section A or Section B?, (a) 144, , (b), , 128, , (c) 576, , (d) 272, , 42. If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (64, 72) is, (a) 2, , (b), , 4, , (c) 6, , (d) 8, , 43. 72 can be expressed as a product of its primes as, (a) 23 × 32 (b), , 21 × 33, , (c) 23 × 31, , (d) 20 × 30, , 44. 5 × 13 × 17 × 19 + 19 is a, (a) Prime number , , (b) Composite number, , (c) Neither prime nor composite, , (d) None of the above, , 45. If p and q are positive integers such that p = a2b3 and q = a3b2, where a, b are prime numbers, then the HCF (p, q) is, (a) ab, , (b), , a2b2, , (c) a3b2, , (d) a3b3, , Q 46 - Q 50 are based on case study-II, Case Study-II, Due to heavy storm an electric wire got bent as shown in the figure. It followed a mathematical shape. Answer the following, questions below., y, 6, 5, 4, 3, 2, 1, –6 –5 –4 –3 –2 –1, –1, –2, –3, –4, –5, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8

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Mathematics, , SP-22, , 46., , Name the shape in which the wire is bent, (a) spiral , , 47., , (d), , parabola, , (b) 3 , , (c), , 1 (d), , 0, , (b) –1, 3 , , (c), , 3, 5 , , (d), , –4, 2, , (c), , x2 – 2x – 3 , , (d), , x2 + 2x + 3, , (c), , 18 , , (d), , 0, , What will be the expression of the polynomial?, (a) x2 + 2x – 3 , , 50., , linear , , The zeroes of the polynomial are, (a) –1, 5 , , 49., , (c), , How many zeroes are there for the polynomial (shape of the wire)?, (a) 2 , , 48., , (b) ellipse , , (b) x2 – 2x + 3 , , What is the value of the polynomial if x = –1?, (a) 6 , , (b) –18

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 4, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., SECTION-A, , Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , If x + y = 1, then x3 + y3 + 3xy = ..............., (a) 0 , , 2., , 5., , 6., , 7., , 2 , , (d), , None of these, , (b) (0, 3) , , (c), , (–2, 2) , , (d), , (3, 0), , Two numbers differ by 3 and their product is 54. Find the numbers., (a) 9 and 6 , , 4., , (c), , Find a point on the x-axis which is equidistant from the points (5, 4) and (–2, 3)., (a) (2, 0) , , 3., , (b) 1 , , (b) – 9 and – 6, , (c), , Both (a) and (b) (d), , 9 and – 4, , A railway half -ticket costs half the full fare but the reservation charges are the same on a half ticket as on full ticket. One, reserved first class ticket from station A to station B costs ` 2125. Also, one reserved first class ticket and one reserved half, first class ticket from A to B costs ` 3200. Find the full fare from station A to B and also the reservation charges for a ticket., (a) ` 1100, ` 15 , , (b) ` 2100, ` 25, , (c), , tan θ − cot θ, is equal to, sin θ cos θ, (a) sec2 θ + cosec2 θ, , (b) cot2 θ – tan2 θ (c), , I. The L.C.M. of x and 18 is 36., II. The H.C.F. of x and 18 is 2., What is the number x ?, (a) 1 , (b) 2 , , (c), , ` 1000, ` 25, , (d), , ` 2000, ` 40, , cos2 θ – sin2 θ, , (d), , tan2θ – cot2θ, , 3 , , (d), , 4, , In the figure, ABC is a triangle in which AD bisects ∠A, AC = BC, ∠B = 72° and CD = 1cm. Length of BD (in cm) is, C, , D, , A, , B

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Mathematics, , SP-26, , (a) 1 , 8., , (c), , 5 –1, , 2, , 3 +1, 2, , (d), , (b) – 6 and 2 , , (c), , 6 and – 1, , (d), , 6 and – 2, , If in a lottery, there are 5 prizes and 20 blanks, then the probability of getting a prize is, (a), , 10., , 1, , 2, , C is the mid-point of PQ, if P is (4, x), C is (y, –1) and Q is (–2, 4), then x and y respectively are, (a) – 6 and 1 , , 9., , (b), , 2, , 5, , (b), , 4, (c), 5, , 1, , 5, , (d), , 1, , 3 , , (d), , 4, , If a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and, L.C.M. (a, b, c) = 23 × 32 × 5, then n =, (a) 1 , , 11., , (b) 2 , , (c), , The area of a circular ring formed by two concentric circles whose radii are 5.7 cm and 4.3 cm respectively is, (Take π = 3.1416), (a) 43.98 sq.cm, , 12., , (c), , 47.24 sq.cm, , (d), , 38.54 sq.cm, , The areas of two similar triangles are 81 cm2 and 49 cm2 respectively, then the ratio of their corresponding medians is, (a) 7 : 9, , 13., , (b) 53.67 sq. cm, , If, , (b) 9 : 81 , , (c), , 9 : 7 , , (d), , 81 : 7, , cos θ, cos θ, +, =, 4, then, 1 − sin θ 1 + sin θ, , 1, 3, 1, tan θ =, , (b) sin θ = (c) θ = 60° , (d), 3, 2, 2, The ratio in which the point (2, y) divides the join of (– 4, 3) and (6, 3) and hence the value of y is, , (a) cos θ =, 14., , (a) 2 : 3, y = 3 , , (b) 3 : 2, y = 4 , , (c), , 3 : 2, y = 3, , (d), , 3 : 2, y = 2, , 15. In a number of two digits, unit’s digit is twice the tens digit. If 36 be added to the number, the digits are reversed. The number, is, (a) 36 , 16., , (b) 63 , , (c), , 48 , , (d), , 84, , Two coins are tossed simultaneously. The probability of getting at most one head is, 1, 1, 3, , (b), (c), (d), 1, 4, 2, 4, ∆ABC is an equilateral triangle with each side of length 2p. If AD ⊥ BC, then the value of AD is, , (a), 17., , (a), 18., , 3 , , 3 p (c), , 2p , , (d), , 4p, , 2 , , (d), , 4, , Lowest value of x2 + 4x + 2 is, (a) 0 , , 19., , (b), , (b) –2 , , (c), , Ratio in which the line 3x + 4y = 7 divides the line segment joining the points (1, 2) and (–2, 1) is, (a) 3 : 5 , , (b) 4 : 6 , , (c), , 4 : 9 , , (d), , None of these

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Sample Paper-4, 20., , SP-27, , In the adjoining figure, OABC is asquare of side 7 cm. OAC is a quadrant of a circle with O as centre. The area of the shaded, region is, , (a) 10.5 cm2 , , O, , C, , A, , B, , (b) 38.5 cm2 , , (c), , 49 cm2, , (d), , 11.5 cm2, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , If the value of a quadratic polynomial p(x) is 0 only at x = – 1 and p(– 2) = 2, then the value of p(2) is, (a) 18 , , 22., , (b) 9 , , (c), , 6 , , (d), , 3, , Which of the following is not correct?, (a) If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium., (b) The line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram., (c) If corresponding sides of two similar triangles are in the ratio 4 : 5, then corresponding medians of the triangles must, be in the ratio 4 : 5., (d) None of the above, , 23., , 24., , 25., , 26., , If 5θ and 4θ are acute angles satisfying, sin 5θ = cos 4θ, then 2sin 3θ –, , 3 tan 3θ is equal to, , (a) sin2θ , , 1, , 2, , (a) k = 15 , (b) k = 11 , (c), If α and β are the zeroes of the polynomial, f(x) = x2 – 5x + k such that α – β = 1, the value of K is(a) 12 , (b) 6 , (c), 2 tan 30°, 1 + tan 2 30°, , 1, , 3, , (d), , 0, , 11, 2, , k = 4 , , (d), , k=, , 4 , , (d), , 1, , 1, , 2, , (d), , 3, 2, , (d), , 7290, , (d), , (–2, 0), , is equal to, (b) cos 60° , , (c), , Find the largest number of four digits exactly divisible by 12, 15, 18 and 27., (a) 9720 , , 28., , (c), , Determine the value of k for which the following system of equations becomes consistent :, 7x – y = 5, 21x – 3y = k., , (a) sin 30° , 27., , (b), , (b) 9728 , , (c), , 9270 , , The point on the X-axis which is equidistant from the points A(–2, 3) and B(5, 4) is, (a) (0, 2) , , (b) (2, 0) , , (c), , (3, 0)

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Mathematics, , SP-28, , 29., , The length of the side of a square whose diagonal is 16 cm, is, (a) 8 2 cm , , (b), , 2 8 cm , , (c), , 4 2 cm, , (d), , 2 2 cm, , 7 : 1 , , (d), , 1:7, , 30. If 3x + 4y : x + 2y = 9 : 4, then 3x + 5y : 3x – y is equal to, (a) 4 : 1 , 31., , (b) 1 : 4 , , An urn contains 6 blue and ‘a’ green balls. If the probability of drawing a green ball is double that of drawing a blue ball,, then ‘a’ is equal to, (a) 6 , , 32., , (b) 18 , , (c), , 24 , , (d), , 12, , (b) 1.5 , , (c), , 1.54 , , (d), , 1.45, , If x = 0.7 , then 2x is, (a) 1.4 , , 33., , (c), , The point which divides the line joining the points A (1, 2) and B(–1, 1) internally in the ratio 1 : 2 is,  –1 5 , (a)  ,  ,  3 3, , 1 5, (b)  ,  , 3 3, , (c), , (–1, 5) , , (d), , (1, 5), , 34. x and y are 2 different digits. If the sum of the two digit numbers formed by using both the digits is a perfect square, then, value of x + y is, (a) 10 , 35., , (b) 11 , , If cosec A + cot A =, (a), , 37., , 13, , (c), , 3 cm , , (d), , 6 cm, , (c), , 44, , 117, , (d), , 11, 117, , (d), , (3, 5), , 11, , then tan A, 2, (b), , 15, , 16, , (b) (0, 3) , , (c), , (1, 3) , , A single letter is selected at random from the word “PROBABILITY”. The probability that the selected letter is a vowel is, 2, , 11, , (b), , 3, , 11, , (c), , 4, , 11, , (d), , 0, , On dividing a natural number by 13, the remainder is 3 and on dividing the same number by 21, the remainder is 11. If the, number lies between 500 and 600, then the remainder on dividing the number by 19 is, (a) 4 , , 40., , (d), , The centroid of the triangle whose vertices are (3, –7), (–8, 6) and (5, 10) is, , (a), 39., , (b) 4.5 cm , , 21, , 22, , (a) (0, 9) , 38., , 12 , , The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, then the length of QR is, (a) 4 cm , , 36., , (c), , (b) 6 , , (c), , 9 , , (d), , 13, , (d), , 4:1, , If ∆ABC ~ ∆APQ and ar (∆APQ) = 4 ar (∆ABC), then the ratio of BC to PQ is, (a) 2 : 1 , , (b) 1 : 2 , , (c), , 1 : 4

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Sample Paper-4, , SP-29, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Students of class X make a design such that, the area of an equilateral triangle ABC is 17320.5 cm2. With each vertex of, the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle., (Use p = 3.14 and, , 3 = 1.73205), , A, , B, , C, , Answer the following questions., 41., , Find the length of side of DABC., (a) 200 cm, , 42., , (d), , 200.5 cm, , (b) 20 cm , , (c), , 10 cm , , (d), , 100 cm, , (c), , 4233.3 cm2, , (d), , 522.2 cm2, , (c), , 15700 cm2, , (d), , 31400 cm2, , (c), , 600 cm , , (d), , 300 cm, , (b) 5223.3 cm2 , , Find the area of the shaded region., (a) 17320.5 cm2, , 45., , 210.3 cm, , Find the area of each sector., (a) 5233.3 cm2, , 44., , (c), , Find the radius circle., (a) 200 cm, , 43., , (b) 105.5 cm , , (b) 1620.5 cm2 , , Find the perimeter of DABC., (a) 60 cm, , (b) 400 cm , , Q 46 - Q 50 are based on case study-II, Case Study-II, On school sport day, a sport teacher make a racing track whose left and right ends are semicircular shown in figure.

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Mathematics, , SP-30, , The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide then, answer the following questions., 46., , Find the radius of inner semicircular end., (a) 30 m , , 47., , (d), , 40 m, , (b), , 50 m, , (c) 40 m , , (d), , 70 m, , (b), , 400.57 m, , (d), , 400 m, , (c) 400.32 m, , The distance around the track along its outer edge is:, (a) 462.43 m, , 50., , (c) 10 m , , The distance around the track along its inner edge is:, (a) 423.57 m, , 49., , 60 m, , Find the radius of outer semicircular end, (a) 30 m , , 48., , (b), , (b), , 461.43 m, , (c) 463 m , , (d), , 463.43 m, , 4230 m2, , (c) 2340 m2 , , (d), , 4120 m2, , Find the area of the track., (a) 4320 m2, , (b)

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 5, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , 1., 2., 3., 4., 5., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , (, , (a) 2 , 2., , ), , If three points (0, 0) 3, 3 and (3, λ) form an equilateral triangle, then λ is equal to, (b) – 3 , , (c), , − 3 , , (d), , 3, , If the sum of the zeros of the polynomial f(x) = (k2 – 14) x2 – 2x – 12 is 1, which is one of the possible values of ‘k’?, (a), , 3., , (b) –14 , (c), 2 , (d), ±4, 14 , ABCD is a square. F is the mid-point of AB, BE is one-third of BC. If the area of the ∆FBE is 108 sq. cm find the, length AC., (a), , 4., , 6., , , , 36 2 cm, , (b) 37 2 cm , , (c), , 36 2 cm, , 3178, , 99, , (b), , 3178, (c), 999, , 3178, , 1000, , (d), , 999, 3178, , (a) a rational number , , (b), , an irrational number, , (c) either A or B , , (d), , neither A nor B, , In DABC, AB = AC, P and Q are points on AC and AB respectively such that BC = BP = PQ = AQ. Then,, ∠AQP is equal to (use p =180º), 2π, 3π, 4π, , (b), (c), , 7, 7, 7, If the circumference of a circle increases from 4π to 8π, then its area is, (a) halved , , 8., , (36)2 cm, , The product of two irrationals is, , (a), , 7., , (d), , Express the number 0.3178 in the form of rational number., (a), , 5., , , , (b) doubled , , (d), , 5π, 7, , (c), , tripled , , (d), , quadrupled, , (c), , 2 , , (d), , –1, , (1 + tan θ + sec θ) (1 + cot θ– cosec θ) =, (a) 0 , , (b) 1

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Mathematics, , SP-34, , 9., , If the point P (p, q) is equidistant from the points A (a + b, b – a) and B (a – b, a + b), then, (a) ap = by , , (b) bp = ay (c), , ap + bq = 0 , , (d), , bp + aq = 0, , 10. In a classroom, one-fifth of the boys leave the class and the ratio of the remaining boys to girls is 2 : 3. If further 44 girls, leave the class, then the ratio of boys to girls is 5: 2. How many more boys should leave the class so that the number of boys, equals that of girls?, (a) 16 (b) 24 , (c), 11., , 13., , 36, , In the adjoining figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is, B, C, , (a) 11 cm , 12., , 30 (d), , (b) 18 cm , , O, (c), , A, 25 cm , , (d), , 36 cm, , Let ABC be a triangle and M be a point on side AC closer to vertex C than A. Let N be a point on side AB such that MN is, parallel to BC and let P be a point on side BC such that MP is parallel to AB. If the area of the quadrilateral BNMP is equal, 5, to, of the area of DABC, then the ratio AM/MC equals, 18, 18, 15, (a) 5 , (b) 6 (c), (d), 5, 2, The points A (– 4, – 1), B (–2, – 4), C (4, 0) and D (2, 3) are the vertices of a, (a) Parallelogram , (b) Rectangle , (c), Rhombus , (d), Square, , 14., , For what value of p, the following pair of linear equations in two variables will have infinitely many solutions ?, px + 3y – (p – 3) = 0, 12x + py – p = 0, (a) 6 (b) – 6 , (c), 0 (d), 2, , 15., , 17., , If a circular grass lawn of 35m in radius has a path 7m wide running around it on the outside, then the area of the path is, (a) 1450 m2 , (b) 1576 m2 , (c), 1694 m2 , (d), 3368 m2, 2, 2, 9 sec A – 9 tan A =, (a) 1 , (b) 9 , (c), 8 , (d), 0, Three - digit numbers formed by using digits 0, 1, 2 and 5 (without repetition) are written on different slips with distinct, number on each slip, and put in a bowl. One slip is drawn at random from the bowl. The probability that the slip bears a, number divisible by 5 is, 5, 4, 2, 1, (a), , (b), (c), (d), 9, 9, 3, 3, , 18., , The value of 0.235 is :, , 16., , (a), , 233, , 900, , (b), , 233, (c), 990, , 235, (d), 999, , 235, 990

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Sample Paper-5, 19., , SP-35, , The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point B, C, D and A. The ratio of the perimeter of the outer circle to that of polygon, ABCD is, , π, 3π, π, , (b), , (c), (d), p, 4, 2, 2, Let P be an interior point of a DABC. Let Q and R be the reflections of P in AB and AC, respectively. If Q, A, R are collinear,, then ∠A equals, (a), , 20., , (a) 30° (b) 60° , (c), , 90° (d), , 120°, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , If α and β are the zeros of the polynomial f(x) = x2 + ax – b, find the polynomial having zeros, (a) abx2 + bx – a , , 22., , a, 1, x2  x , b, b, , (b) 4, 3 and 6 , , 1, Area (∆CAE) , 2, (c) Area (∆ABD) = 3 Area (∆CAE) , , 26., , 27., , b, 1, x2  x , a, a, , (c), , 4, 3 and 7 , , (d), , 7, 4 and 3, , (b), , Area (∆ABD) = Area (∆CAE), , (d), , 2Area (∆ABD) = Area (∆CAE), , (b) 3 , , (c), , more than 3 , , (d), , more than 5, , Find a point on the x-axis which is equidistant from the points (5, 4) and (–2, 3)., (a) (2, 0) , (b) (0, 3) , (c), (–2, 2) , (d), (3, 0), A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many, tickets has she bought?, (a) 40 (b) 240 , (c), 480 (d), 750, 2 tan 30°, =, 1 − tan 2 30°, (a) cos 60° , , 28., , (d), , A polynomial of degree 7 is divided by a polynomial of degree 4. Degree of the quotient is, (a) less than 3 , , 25., , abx2 – bx + a , , Triangle ABC is an isosceles triangle right angled at B. ∆ADB and ∆AEC are equilateral triangle then., (a) Area (∆ABD) =, , 24., , (c), , A class of 20 boys and 15 girls is divided into n groups so that each group has x boys and y girls. Values of x, y and n, respectively are, (a) 3, 4 and 8 , , 23., , (b), , 1, 1, and .,  , , (b) sin 60° , , (c), , tan 60° , , (d), , sin 30°, , If the value of a quadratic polynomial p(x) is 0 only at x = –1 and p(–2) = 2, then the value of p(2) is, (a) 18 , (b) 9 , (c), 6 , (d), 3

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Mathematics, , SP-36, , 29., , If the sector of a circle of diameter 10 cm subtends an angle of 144° at the centre, then the length of the arc of the sector is, (a) 2π cm , , (b) 4π cm , , (c), , 5π cm , , (d), , 6π cm, , 30. x and y are two non-negative numbers such that 2x + y = 10. The sum of the maximum and minimum values of (x + y) is, 31., , 32., , (a) 6 (b) 9 , (c), sin 2A = 2 sin A is true when A =, (a) 0° , (b) 30° , (c), Given that, , 15, , 45° , , 60°, , (d), , 1, = 0.142857 , which is a repeating decimal having six different digits. If x is the sum of such first three positive, 7, , integers n such that, , 1, = 0.abcdef , where a, b, c, d, e and f are different digits, then the value of x is, n, , (a) 20 , 33., , 10 (d), , (b) 21 , , (c), , 41 , , (d), , 42, , (c), , 0 < q < 1 , , (d), , None of these, , For an event E, P (E) + P ( E ) = q, then, (a) 0 ≤ q < 1 , , (b), , 0 < q ≤ 1 , , 34. A boat travels with a speed of 15 km/hr in still water. In a river flowing at 5 km/hr, the boat travels some distance downstream, and then returns. The ratio of average speed to the speed in still water is, (a) 8 : 3 , 35., , 36., , (c), , 8 : 9 , , (d), , 9:8, , (d), , 0, , Which of the following relationship is the correct ?, (a) P (E) + P ( E ) = 1 , , (b), , P ( E ) – P(E) = 1, , (c) P(E) = 1 + P ( E ) , , (d), , None of these, , (c), , sin 45° , , 1 − tan 2 45°, 1 + tan 2 45°, , =, , (a) tan 90° , 37., , (b) 3 : 8 , , (b) 1 , , The sum of two numbers is 528 and their H.C.F. is 33, then find the number of pairs of numbers satisfying the above, conditions., (a) 4 (b) 5 , (c), , 6 (d), , 2, , 38. A man can row a boat in still water at the rate of 6 km per hour. If the stream flows at the rate of 2 km/hr, he takes half the, time going downstream than going upstream the same distance. His average speed for upstream and down stream trip is, , 39., , (a) 6 km/hr , , (b), , 16/3 km/hr, , (c) Insufficient data to arrive at the answer , 2 tan 30°, =, 1 + tan 2 30°, , (d), , none of the above, , (a) sin 60° , , (c), , tan 60° , , (b) cos 60° , , (d), , sin 30°

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Sample Paper-5, 40., , SP-37, , The unit digit in the expression 55725 + 735810 + 22853 is, (a) 0 (b) 4 , (c), , 5 (d), , 6, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Soniya and Anuj are students of class X and they given a polynomial such that “If one zero of the polynomial, 3x2 – 8x + 2k + 5 is four times the other 4x2 – 12x + 3k + 8., Then, answer the following questions., 41. Find the sum of zeroes., (a) 3 (b), , 12, (d), 3, , 4 (c), , 12, 5, , 42. For quadratic polynomial ax2 + bx + c, a ≠ 0, write the formula to find product of zeroes., (a), , b, , a, , (b), , –, , b, (c), a, , –, , c, (d), a, , c, a, , 43. If α and β be the zeroes of given polynomial. Then, what is the relation between α and β?, (a) α + β = 4, , (b), , αβ = 4 , , β = 4α (d), , (c), , α2 = 16β, , 44. If α and β be the zeroes of the given polynomial, then find value of α., , 1, 7, 2, (b), (c), (d), 5, 4, 5, 45. Find the value of k. If α and β be the zeroes of given polynomials., (a), , (a), , 56, , 75, , (b), , –, , 56, (c), 75, , 75, (d), 56, , Q 46 - Q 50 are based on case study-II, Case Study-II, In a classroom, 4 friends are seated at the points P, Q, R and S as shown in figure., Then answer the following questions., 10, , 9, 8, Q, , 7, 6, Rows 5, , R, , P, , 4, 3, 2, , S, , 1, 1, , 2, , 3, , 4 5 6, Columns, , 7, , 8, , 9 10, , 3, 5, 65, 75

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Mathematics, , SP-38, , 46., , The coordinate of P is :, (a) (4, 3) , , 47., , (d), , (6, 7), , (b) 4 unit , , (c), , 2 3 unit, , (d), , 6 unit, , (b), , 6 2 unit , , (c), , 6 unit , , (d), , 5 unit, , (b) Rectangle , , (c), , Rhombus, , (d), , Parallelogram, , (b) (7, 4) , , (c), , (6, 2) , , (d), , (6, 4), , The name of quadrilateral is :, (a) Square , , 50., , (6, 1) , , The distance of PR is :, (a) 7 unit , , 49., , (c), , The distance of PQ is :, (a) 3 2 unit , , 48., , (b) (3, 4) , , The mid point of QS is :, (a) (5, 4)

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Page for Rough Work

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Sample Paper, , 6, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , Let P(x) be a polynomial of degree 3 and P(n) =, (a) 0, , 2., , (b), , (b) 10 cm2, , (c) 5 2 cm2, , (d) 25 cm2, , (b) (– 4, 2), , (c) (– 6, 5), , (d) (6, 4), , (b) 72, , (c) 63, , (d) 36, , (b) 9728, , (c) 9270, , (d) 7290, , A circle passes through the vertices of a triangle ABC. If the vertices are A(–2, 5), B(–2, –3), C(2, –3), then the centre of, the circle is, (a) (0, 0), , 7., , 3, 5, , Find the largest number of four digits exactly divisible by 12, 15, 18 and 27., (a) 9720, , 6., , (d), , The sum of the digits of a two-digit number is 9. If 27 is added to it, the digits of the number get reversed. The number is, (a) 25, , 5., , 2, 5, , Which of the following points is 10 units from the origin?, (a) (– 6, 8), , 4., , (c) −, , If the area of a square inscribed in a semicircle is 2cm2, then the area of the square inscribed in a full circle of the same, radius is ______, (a) 5 cm2, , 3., , 1, 5, , 1, for n = 1, 2, 3, 4. Then the value of P(5) is, 2, , (b) (0, 1), , (c) (–2, 1), , (d) (0, –3), , The value of (sin 45° + cos 45°) is, (a), , 1, 2, , (b), , 2, , (c), , 3, 2, , (d) 1

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Mathematics, , SP-42, , 8., , In a right angled triangle ∆ABC, length of two sides are 8 cm and 6 cm, then which among the given statements is/are correct?, A, , B, , 9., , (b) ∠ACB = 45°, , (c) ∠BAC = 45° , , (d) Pythagoras theorem is not applicable here., , Product of two co-prime numbers is 117. Their L.C.M. should be, (a) 1, , 10., , (b) 117, , (b) (–3, –2), , (d) Lies between 1 to 117, , (c) (3, – 2), , (d) (–3, 2), , (c) odd numbers, , (d) even numbers, , Let a and b be co-prime, thus a2 and b2 are:, (a) co-prime, , 12., , (c) equal to their H.C.F., , The centre of the circle passing through the ponts (6, – 6), (3, – 7) and (3, 3) is, (a) (3, 2), , 11., , C, , (a) Length of greatest side is 10cm, , (b) not co-prime, , Which among the following is/are correct?, (I) If the altitudes of two similar triangles are in the ratio 2 : 1, then the ratio of their areas is 4 : 1., (II) PQ || BC and AP : PB = 1 : 2., , Then,, , area ( ∆APQ ), area ( ∆ABC ), , =, , 1, 4, , (III) The areas of two similar triangles are respectively 9cm2 and 16cm2. The ratio of their corresponding sides is 3 : 16., (a) I, 13., , (b) II, , 3 14 m, , (c) 28 3 m, , The pair of equations 5x – 15y = 8 and 3 x – 9 y =, (a) one solutio , , The value of, (a), , 16., , 17., , 1, 2, , (d) 7 3 m, , 24, has, 5, (b) two solutions, , (c) infinitely many solutions, 15., , (d) None of these, , If Anish is moving along the boundary of a triangular field of sides 35 m, 53 m and 66 m and you are moving along the, boundary of a circular field whose area is double the area of the triangular field, then the radius of the circular field is (Take, 22, π= ), 7, (a) 14 3 m (b), , 14., , (c) III, , (d) no solution, , tan 30°, is, cot 60°, (b), , 1, , (c), , 3, , The decimal expansion of the rational number, , 33, 22.5, , 3, , (d) 1, , will terminate after, , (a) one decimal place , , (b) two decimal places, , (c) three decimal places , , (d) more than 3 decimal places, , Which among the following is/are correct?, (a) The ratios of the areas of two similar triangles is equal to the ratio of their corresponding sides., (b) The areas of two similar triangles are in the ratio of the corresponding altitudes.

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Sample Paper-6, , SP-43, , (c) The ratio of area of two similar triangles are in the ratio of the corresponding medians., (d) If the areas of two similar triangles are equal, then the triangles are congruent., 18., , A bag contains card numbers 3, 4, 5, 6, 7....27. One card is drawn, then probability of prime number card is, (a), , 19., , 9, 25, , (c), , 8, 25, , (d), , 1, 5, , 3, . The coordinates, 5, of the point, which lies in the fourth quadrant at a unit distance from the origin and on perpendicular to l, are, A line l passing through the origin makes an angle q with positive direction of x-axis such that sin θ =, ,  3 4, (a)  , − ,  5 5, 20., , 8, 27, , (b), ,  4 3, (b)  ,− ,  5 5, , (c) (3, –4), , (d) (4, –3), , The area of a circular path of uniform width ‘d’ surrounding a circular region of radius ‘r’ is, (a) πd(2r + d), , (b) π(2r + d) r, , (c) π(d + r)d , , (d), , π(d + r)r, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , If ∆ABC is an equilateral triangle such that AD ­⊥ BC, then AD2 =, 3a 2, 4, (a) A and C, A., , 22., , (b) 8 km/hr, , 1, 2, , (b), , 1, 5, , (c) 6 km/hr, , (d) 5 km/hr, , (c), , 2, 5, , (d), , 1, 3, , P is a point on the graph of y = 5x + 3. The coordinates of a point Q are (3, –2). If M is the mid point of PQ, then M must, lie on the line represented by, (a) y = 5x + 1 , (c) y =, , 25., , 3, a, 2, (d) B and C, , D., , From the data (1, 4, 7, 16, 27, 29) if 29 is removed, the probability of getting a prime number is, (a), , 24., , 3, BC2, 4, (c) D, C., , A boat takes 3 hours to travel 30 km downstream and takes 5 hours to return to the same spot upstream. Find the speed of, the boat in still water. (km/hr), (a) 10 km/hr, , 23., , 3a 2, 2, (b) A, , B., , 5, 7, x – , 2, 2, , (b) y = 5x – 7, (d) y =, , 5, 1, x+, 2, 2, , If the perimeter of a semi-circular protractor is 36 cm, then its diameter is, (a) 10 cm, , (b) 14 cm, , (c) 12 cm, , (d) 16 cm, , 26. The polynomial, f(x) = (x – 1)2 + (x – 2)2 + (x – 3)2 + (x – 4)2 has minimum value, when x = ..................., (a) 40, 27., , (b) 20, , (c) 10, , (d) 2.5, , In village Madhubani 8 women and 12 girls can paint a large mural in 10 hours. 6 women and 8 girls can paint it in 14 hours., The number of hours taken by 7 women and 14 girls to paint the mural is, (a) 10, , (b) 15, , (c) 20, , (d) 35

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Mathematics, , SP-44, , 28., , In a triangle ABC, ∠BAC = 90°; AD is the altitude from A on to BC. Draw DE perpendicular to AC and DF perpendicular, to AB. Suppose AB = 15 and BC = 25. Then the length of EF is, (a) 12, , 29., , 1 1, + =2, a b, , (b), , 1 1, − =1, a b, , (b) 0, , (c), , 1 1, − =2, a b, , (d), , 1 1, + =1, a b, , (c) 1, , (d) 2, , If one zero of the quadratic polynomial x2 + 3x + k is 2, then the value of k is, (a) 10, , 32., , (d) 5 5, , The value of (sin 30° + cos 30°) – (sin 60° + cos 60°) is, (a) –1, , 31., , (c) 5 3, , If the points (a, 0), (0, b) and (1, 1) are collinear then which of the following is true :, (a), , 30., , (b) 10, , (b) –10, , (c) 5, , (d) –5, , A box contains four cards numbered as 1, 2, 3 and 4 and another box contains four cards numbered as 1, 4, 9 and 16. One, card is drawn at random from each box. What is the probability of getting the product of the two numbers so obtained , more, than 16?, (a), , 5, 8, , (b), , 1, 2, , (c), , 3, 8, , (d), , 1, 4, , 33. The distances of a point from the x-axis and the y-axis are 5 and 4 respectively. The coordinates of the point can be, (a) (5, 4), 34., , 1 + tan 2 A, , (c) (0, 4), , (d) (4, 5), , (b), , (c) cot2 A, , (d) tan2 A, , =L, , 1 + cot 2 A, (a) sec2 A, , 35., , (b) (5, 0), , –1, , Consider the following two statements:, I. Any pair of consistent linear equations in two variables must have a unique solution., II. There do not exist two consecutive integers, the sum of whose squares is 365., Then,, , 36., , (a) both I and II are true , , (b) both I and II are false, , (c) I is true and II is false , , (d) I is false and II is true, , If the radius of a circle is diminished by 10%, then its area is diminished by, (a) 10%, , (b) 19%, , (c) 36%, , (d) 20%, , 37. Let D be a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. If AC = 21 cm, then the side of an equilateral, triangle whose area is equal to the area of the rectangle with sides BC and DC is, 1/2, , (a) 14 × 3, 38., , –1/2, , 3/4, , (c) 14 × 3, , 1/2, , (d) 42 × 3, , If one of the zeroes of the quadratic polynomial (k –1) x2 + kx + 1 is –3, then the value of k is, (a), , 39., , (b) 42 × 3, , 4, 3, , (b), , −4, 3, , (c), , 2, 3, , (d), , −2, 3, , (sec A + tan A) (1 – sin A) =, (a) sec A, , (b) sin A, , (c) cosec A, , (d) cos A

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Sample Paper-6, 40. The equations, (a) 10, 5, , SP-45, , 1 1, 1 1, + = 15 and - = 5 are such that ax = 1 and by = 1. The values of ‘a’ and ‘b’ respectively are, x y, x y, (b) 10, –5, , (c) –5, 10, , (d) 5, 10, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Class X students of a secondary school in Krishnagar have been allotted a rectangular plot of a land for gardening activity. Saplings, of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as, shown in the fig. The students are to sow seeds of flowering plants on the remaining area of the plot., C, , B, P, R, , Q, A, , D, , Considering A as origin, answer question (i) to (v), 41. Considering A as the origin, what are the coordinates of A?, (a) (0, 1) , (b) (1, 0) , (c), (0, 0) , (d) (–1, –1), 42. What are the coordinates of P?, (a) (4, 6) , (b) (6, 4) , (c), (4, 5) , (d) (5, 4), 43. What are the coordinates of R?, (a) (6, 5) , (b) (5, 6) , (c), (6, 0) , (d) (7, 4), 44. What are the coordinates of D?, (a) (16, 0) , (b) (0, 0) , (c), (0, 16) , (d) (16, 0), 45. What are the coordinate of P if D is taken as the origin?, (a) (12, 2), (b) (–12, 2), (c) (12, 3) , (d), (6, 10), Q 46 - Q 50 are based on case study-II, Case Study-II, Rakesh and Mohit playing a card game. Rakesh picked up a card from properly mixed cards numbered from 1 to 25., Then answer the following questions :

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Mathematics, , SP-46, , 46., , 47., , 48., , 49., , The probability of getting prime numbers is :, (a), , 9, 10, (b), 25, 25, The probability of getting multiple of 3 is :, , (c), , 7, 25, , (d), , 8, 25, , (a), , 7, 8, (b), 25, 25, The probability of getting multiple of 2 is :, , (c), , 6, 25, , (d), , 9, 25, , (a), , 10, 13, (b), 25, 25, The probability of getting multiple of 2 and 3 is :, , (c), , 12, 25, , (d), , 11, 25, , 3, 4, (b), 25, 25, The probability of getting multiple of 2 or 3 is :, , (c), , 2, 25, , (d), , 16, 25, , (c), , 3, 25, , (d), , 10, 25, , (a), 50., , (a), , 16, 25, , (b), , 4, 25

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 7, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., 2., , The distance between which of the following two points is 2 units?, (a) (–2, –3) and (–2, –4), (b) (0, 4) and (0, 6), (c) (7, 2) and (6, 2), Which of the following is/are a polynomial?, , (d), , 1, (b) 2 x 2 – 3 x + 1, (c) x3 – 3x + 1, (d), x, In Fig. DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x., (a) x 2 +, , 3., , (4, –3) and (2, 3), 3, 2x 2, , C, E, , D B, , A, , 4., , (a) 4, (b) 7, (c), Two dice are rolled, then probability of getting a total of 9 is, (a), , 5., , 1, 3, , (b), , 1, 9, , (c), , Which of the following statement(s) is/are always true?, (a) The sum of two distinct irrational numbers is rational., (b) The rationalising factor of a number is unique., (c) Every irrational number is a surd., (d) None of these, , 5, , (d), , 2, , 9, 10, , (d), , 8, 9, , – 5x

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Mathematics, , SP-50, , 6., , 7., , 5, 3, II. The system of equations 3x + 2y = a and 5x + by = 4 has infinitely many solutions for x and y, then, a = 4, b = 3, x y, III. If + =, 2 and ax – by = a2 – b2, then x = a, y = b, a b, Which is true?, (a) I only, (b) II only, (c) III only, (d) None of these., 2sin θ cos θ, If 13 tan q = 12, then find the value of, cos 2 θ − sin 2 θ, , I., , If x – y = xy = 1 – x – y, then x + y is, , (a), , 25, 312, , (b), , 1, 25, , (c), , 12, 31, , (d), , 312, 25, , 8., , From a bag containing 100 tickets numbered 1, 2, 3, ........., 100 one ticket is drawn. If the number on this ticket is x, then, 1, the probability that x + > 2 is ......, x, (a) 0, (b) 0.99, (c) 1, (d) None of these, , 9., , A right triangle has hypotenuse of length p cm and one side of length q cm. If p – q = 1, find the length of the third side of, the triangle., 2q + 1cm, , (a), 10., , 11., , 13., , 15., , (c), , 2q + 1cm, , 2q + q 2 cm, , (d), , 73, , is a non-terminating repeating decimal., 54, (b) If a= 2 + 3 and b = 2 – 3 , then a + b is irrational., (c) If 19 divides a3, then 19 divides a, where a is a positive integer., (d) Product of L.C.M. and H.C.F. of 25 and 625 is 15625., Which of the following given options is/are correct?, (a) Degree of a zero polynomial is ‘0’., (b) Degree of a zero polynomial is not defined., (c) Degree of a constant polynomial is not defined., (d) A polynomial of degree n must have n zeroes., (2 + 2sin θ) (1 − sin θ),  15 , If cot θ =   , then evaluate,  8, (1 + cos θ) (2 − 2 cos θ), 156, 225, (c), 7, 64, A coin is tossed. Then the probability of getting either head or tail is, 1, 1, (a) 1, (b), (c), 3, 2, Which of the following is / are not correct ?, Three points will form :, (a) an equilateral triangle, if all the three sides are equal., (b) an isosceles triangle, if any two sides are equal., (c) a collinear or a line, if sum of two sides is equal to third side., (d) a rhombus, if all the four sides are equal., (a) 1, , 14., , 2(q + 1) cm, , Suppose we have two circles of radius 2 each in the plane such that the distance between their centers is 2 3 . The area of, the region common to both circles lies between, (a) 0.5 and 0.6, (b) 0.65 and 0.7, (c) 0.7 and 0.75, (d) 0.8 and 0.9, Which of the following statement(s) is/are not correct?, (a), , 12., , (b), , (b), , (d), , –1, , (d), , 1, 4

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SP-51, , Sample Paper-7, 16., , 17., , A circle is inscribed in a right angled triangle of perimeter 7p . Then the ratio of numerical values of circumference of the, circle to the area of the right angled triangle is, (a) 4 : 7, (b) 3 : 7, (c) 2 : 7, (d) 1 : 7, In the given figure, S and T trisect the side QR of a right triangle PQR. Then which of the following is correct?, P, , y, , Q, , 18., 19., , S, , x, , (a) 8PT 2 = 3PR2 + 5PS2 (b) 8PR 2 = 8PT2 + 8PS2, The product of unit digit in (795 – 358) and (795 + 358) is, (a) 8, (b) lies between 3 and 7, Which of the following given options is/are correct?, 2, (a), + 3 is a polynomial , x, (c), , 20., , x, , 2, is a polynomial , 3x – 4, , x, , T, , R, , (c), , 8PT 2 – 4PR2 = 6PS2, , (d), , 8PT 2 = 7RP2 – 6PS2, , (c), , 6, , (d), , lies between 3 and 6, , (b), , x + 5 is a polynomial, , (d), , 5x2 +, , (c), , 1, 3, , 1, 3, x + is a polynomial, 2, 7, , If 5θ and 4θ are acute angles satisfying, sin 5θ = cos 4θ, then 2sin 3θ –, (a) sin2θ, , (b), , 3 tan 3θ is equal to, 1, 2, , (d), , 0, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , Which of the following is / are correct?, Four points will form :, (a) a rectangle, if opposite sides and diagonals are not equal., (b) a parallelogram, if opposite sides are not equal., (c) a square, if all the four sides and diagonals are equal., (d) a right angle triangle, if sum of squares of any two sides is equal to square of third largest side., , 22., , Two dice are rolled simultaneously. Find the probability that they show different faces., (a), , 3, 4, , (b), , 1, 6, , (c), , 1, 3, , (d), , 5, 6

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Mathematics, , SP-52, , 23., , In the given figure PA, QB and RC, each are perpendicular to AC., R, P, Q, , z, , x, y, A, , B, , C, , Which of the following is correct ?, , 24., 25., , 1, 1, 1 1 1, =, =, +, (c), (d) None of these, x+ z y, y x z, If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then the values of a and b are, respectively., (a) y + z = x, , (b), , (a) 3 and 5, , (b), , (d), , – 1 and – 3, , (b), , 2, , (c), , –1, , (d), , 1, , (b), , 20 minutes, , (c), , 10 minutes, , (d), , 25 minutes, , 2π, radians?, 3, , The value of c for which the pair of equations cx – y = 2 and 6x + 2y = 3 will have infinitely many solutions is, (a) 3, , 28., , 3 and 1, , How much time the minute hand of a clock will take to describe an angle of, (a) 15 minutes, , 27., , (c), , If the distance between the points (2, –2) and (–1, x) is 5, one of the values of x is, (a) –2, , 26., , 5 and 3, , (b), , –3, , (c), , – 12, , (d), , no value, , Which of the following is/are not correct?, (a) If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium., (b) The line segments joining the mid-points of the adjacent sides of a quadrilateral form a parallelogram., (c) If corresponding sides of two similar triangles are in the ratio 4 : 5, then corresponding medians of the triangles must, be in the ratio 4 : 5., (d) None of the above, , 29., , A line is of length 10 units and one end is (2, –3). If the abscissa of the other end is 10, what is the ordinate?, (a) 3 or 9, , 30., , (c), , 3 or –9, , (d), , –3 or 9, , (b), , negative, , (c), , zero, , (d), , one, , (d), , 3, , If sin A + sin2A = 1, then the value of the expression (cos2A + cos4A) is, (a) 1, , 32., , –3 or –9, , The probability of an event can not be, (a) positive, , 31., , (b), , (b), , 1, 2, , (c), , 2, , Which of the following statement(s) is/are not correct?, (a) There are infinitely many even primes., (b) Let ‘a’ be a positive integer and p be a prime number such that a2 is divisible by p, then a is divisible by p., (c) Every positive integer different from 1 can be expressed as a product of non-negative power of 2 and an odd number., (d) If ‘p’ is a positive prime, then, , 33., , If the radius of a circle is, (a), , 49 2, cm, π, , p is an irrational number., , 7, cm, then the area of the circle is equal to, π, (b), , p cm2, , (c), , 154 cm2 , , (d), , 49 cm2

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SP-53, , Sample Paper-7, 34., , 35., , The zeroes of the quadratic polynomial x2 + 99x + 127 are, (a) both positive , , (b), , both negative, , (c) one positive and one negative , , (d), , both equal, , Which of the following points will be collinear with the points (–3, 4) and (2, –5)?, (a) (0, 0), , 36., , (7, –14), , (c), , (0, –1), , (d), , (3, 1), , a, Given that sin θ = , then cos θ is equal to, b, (a), , 37., , (b), , b, 2, , b −a, , 2, , b, a, , (b), , b2 − a 2, b, , (c), , (d), , a, 2, , b − a2, , Which of the following statement(s) is/are not correct?, (a) Every integer is a rational number., (b) The sum of a rational number and an irrational number is an irrational number., (c) Every real number is rational., (d) Every point on a number line is associated with a real number., , 38., , A die is thrown once then,, 2, 3, (c) the probability of getting a prime number is 2/3, (a) the probability of getting an odd number is, , 39., , 40., , (b), , the probability of getting multiple of 3 is 1/3, , (d), , the probability of getting number greater than 5 is 1/3, , (a) their corresponding angles are equal., , (b), , their corresponding sides are equal., , (c) both are right triangle. , , (d), , None of the above, , Two triangles are similar if, ,  13 , A circle drawn with origin as the centre passes through  , 0  . The point which does not lie in the interior of the circle is, 4 ,  −3 , (a)  ,1,  4 , , (b), ,  7,  2, ,  3, , (c), ,  −1 ,  3, ,  2 , , (d), , 5, ,  −6, , 2, , , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the, string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the, tip of her rod to the fly) is taut shown in figure., , 1.8 m, , 2.4 m

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Mathematics, , SP-54, , Answer the following questions., 41., , How much string does she have out?, (a) 1 m, , 42., , 3m, , (d), , 4m, , (b), , 1.2 m, , (c), , 1.5 m, , (d), , 2m, , 1.2 m, , (c), , 1m, , (d), , 0.8 m, , (c), , ASA, , (d), , SAS, , (d), , 40 sec., , (b), , Both triangles are similar by similarity criterion is:, (a) AAA, , 45., , (c), , Find the length of her fishing rod., (a) 1.5 m, , 44., , 2m, , Find the length of CD., (a) 1 m, , 43., , (b), , (b), , SSS, , If she pulls in the string at the rate of 5 cm per second, then time taken to pulls all string., , (a) 1 min., (b) 30 sec., Q 46 - Q 50 are based on case study-II, , (c), , 30 min., , Case Study-II, A compound angle is that which is made of up of algebraic sum of two or more angles., sin(A + B) = sin A cos B + cos A sin B, tan(A + B) =, , tan A – tan B, tan A + tan B, , tan (A – B) =, 1 + tan A ⋅ tan B, 1 – tan A tan B, , cos (A + B) = cos A cos B – sin A sin B, cos (A – B) = cos A cos B + sin A sin B, 46., , The value of sin 75° is, (a), , 47., , (c), , 2– 3, 2 2, , (c), , 1, 1– 3, , (c), , 3 –1, 2 2, , (c), , 3 –1, 2 2, , (c), , (d), , 3 +1, 2 2, , (d), , 2+ 3, , (d), , 3 +1, 3 –1, , 3+2, 2, , (d), , 3–2, 2, , 3 +1, 3 –1, , (d), , 3 –1, 3 +1, , 0, , 3 +1, 3 –1, , (b), , 3 –1, 3 +1, , 2, 2– 3, , (b), , 0, , The value of cos 15° is, (a), , 50., , 3+ 3, 2 2, , The value of tan 75° is, (a), , 49., , (b), , The value of tan 15° is, (a), , 48., , 3 –1, 2 2, , 3 +1, 2 2, , (b), , The value of cos 75° is, (a), , 3 +1, 2 2, , (b)

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 8, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., , 1., 2., 3., 4., 5., , SECTION-A, Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , A boat goes 12 km. upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the, same time. Find the speed of the boat in still water and the speed of the stream., (a) 4 km/hr, 5 km/hr, (b) 3 km/hr, 1 km/hr, (c), 6 km/hr, 2 km/hr (d), 7 km/hr, 2 km/hr, , 2., , Find the distance between the points, (a), , 3., , 3 , , (b), , (, , ), , 3 + 1, 2 -1 and, , (, , 2 3 , , ), , 3 -1, 2 + 1 ., 2 , , (c), , (d), , 2 2, , If in fig. O is the point of intersection of two chords AB and CD such that OB = OD, then triangles OAC and ODB are, , A, D, 45°, , O, , C, , 4., 5., , B, , (a) equilateral but not similar (b), , isosceles but not similar, , (c) equilateral and similar (d), , isosceles and similar, , If the H.C.F of 210 and 55 is expressible in the form 210 × 5 + 55y, find y., (a) 20 , (b) 19 , (c), – 91 , , (d), , – 19, , (d), , 3, , A child has a die whose six faces show the number as given below:, 1 2 2 3 4 6, The die is thrown once. What is the probability of getting an even number?, (a), , 1, , 6, , (b), , 2, , 3, , (c), , 0

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Mathematics, , SP-58, , 6., , Which of the following is/are not graph of a quadratic polynomial ?, , (a) X, , A, , B, , O, , X, , (b), , (c), , B, , X A, , O, , (d), , X, , O, Y, , The two opposite vertices of a square are (–1, 2) and (3, 2). Find the co-ordinates of the other two vertices., (a) (1, 0), (1, 2) , , 8., , X, , Y, , Y, , 7., , Y, , Y, , Y, , I., , (b) (1, 0), (2, 1) , , (c), , (1, 4), (1, 0), , (d), , (4, 1), (1, 0), , (c), , III , , (d), , None of these, , If 3x – 5y = –1 and x – y = – 1, then x = –2, y = –1, , II. 2x + 3y = 9, 3x + 4y = 5 ⇒ x = –21, y = 17, 2x y, x y, 2, =, –, 4 ⇒ x = 2a, y = 2b, III., =, +, a b, a b, Which is true?, (a) I , 9., , (b) II , , In figure given below , O is a point inside, ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm. If PQ = 24 cm, QR = 26 cm. Then, , 24, , cm, , 6 cm, , P, , O, , 8 cm, Q, (a) ∠QRP = 90° , 10., , (c), , R, ∠QPR = 90°, , (d), , ∆PQR is an isosceles, , If the ratio of the areas of the two circles is 25 : 16, then the ratio of their circumferences is, (a), , 11., , (b) ∠PRQ = 90° , , 26 cm, , 25, , 16, , (b), , 4, , 5, , (c), , 5, , 4, , (d), , 500, 625, , (d), , 2:5, , p, is a terminating decimal, what can you say about q ?, q, (a) q must be in the form 2n, , If, , (b) q must be in the form 5m, (c) q must be in the form 2n.5m, (d) q must be in the form 2n.5m, where n and m are non negative integers., 12., , Identify the ratio in which the line joining (4, 5) and (– 10, 2) is cut by the Y-axis., (a) – 5 : 2 , , (b) 3 : 5 , , (c), , – 5 : 3 , , 13., , From a normal pack of cards, a card is drawn at random, find the probability of getting a jack or a king., , 14., , 7, 4, , (b), , 52, 13, The graph of y = x2 – 6x + 9 is :, (a), , (c), , 2, , 13, , (a) a parabola open upward , , (b), , a parabola open downward, , (c) a straight line , , (d), , None of these, , (d), , 3, 13, , X

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Sample Paper-8, 15., , 16., , 17., , Identify the incorrect statement., (a) A right angled triangle may have 1, 1 and 2 as its sides., (b) 1, 2, 3 are the sides of a right angled triangle., (c) The ratio of corresponding sides of two squares whose areas are in the ratio 4 : 1 is 2 : 1, (d) 17, 8 and 15 are the sides of a right angled triangle., Two dice are thrown at a time, then find the probability that the difference of the numbers shown on the dice is 1., 5, 3, 7, 7, (a), , (b), , (c), , (d), 18, 16, 36, 18, Which of the following is not a rational number?, (a), , 18., , 19., , 20., , SP-59, , 2 , , (b), , 4 , , (c), , 9 , , (d), , 16, , If the sector of a circle of diameter 14cm subtends an angle of 30° at the centre, then its area is, 49π, 121, 242, (a) 49π , (b), , (c), , (d), 12, π, 3π, What is a system of simultaneous equations called if it has no solution?, (a) Consistent system , (b), Independent system, (c) Inconsistent system , (d), Dependent system, Find the probability for a randomly selected number of 1, 2, 3, 4,.....25 to be a prime number., (a), , 4, , 25, , (b), , 7, , 25, , (c), , 8, , 25, , 9, 25, , (d), , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , If a and b are the zeroes of the quadratic polynomial f (x) = ax2 + bx + c then evaluate, (a) a2 – b2 , , (b), , 3abc – b3, 3, , , , (c), , –b, , a, , (d), , 1, α, , 3, , +, , 1, β3, c, a, , ., , 22., , c, Find the chance that a non-leap year contains 53 Saturdays., , 23., , 1, 2, 3, 5, , (b), , (c), (d), 7, 7, 7, 7, What is the value of ‘x’ if (4, 3) and (x, 5) are points on the circumference of a circle with centre O(2, 3)?, (a), , (a) 4 , , (b) 2 , , (c), , –2 , , 24., , Which of the following is not correct?, , 25., , 1, is rational having non-terminating is repeating decimal fraction., 7, 11, (b), is rational non-terminating repeating decimal., 30, 31, (c), is rational having non-terminating repeating decimal., 91, 13, (d), is rational having non-terminating repeating decimal., 125, In DABC, ∠B = 90° and D is the midpoint of BC. Then, , (d), , 0, , (a), , 26., , (a) AC2 = AD2 + 3CD2 , , (b), , AC2 + AD2 = CD2, , (c) 3AC2 = AD2 + CD2 , 3 4, 4 2 11, Solve for x and y : + = 1; + =, x y, x y 12, , (d), , AD2 = CD2 = 3AC2, , (a) x = 1, y = 2 , , (c), , x = 4, y = 5, , (b) x = 6 , y = 8 , , (d), , x = 7, y = 3

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Mathematics, , SP-60, , 27., , 28., , Which of the following statement is/are not correct?, (a) A chord divides the interior of a circle into two parts., (b) An arc of a circle whose length is less than that of a semicircle of the same circle is a called a minor arc., (c) Circles having the same centre but different radii are called concentric circles., (d) A line segment joining any two points of a circle is called an arc., When two dice are thrown, find the probability of getting a number always greater than 4 on the second dice., (a), , 29., 30., , 2, , 3, , (b), , 1, , 3, , 3, , 5, , (c), , (d), , 2, 5, , 3, 2, Find a and b if x + 1 and x + 2 are factors of p (x) = x + 3 x − 2αx + β, , (a) 3, –1 , (b) –1, 0 , (c), 0, –3 , (d), 5, 6, A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same, point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street., E, , D, 15m, , 15m, 9m, , 31., 32., 33., , 12 m, , A, C, B, (a) 21 m , (b) 18 m , (c), 22 m , (d), 12 m, If a pair of linear equations is inconsistent, then the lines will be, (a) parallel , (b) always coincident, (c), intersecting, (d), coincident, If ABC and EBC are two equilateral triangles such that D is mid-point of BC, then the ratio of the areas of triangles ABC, and BDE is, (a) 2 : 1 , (b) 1 : 2 , (c), 1 : 4 , (d), 4:1, If the mid-point of the line segment AB (shown in the adjoining figure) is (4, –3), then the coordinates of A and B are, Y, A, , O, , 34., , (a) (8, 0) and (– 6, 0) , (c) (0, 8) and (– 6, 0) , For what value of ‘x’ does 6x end with 5?, (a) 0 , , 35., , B, , X, , (b) (8, 0) and (0, – 6), (d) (0, 8) and (0, – 6), , (b) 1 , , (c), , 5 , , (d), , Never ends with 5, , Which of the following is/are not correct?, 132 2, cm ., 14, (b) If a chord of circle of radius 14 cm makes an angle of 60° at the centre of the circle, then area of major sector is, 512.87 cm2., (a) Area of a circle with radius 6 cm, if angle of sector is 60°, is, , (c) The ratio between the circumference and area of a circle of radius 5 cm is 2 : 5., (d) Area of a circle whose radius is 6 cm, when the length of the arc is 22 cm, is 66 cm2.

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Sample Paper-8, 36., , SP-61, , In the given figure, DE || BC and AD : DB = 5 : 4 then ar (DDFE) : ar(DCFB)., A, , D, , E, F, , B, (a) 25 : 81 , 37., , 38., , 39., , 40., , (b) 5 : 81 , , C, (c), , 81 : 25 , , (d), , 4, is a root of the polynomial f (x) = 6x3 – 11x2 + kx – 20, then find the value of k., 3, (a) 10 , (b) 19 , (c), – 5 , (d), , 22 : 88, , If x =, , 3, , For what values of k, do the equations 3x – y + 8 = 0 and 6x – ky = –16 represent coincident lines?, (a) solution of 3k – 9 = 0 , , (b), , solution of 2k – 8 = 0, , (c) 2 , , (d), , 3, , A line intersects the y-axis and x-axis at the points P and Q respectively. If (2, –5) is the mid point of PQ, then the coordinates, of P and Q are respectively, (a) (0, –5) and (2, 0) , , (b), , (0, 10) and (–4, 0), , (c) (0, 4) and (–10, 0) , , (d), , (4, 0) and (0, 10), , The decimal expansion of, (a) terminating, , 21, is :, 45, , (b) non-terminating and repeating, (c) non-terminating and non-repeating, (d) none of these, SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Two unbiased coins are tossed simultaneously., The word ‘unbiased’ means each outcome is equally likely to occure., 41., , The probability of getting two heads is, , 42., , 1, (b), 2, The probability of getting one tail is, (a), , (a), , 1, (b), 2, , 1 , , (c), , 1, , 3, , (d), , 1, 4, , 1 , , (c), , 1, , 3, , (d), , 1, 4

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Mathematics, , SP-62, , 43., , 44., , 45., , The probability of getting no head is, 1, (a), (b), 2, , 1 , , The probability of getting at most one head., 1, 1, (a), (b), , 2, 4, The probability of getting at least one head, 1, 3 , (a), (b), 4, 4, , (c), , 1, , 3, , (d), , 1, 4, , (c), , 3, , 4, , (d), , 1, , (c), , 9, , 2, , (d), , 1, , Q 46 - Q 50 are based on case study-II, Case Study-II, A horse is tied to a peg at one corner of a square shaped grass field of side 15m. (Use p = 3.14), , 46., , If rope of horse is 5m long then the area of that part of the field in which the horse can graze is :, (a) 19.625m2, , 47., , 18.625m2, , (b), , 78.5m2, , (c) 58.5m2 (d), , 73.5m2, , (b), , 58m2, , (c) 57.875m2 , , (d), , 68.87 m2, , If rope of horse is 5 m long then the area of that part of the field in which the horse can not graze is:, (a) 204.37m2, , 50., , (c) 19 m2 (d), , The increase in the grazing area if the rope were 10m long instead of 5m., (a) 58.875m2, , 49., , 29.625m2, , If rope of horse 10 m long then the area of that part of the field in which the horse can graze is:, (a) 68.5m2 , , 48., , (b), , (b), , 200.37m2, , (c) 205.37m2 , , (d), , 205m2, , If rope of horse 10m long then the area of that part of the field in which the horse can not graze is :, (a) 146.5 m2, , (b), , 205.37m2, , (c) 46.5m2 (d), , 146 m2

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 9, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., SECTION-A, , Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , 2., , The height of mountains is found out using the idea of indirect measurements which is based on the, (a) principal of congruent figures, , (b), , principal of similarity of figures, , (c) principal of equality of figures, , (d), , none of these, , Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively., (a) x2 – 3x – 2, , 3., , (b) x2 + 3x + 2, , (c), , x2 – 3x + 2, , (d) x2 + 3x – 2, , The figure given shows a rectangle with a semicircle and 2 identical quadrants inside it., 28 cm, , 16 cm, , 23 cm, , What is the shaded area of the figure?, , (a) 363 cm2, 4., , 6., , (c), , 305 cm2, , (d) 216 cm2, , A lady has 25 p and 50 p coins in her purse. If in all she has 40 coins totalling `12.50, find the number of coins of each type, she has., (a) 10, 15, , 5., , 22, ), 7, (b) 259 cm2, (Use p =, , (b) 30, 10, , (, , ), , (c), , 20, 30, , (d) 10, 10, , The points (a, a) (–a, –a) and − 3a, 3a are the vertices of, (a) a scalene triangle , , (b), , a right angled triangle, , (c) an isosceles right angled triangle, , (d), , an equilateral triangle, , (c), , common factor, , H.C.F. of pair of co-primes is _________., (a) one, , (b) product of numbers, , (d) lowest common factor

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Mathematics, , SP-66, , 7., , What is the maximum value of, (a) 0, , 1, ?, secθ, , (b) 1, , (c), , –1, , (d) –2, , 8., , If in an isosceles triangle ‘a’ is the length of the base and ‘b’ is the length of one of the equal side, then its area is equal to, , 9., , a2 (, 4b – a 2 ), 4, The zeroes of the polynomial are, (a) a 2 b 2 – 4b 2, , , , a, 4b 2 – a 2, 4, , (d), , (b) 5, 15, , (c), , 15, – 5, , (d) – 5, – 15, , (c), , equilateral triangle, , (d) scalene triangle, , The points (– 4, 0), (4, 0), (0, 3) are the vertices of a, (a) right triangle, , 11., , 1 2, a + b2, 4, , (c), , p(x) = x2 – 10x –75, , (a) 5, – 15, 10., , (b), , (b) isosceles triangle, , Arjun drew a figure as shown in figure, where a circle is divided into 18 equal parts. He then shaded some of the parts. (Take, p = 3.14), , 8 cm, Find the total area the Arjun shaded., (a) 25.12 cm2, 12., , (b) 29.25 cm2, , (c), , 36.4 cm2, , (d) 45.2 cm2, , sum, , (d) none of these, , L.C.M = ____________ of highest powers of all the factors., (a) product, , (d) difference, , (c), , 13., , When two dice are thrown, find the probability of getting same numbers on both dice., (d), , 14., , 2, 1, 1, (b), (c), 3, 12, 6, The points A (9, 0), B (9, 6), C (– 9, 6) and D (– 9, 0) are the vertices of a, (a) square, , (d) trapezium, , (a), , 15., , (b) rectangle, , rhombus, , A man steadily goes 10 m due east and then 24 m due north. then his distance from the starting point is, (a) 28 m, , 16., , (c), , 1, 9, , (b) 26 m, , (c), , 25 m, , (d) 18 m, , The perimeter of a rectangle is 40 cm. The ratio of its sides is 2 : 3. Find its length and breadth., (a) l = 10 cm,, , b = 8 cm, , (b) l = 12 cm,, , b = 18 cm, , (c) l = 12 m,, , b=8m, , (d) l = 40 m,, , b = 30m, , 3, then, what is the value of sin A?, 4, , 17., , If tan A =, , 18., , 3, 4, (b) 1, (c), 5, 3, Which of the following numbers has the terminal decimal representation?, (a), , (a), , 1, 7, , (b), , 1, 3, , (c), , 3, 5, , (d) 0, , (d), , 17, 3

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Sample Paper-9, 19., , SP-67, , If A(2, 2), B(–4, –4) and C(5, –8) are the vertices of a triangle, then the length of the median through vertex C is, (a), , (b), , (d), 113, 85, 20. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, Find the, number of blue balls in the bag., 65, , (a) x = 10, , (c), , 117, , (b) x = 12, , (c), , x=9, , (d) x = 8, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., , Two coins are tossed simultaneously. Find the probability of getting atmost one head., , 22., , 1, 3, 2, (b), (c), 4, 4, 3, which of the following is true if following pair of equations has unique solution?, (a), , (d), , 1, 2, , 3x – 2y = – 8, (2m – 5)x + 7y – 6 = 0, 11, 4, , 11, 11, (d) m ≠, 4, 4, 23. A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow, 16 metres long. Find the height of the telephone pole., (a) m =, , (a) 40 cm, 24., , m=–, , 11, 4, , (b) 24 cm, , (c), , m≠–, , (c), , 101 cm, , (d) 10 cm, , The graph of y = p(x) is given in fig. below, for a polynomial p(x). The number of zeroes of p(x),is/are, , (a) 4, 25., , (b), , (b) 3, , (c), , no zero, , (d) 2, , (c), , 1, , (d) None of these, , Given that sin q + 2 cos q = 1, then 2 sin q – cos q =, (a) 0, , (b) 2, , 26. What is the condition that a system of simultaneous equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 must satisfy to have, exactly one solution?, (a), 27., , a1 b1, =, a2 b2, , a1 b1, ≠, a2 b2, , (c), , a1 c1, =, a2 c2, , (d), , b1 c1, =, b2 c2, , The least number which is a perfect square and is divisible by each of 16, 20 and 24 is, (a) 240, , 28., , (b), , (b) 1600, , (c), , 2400, , (d) 3600, , If the end points of a diameter of a circle are A (–2, 3) and B (4, –5), then the coordinates of its centre are, (a) (2, –2), , (b) (1, –1), , (c), , (–1, 1), , (d) (–2, 2)

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Mathematics, , SP-68, , 29., , The graph of y = f(x) is shown in the figure., What type of polynomial f(x) is?, Y, , X, , O, , , , Y, , (a) cubic, 30., , If 1 +, , X, , sin2, , (b) quadratic, , (c), , linear, , (d) none of these, , q = 3 sin q cos q, then tan q can have values, 3 1, ,, 4 3, , (d) 1, 1, 2, 31. ABC is a right-angled triangle right angled at A. A circle is inscribed in it and the lengths of the two sides containing the, right angle are 6 cm and 8 cm. Find the radius of the circle., 32., , (a) 4, 0, , (b), , (c), , None of these, , (a) 1.5 cm, , (b) 2.2 cm, , (c), , 3 cm, , (a) any positive, , (b), , any negative integer, , (c) any odd natural number, , (d), , any even natural number, , (d) 2 cm, , If (–1)n + (–1)4n = 0, then n is, , 33. A chord of a circle of radius 28 cm subtends an angle of 45° at the centre of the circle. Then the area of the minor segment, is, (a) 30.35 cm2, 34., , 36., , 37., , 30.45 cm2, , (c), , (d) 30.25cm2, , In what ratio is the line segment joining the points (3, 5) & (–4, 2) divided by y–axis?, (a) 3 : 2, , 35., , (b) 30.81 cm2, (b) 3 : 4, , (c), , 2:3, , (d) 4 : 3, , What is a system of simultaneous equations called if its graph has intersecting lines?, (a) Inconsistent system, , (b), , Consistent system, , (c) Dependent system, , (d), , Independent system, , tan θ + sec θ − 1, =, tan θ − sec θ + 1, 1 + sin θ, 1 + cos θ, (a), (b) cos q + sin q, (c), cos θ, sin θ, Choose the zeros of the polynomial whose graph is given., Y, , (d) cos2q – sin2q, , 3, 2, 1, X, , X', , 0 1, –1, –2, , –2 –1, , 2, , 3, , Y', , (a) 1, –1, 2, , (b) –2, 1, 3, , (c), , –2, 0, 3, , (d) –2, 2, 3

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Sample Paper-9, 38., , SP-69, , In DABC, D is the mid point of BC and AE^ BC. If AC > AB, then, (a) AB2 = AD2 – BC2 + BC2, (b) AB2 = AD2 – BC ⋅ DE +, , 1, BC2, 4, , 1, BC2 – BC ⋅ DE, 4, (d) All of the above, (c) AD2 = AB2 +, 39., , Find the H.C.F. of, 23 × 32 × 5 × 74, 22 × 35 × 52 × 73, 23 × 53 × 72., (a) 980, , 40., , (b), , 890, , (c) 900, , (d), , 809, , The perimeter of a sector of a circle with central angle 90° is 25 cm. Then the area of the minor segment of the circle is., (a) 14 cm2, (d), , (b), 24 cm2, , 16 cm2, , (c), , 18 cm2, , SECTION-C, Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground ., , Answer the following questions., 41., , The length of her shadow after 4 seconds is :, (a) 4.8 m, , 42., , (d), , 2m, , (b), , 1.6 m, , (c) 4 m, , (d), , 3m, , (b), , 1.8 m, , (c) 5.4 m, , (d), , 3.2 m, , (b), , SSS, , (c) SAS, , (d), , ASA, , (b), , ∠B = ∠D, , (c) ∠A = ∠D (d), , Similarity criterion of DABE and DCDE is :, (a) AA , , 45., , (c) 4 m, , Distance between their tops is :, (a) 4 m, , 44., , 1.6 m, , Distance travel by girl after 4 second is :, (a) 4.8 m, , 43., , (b), , Which of the following is true ?, (a) ∠B = ∠C, , ∠A = 90°

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Mathematics, , SP-70, , Q 46 - Q 50 are based on case study-II, Case Study-II, A die has two faces each with number ‘1’, three faces each with number ‘2’ and one face with number ‘3’. Die is rolled once., 46. The probability of obtaining the number 2 is, 1, (b), 2, , (c), , 1, 3, , (d), , None of these, , (c), , 1, 2, , (d), , None of these, , 47., , 1, , 6, The probability of getting the number 1 or 3 is, , 48., , 1, 1, (b), , 6, 3, The probability of not getting the number 3 is, 5, , 6, , (c), , 1, 2, , (d), , None of these, , 49., , 1, (b), 6, Probability of getting prime number, 1, (b), 6, , 1, , 2, , (c), , 1, 3, , (d), , 1, , 1, , 6, , (c), , 1, 3, , (d), , 0, , (a), , (a), , (a), , (a), 50., , Probability of getting odd number, (a), , 1, (b), 2

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 10, , Time : 90 Minutes , , Max Marks : 40, , General Instructions, 1., 2., 3., 4., 5., , The question paper contains three parts A, B and C., Section A consists of 20 quesions of 1 mark each. Any 16 quesitons are to be attempted., Section B consists of 20 quersions of 1 mark each. Any 16 quesions are to be attempted., Section C consists of 10 quesions based two Case Studies. Attempt any 8 questions., There is no negative marking., SECTION-A, , Section A consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 1., , Find a quadratic polynomial whose zeroes are 8 and 10., (a) k(x2 + 10x + 80), , 2., , 3., , ), , (d) k(x2 + 6x + 9), , 3, −3 3 as vertices?, , (b) An equilateral triangle, , (c) An isosceles triangle , , (d) A right triangle, , The difference between two numbers is 26 and one number is three times the other. Find them., (b) 41, 67, , (c) 96, 70, , (d) 52, 26, , A copper wire when bent in the form of an equilateral triangle has area 121 3 cm 2 . If the same wire is bent into the form, of a circle, find the area enclosed by the wire., (b) 346.5 cm2, , (c) 342.5 cm2, , (d) 340.25 cm2, , Three wheels can complete respectively 60, 36, 24 revolutions per minute. There is a red spot on each wheel that touches, the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?, (a) 3 second, , 6., , ( −3, , (a) A scalene triangle , , (a) 345.5 cm2, 5., , (c) k(x2 – 18x + 80), , What type of a triangle is formed with points (3, –3), (–3, 3) and, , (a) 39, 13, 4., , (b) k(x2 – 2x + 1), , If , sin θ =, , 2, , (b) 4 second, , (c) 5 second, , (d) 7 second, , 2, , a −b, then find cosec q + cot q., a 2 + b2, , a+b, a, b+a, a2, (b), (c), (d), a−b, b−a, a+b, a+b, The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1 : 2 internally lies in the, (a), , 7., , (a) I quadrant, 8., , (c) III quadrant, , (d) IV quadrant, , An unbiased die is rolled twice. Find the probability of getting the sum of two numbers as a prime, 3, 5, , 5, 7, 4, (c), (c), 12, 12, 5, Given DABC ~ DDEF. If AB = 2DE and area of DABC is 56 cm2 find the area of DDEF., (a), , 9., , (b) II quadrant, , (a) 14 sq.cm, , (b), , (b) 5 sq.cm, , (c) 18 sq.cm, , (d) 56 sq.cm

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Mathematics, , SP-74, , 10., , A sheet is 11 cm long and 2 cm wide. Circular pieces of diameter 0.5 cm are cut from it to prepare discs. Calculate the, number of discs that can be prepared., (a) 114, , 11., , If two positive integers a and b are written as a =, (a) xy, , 12., , (b) 113, , If, , (b), , xy2, , (c) 110, x3y2, , and b =, (c), , xy3;, , (d) 112, x, y are prime numbers, then HCF (a, b) is, , x3y3, , (d) x2y2, , tan q, cot q, –k, +, =, + sec q cosec q, 1 − cot q 1 − tan q, 2, , Find the value of k., (a) 1, 13., , (c) 3, , (d) 2, , Five years ago Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and, Sonu?, (a) 50 yrs, 20 yrs, , 14., , (b) 0, , (b) 40 yrs, 30 yrs, , (c) 60 yrs, 40 yrs, , (d) 45 yrs, 15 yrs, , ABC is an isosceles triangle in which AB = AC = 10 cm. BC = 12 cm. PQRS is a rectangle inside the isosceles triangle., Given PQ = SR = y cm. and PS = QR = 2x cm. then x =, 4y, 3y, 7x + 8y, (b) 6 + 6y, (c) 6 +, (d), 3, 4, 4, 2, 2, 2, If f(x) = x + 5x + p and g(x) = x + 3x + q have a common factor, then (p – q) = _________, , (a) 6 −, 15., 16., , 17., , (a) 2(5p – 3q), (b) 2(3p – 5q), (c) 3p – 5q, (d) 5p – 3q, A month is randomly selected from a year. An event X is defined as ‘the month with 30 days’. Identify the number of, outcomes of event X., (a) 1, (b) 6, (c) 3, (d) 4, 5, 2, If x = , then find whether the variable x is rational or irrational, 9, (a) Rational, , 18., 19., , 20., , (b) Irrational, , (c) Composite, , (d) Integer, , If P = (2, 5), Q = (x, –7) and PQ = 13, what is the value of ‘x’?, (a) 5, (b 3, (c) –3, (d) –5, In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Then, which of the following is true ?, PB.PB, (a) PA.PB = PC2, (b) PA.PB = PC.PD, (c) (PA)2 =, (d) PC × PC = PD, 2, a sin φ, b sin q, a, If tan q =, and tan φ =, , then, =, 1 − a cos φ, 1 − b cos q, b, (a), , sin q, 1 − cos φ, , (b), , sin q, 1 − cos φ, , (c), , sin φ, sin q, , (d), , sin q, sin φ, , SECTION-B, Section B consists of 20 questions of 1 mark each. Any 16 quesions are to be attempted., 21., 22., , 23., , xn + yn is divisible by (x + y) when ‘n’ is _________, (a) an even number, (b) an odd number, (c) a prime number, (d) a natural number, DABC is an isosceles triangle right angled at B. Similar triangles ACD and aBE are constructed on sides AC and AB. ratio, between the areas of DABE and DACD is, (a) 1 : 4, (b) 2 : 1, (c) 1 : 2, (d) 4 : 3, in the given figure, a circle with centre B overlaps another circle with centre A and a square. The ratio of areas of P and Q, 1, is 5 : 4 and the area of Q is the area of circle B. The radii of circle A and circle B are 10 cm and 8 cm respectively., 8

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Sample Paper-10, , SP-75, , A, , 24., , 25., 26., 27., , 7 cm, , Find the area of the unshaded part of the figure. (Take p = 3.14), (a) 449.75 cm2, (b) 520.60 cm2, (c) 563.72 cm2, The set of real numbers does not satisfy the property of, (a) multiplicative inverse , (b) additive inverse, , (d) 507.44 cm2, , (c) multiplicative identity , (d) none of these, The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0)is, (a) 5, (b) 12, (c) 11, (d) 7 + 5, Divide 62 into two parts such that fourth part of the first and two-fifth part of the second are in the ratio 2 : 3., (a) 24, 38, (b) 32, 30, (c) 16, 46, (d) 40, 22, 1, , 1, , For the equations (p + 2)  q −  = pq – 5 and (p – 2)  q −  = pq – 5, find the solution set (p, q)., , 2, 2, , , 1, , (a)  −10, − , 2, , , 28., , P B Q, , 1, , (b)  −10, , 2, , , 1, , (c) 10, − , 2, , ,  1, (d) 10, ,  2, , (b) sin q, , (c) 2 cos q, , (d) 2 sin q, , If cos q + 3 sin, =, q 2sin q, Then sin q − 3 cos q, (a) cos q, , 29., 30., , 31., 32., 33., 34., , 35., , Which of the following is the quadratic polynomial whose zeros are, , 1, −2, and, ?, 3, 5, , (a) 15x2 + x – 2, (b) 15x2 + 5x – 6, (c) 15x2 – 5x + 6, Two fair dice are thrown. Find the probability that both dice show different numbers., 1, 5, 32, (a), (b), (c), 6, 6, 36, , (d) 15x2 – x + 2, (d), , 29, 36, , The coordinates of the mid points of the line segment joining the points (3p, 4) and (–2, 2q) are (5, p). Then, (a) p = 4, q = 2, (b) q = 6, p = 2, (c) p + q = 8, (d) p – q = –2, The sum of a rational and an irrational number is _______., (a) an irrational number (b) a rational number, (c) an integer, (d) a whole number, Solve for q, , cos 2 q, , = 3; (q < 90º ) :, cot 2 q − cos 2 q, (a) 30°, (b) 90°, (c) 0°, (d) 60°, Two poles of heights 6 metres and 11 metres stand vertically on a plane ground. If the distance between their feet is 12, metres, what will be the distance between their tops?, (a) 10 m, (b) 12 m, (c) 13 m, (d) 15 m, In the given figure, O is the centre of the circle whose diameter is 14 cm. , 35 m, 22, Find the perimeter of the figure. (Use p =, ), 7, (a) 134 cm, O, (b) 124 cm, (c) 112 cm, , 36., , (d) 160 cm, Twice the product of the zeroes of the polynomial 23x2 – 26x + 161 is 14p. Then p is, 5, (a) 3, (b) 1, (c), 2, , (d) (–1)

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Mathematics, , SP-76, , 37., 38., 39., 40., , In what ratio does the point (–2, 3) divide the line–segment joining the points (–3, 5) and (4, –9) ?, (a) 2 : 3, (b) 1 : 6, (c) 6 : 1, (d) 2 : 1, The sum of three non-zero prime number is 100. One of them exceeds the other by 36. Find the largest number., (a) 73, (b) 91, (c) 67, (d) 57, If DABC ~ DDEF such that BC = 2.1 cm and EF = 2.8 cm. If the area of triangle DEF is 16 cm2, then the area of triangle, ABC (in sq. cm) is, (a) 9, (b) 12, (c) 8, (d) 13, The value of k for which the system of equation kx – y = 2, 6x – 2y = 3 has unique solution is, (a) not equal to one, (b) equal to three, (c) not equal to zero, (d) not equal to three, SECTION-C, , Case Study Based Questions:, Section C consists of 10 quesions of 1 mark each. Any 8 quesions are to be attempted., Q 41. - Q 45 are based on case study-I, Case Study-I, Situation-1, H.C.F. × L.C.M. = Product of two integers., 41. The H.C.F. of two numbers is 16 and their product is 3072. Find their L.C.M., (a) 182, (b) 121, (c) 192, (d) 3647, 42. The sum of two numbers is 135 and their H.C.F. is 27. If their L.C.M. is 162, the numbers are, (a) 108, 27, (b) 72, 54, (c) 81, 54, (d) 99, 36, Situation-2, HCF of natural numbers is the largest factor which is common to all the number and LCM of natural numbers is the smallest, natural number which is multiple of all the numbers., 43. If p and q are two co-prime natural numbers, then their HCF is equal to, (a) p, (b) q, (c) 1, (d) pq, 44. The LCM and HCF of two rational numbers are equal, then the numbers must be, (a) prime, (b) co-prime, (c) composite, (d) equal, 45. If two positive integers a and b are expressible in the form a = pq2 and b = p3q; p, q being prime number, then LCM (a, b), is, (a) pq, (b) p3q3, (c) p3q2, (d) p2q2, Q 46 - Q 50 are based on case study-II, Case Study-II, A chord of a circle of radius 10 cm subtends a right angle at the centre., 46. The area of minor sector is, (a) 78 cm2, (b) 79 cm2, 2, (c) 78.5 cm, (d) 77 cm2, 47. The area of minor segment is, (a) 28.5 cm2, (b) 27 cm2, 2, (c) 26 cm, (d) 30 cm2, 48. The area of major sector is, (a) 236 cm2, (b) 234 cm2, 2, (c) 237 cm, (d) 235.5 cm2, 49. The area of major segment is, (a) 285.5 cm2, (b) 286 cm2, 2, (c) 287 cm, (d) 288 cm2, 50. The length of arc APB is, (a) 17.15 cm, (b) 15.71 cm, (c) 25 cm, (d) 15 cm, , Q, , O, 90°, , A, , P, , B

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OMR ANSWER SHEET, Sample Paper No –, , Use Blue / Black Ball pen only., Please do not make any atray marks on the answer sheet., Rough work must not be done on the answer sheet., Darken one circle deeply for each question in the OMR Answer sheet, as faintly darkend / half darkened circle might by rejected., , , , , , , Start time : ____________________ End time ____________________ Time taken ____________________, 1., , Name (in Block Letters), , 2., , Date of Exam, , 3., , Candidate’s Signature, , SECTION-A, 1., , a, , 2., , a, , 3., , a, , 4., , a, , 5., , a, , 6., , a, , 7., , a, , 8., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 9., , a, , d, , 10., , a, , d, , 11., , a, , d, , 12., , a, , d, , 13., , a, , d, , 14., , a, , d, , 15., , a, , d, , 16., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 17., , a, , d, , 18., , a, , d, , 19., , a, , d, , 20., , a, , d, , 37., , a, , d, , 38., , a, , d, , 39., , a, , d, , 40., , a, , d, , 49., , a, , d, , 50., , a, , d, ,   ,   ,   , , b, b, b, ,   , , b, ,   , , b, ,   ,   ,   , , c, c, c, ,   , , c, ,   , , c, ,   ,   ,   , , d, d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-B, 21., , a, , 22., , a, , 23., , a, , 24., , a, , 25., , a, , 26., , a, , 27., , a, , 28., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, , 29., , a, , d, , 30., , a, , d, , 31., , a, , d, , 32., , a, , d, , 33., , a, , d, , 34., , a, , d, , 35., , a, , d, , 36., , a, ,   ,   ,   ,   ,   ,   ,   ,   , , b, b, b, b, b, b, b, b, ,   ,   ,   ,   ,   ,   ,   ,   , , c, c, c, c, c, c, c, c, ,   ,   ,   ,   ,   ,   ,   ,   , , d, ,   ,   , , b, b, ,   , , b, ,   , , b, ,   ,   , , c, c, ,   , , c, ,   , , c, ,   ,   , , d, d, ,   , , d, ,   , , d, , d, d, d, , SECTION-C, 41., , a, , 42., , a, , 43., , a, , 44., , a, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , No. of Qns. Attempted, , c, c, c, c, ,   ,   ,   ,   , , d, , 45., , a, , d, , 46., , a, , d, , 47., , a, , d, , 48., , a, , Correct, ,   ,   ,   ,   , , b, b, b, b, ,   ,   ,   ,   , , c, c, c, c, ,   ,   ,   ,   , , d, ,   , , b, ,   , , d, , Incorrect, , Marks, , c, ,   , , d

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Sample Paper, , 1, , ANSWERKEY, 1, 11, 21, 31, 41, , (a), (a), (c), (a), (c), , 1., (a), , ∴, 2., (a), 3., (d), , 2, 12, 22, 32, 42, , (a), (d), (a), (c), (d), , 3, 13, 23, 33, 43, , (d), (d), (d), (a), (c), , 4, 14, 24, 34, 44, , (b), (b), (c), (b), (d), , 5, 15, 25, 35, 45, , (a), (b), (d), (a), (c), , Let the required numbers be 15x and 11x., Then, their H.C.F. is x. So, x = 13., The numbers are 5 × 13 and 11 × 13 i.e., 195 and 143., Terminating, (a) is false Q π is the ratio of the circumference of, a circle to the length of the diameter., , 22, but not exactly., 7, , (b) False [Q real numbers can be irrationals also], , (c) False [Q non-terminating decimal can be, recurring or non-recurring], , (d) Since 0.2 < 0.21 < 0.3, 4., (b) Degree of quotient = degree of dividend – degree of, divisor, , Degree of quotient = 7 – 4 = 3., 5., (a) 1 is zero of p(x), ⇒ p (1) = 0, ⇒ a(1)2 – 3 (a – 1) (1) –1, = 0, ⇒ –2a + 2 = 0, ⇒ a = 1, , , 6., , It is nearly equal to, , (c) Here, the two triangles are similar., , Ratio of areas of two similar triangles is equal to the, ratio of squares of their corresponding altitudes., , , h12, , 25, So,, =, 2, h2 36, ∴, , 7., (b), , , , , h1 5, =, h2 6, , n(S) = 6 × 6 = 36, E= {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5),, (5, 4), (5, 6), (6, 5)}, n(E) = 10, , 6, 16, 26, 36, 46, , (c), (c), (d), (d), (a), , 7, 17, 27, 37, 47, , \, , (b), (a), (a), (c), (c), , P(E), =, , 8, 18, 28, 38, 48, , (c), (b), (b), (d), (a), , 9, 19, 29, 39, 49, , (a), (a), (b), (b), (b), , 10, 20, 30, 40, 50, , (d), (a), (c), (d), (b), , n(E) 10 5, = =, n(S) 36 18, , 8., , (c) For reflection of a point with respect to x-axis change, sign of y-coordinate and with respect to y-axis change, sign of x-coordinate., , 9., , (a) It is given that AD is the bisector of ∠A., AB BD, 6×3, ⇒ AC = = 4.5 cm, =, AC DC, 4, , 10., , (d) Given, tan θ =, , a, b, , a sin θ − b cos θ a tan θ − b a 2 − b 2, =, =, a sin θ + b cos θ a tanθ + b a 2 + b 2, (a) Let the age of father be ‘x’ years and the age of son, be ‘y’ years, , ∴, 11., , , , According to question, x + y = 65 , , ...(i), , , , and 2(x – y) = 50 ⇒ x – y = 25 , , ...(ii), , , , Adding eqs. (i) and (ii), we get, 2x = 90 ⇒ x = 45, , , , Hence, the age of father = 45 years, ,  4 × 3 + 1× 6 3 × y + 1× 5 , ,, (d) P(6, 2) = , , 3 +1 ,  3 +1, 18, Q 6≠, , (Question is wrong), 4, 3y + 5, , 2=, ⇒ 3y + 5 = 8, 4, 3y = 3 ⇒ y = 1, 12.

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Mathematics, , S-2, , 13., , x, (d) We know that sec2θ – tan2θ = 1 and sec θ = ,, p, y, tan θ =, q, ∴, , 14., , x2q2 – p2y2 = p2q2, , (b) Substitute x = 1 in f (x) and x = –2 in g(x), and add, , f (1) = 2(1) – 6(1) + 4(1) – 5 = –5 ⇒ g(–2) = 3(4) – 9 = 3, f (1) + g(–2) = –2, 15. (b) S = {1, 2, 3, ......, 100}, , n(S) = 100, , E = {11, 22, 33, 44, 55, 66, 77, 88, 99}, , n(E) = 9, 9, \ P(E) =, 100, , 20. (a), , , , , \, 21. (c), , Let the numbers be 37a and 37b. Then, 37a × 37b = 4107 ⇒ ab = 3, Now, co-primes with product 3 are (1, 3), So, the required numbers are, (37 × 1, 37 × 3) i.e., (37, 111)., Greater number = 111, We have ABCD is square,, , , , AF = BF, BE =, , 1, BC,, 3, , Area ∆FBE = 108 sq cm., , , Let AB = x ⇒ BF =, , x, x, and BE =, 2, 3, , D, , C, , 16., 17., , (c), (a) Let the speeds of the cars starting from A and B be x, km/hr and y km/hr respectively, , According to problem,, 9x – 90 = 9y , .... (i), 9, 9, 90, x+ y =, 7, 7, , , , A, .... (ii), , , Solving we get x = 40 km/hr, y = 30 km/hr,, , speed of car A = 40 km/hr, , & speed of car B = 30 km/hr, 18. (b) Given (x)2 + (x + 2)2 = 290, ⇒ x2 + x2 + 4x + 4 = 290, ⇒ 2x2 + 4x – 286 = 0, ⇒ x2 + 2x – 143 = 0, ⇒ x2 + 13x – 11x – 143 = 0, ⇒ (x + 13) (x – 11) = 0, ⇒ x = –13, x = 11, x cannot be negative, discard x = –13, so x = 11, Hence the two consecutive positive integers are 11, 13, 19. (a) Given equations are :, 7x – y = 5 and 21x – 3y = k, Here a1 =, 7, b1 =, −1, c1 =, 5, 21, b2 =, −3, c2 =, k, a2 =, , We know that the equations are consistent with unique, solution, a1 b1, ≠, if, a2 b2, , , Also, the equations are consistent with many solutions, , a1 b1 c1, if = =, a2 b2 c2, 7 −1 5, 1 5, =, = ⇒ = ⇒ k = 15, 21 −3 k, 3 k, Hence, for k = 15, the system becomes consistent., , ∴, , , E, x/3, x, , , Area of DFBE =, , F, , x/2, , B, , 1, BF × BE, 2, , 1 x x x2,   , 2 2 3 12, = 108 sq. cm. (Given), ⇒ x2 = 12 × 108, ⇒ x2 = 12 × 12 × 3 × 3, ⇒ x = 12 × 3 = 36 cm, In rt. ∆ABC, AC = AB2  BC2, (By Pythagoras Theorem), , =, , , , , , = 362 + 362 = 36 2 cm, , 22. (a) Let D be the window at a height of 9m on one side, of the street and E be the another window at a height, of 12 m on the other side., In ∆ADC, AC2 = 152 – 92, , = 225 – 81, , AC = 12 m, In ∆ECB, CB2 = 152 – 122, , = 225 – 144, , CB = 9 m, , Width of the street = (12 + 9)m = 21 m, 23. (d) x – y = 2 , ... (i), , kx + y = 3 , ... (ii), , by adding (i) and (ii), , kx + x = 5

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S-3, , Solutions, , , x(k + 1) = 5, 5, x=, k, +1, , , putting value of x in equation (i), 5, −y=, 2, k, + 1, 5, −2=, y, k, + 1, 5 − 2k − 2, =y, k + 1, 3 − 2k, y=, k +1, , , y should be positive as they intersect in 1st quadrant, therfore, , y>0, 3 − 2k, 2k − 3, >0⇒, <0, k +1, k +1, + . − . +, −, ∞, −1, ∞, 3/ 2, , , ∴ k should lie between – 1 and 3/2, 24. (c) We have, sin x + cosec x, 1, ⇒ sin x +, sin x, , ∴ We know that sum of the number and its reciprocal, is greater than or equal to 2., 25. (d) We have, sin 5θ = cos 4θ, ⇒ 5θ + 4θ = 90°, , [ sin α = cos β, than α + β = 90°], ⇒ 9θ = 90° ⇒ θ = 10°, , Now, 2 sin 3θ – 3 tan 3θ, = 2sin 30° – 3 tan 30°, 1, 1, = 2× − 3 ×, = 1−1 = 0, 2, 3, , 26., 27., , (d) If two similar triangles have equal area then triangles, are necessarily congruent., (a) Here, the two lines are 2x + 3y = 7 and 2ax + (a + b), y = 28. The above lines are coincident., Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are, a1 b1 c1, =, coincident if =, a2 b2 c2, , 2, 3, −7, So, =, =, 2a a + b 28, ⇒ a = 4, b = 8, \ b = 2a, 28. (b) 2ax – 2by + a + 4b = 0 , ........ (i), , and 2bx + 2ay + b – 4a = 0, ........ (ii), , Multiplying eq. (i) with b and eq. (ii) with a, we get, 2abx – 2b2y + ab + 4b2 = 0, ........ (iii), , and 2abx + 2a2y + ab – 4a2 = 0 ........ (iv), , Subtracting (iv) from (iii), we get, , – (2b2 + 2a2) y + 4b2 + 4a2 = 0, , ⇒ – (2b2 + 2a2) y = – 4b2 – 4a2 ⇒ y = 2, , Substituting y = 2 in eq. (i), we get, 2ax – 2b × 2 + a + 4b = 0, ⇒ x = – 1/2, ∴ x = – 1/2, y = 2, 29. (b) Put x + 1 = 0 or x = – 1 and x + 2 = 0 or, x = – 2 in p (x), , Then, p (–1) = 0 and p (–2) = 0, 3, 2, ⇒ p (–1) = ( −1) + 3 ( −1) − 2α ( −1) + β = 0, ⇒ −1 + 3 + 2α + β = 0 ⇒ β = −2α − 2 .... (i), 3, 2, p ( −2) = ( −2) + 3( −2) − 2α ( −2) + β = 0, ⇒ −8 + 12 + 4α + β = 0 ⇒ β = − 4α − 4 .... (ii), , By equalising both of the above equations, we get, −2α − 2 = − 4α − 4, , ⇒ 2α = −2 ⇒ α = −1, put α = −1 in eq. (i), ⇒ β =−2 ( −1) − 2 =2 − 2 =0, , 30., , Hence, α = −1, β = 0, , 5, (c) Let a, b be two zeroes of 2x2 – 8x – m, where α = ., 2, \ α+β =, , ( −Coefficient of x ), Coefficient of x 2, , 5, 8, +β =, 2, 2, , 8 5 3, − = ., 2 2 2, 31. (a) Let f (x) = 2 x 3 – 5x2 + ax + b, f (2) = 2 (2)3 – 5 (2)2 + a (2) + b = 0, ⇒ 16 – 20 + 2a + b = 0 ⇒ 2a + b = 4, f (0) = 2 (0)3 – 5 (0)2 + a (0) + b = 0 ⇒ b = 0, ⇒ 2a = 4 ⇒ a = 2, b = 0, 32. (c) cos A =3 ⇒ sin A = 1 − 9 =4, 5, 25 5, ⇒, , ⇒, , β=, , consider, 9 cot 2=, A −1,  , , 9 cos 2 A, 9 cos 2 A − sin 2 A, =, −, 1, sin 2 A, sin 2 A, ,  9   16 , 9  −  ,  25   25 , =, =, 16, , 25, , (81 − 16) ×=, 25, 25, , 16, , 65, 16, , (a) All isosceles triangles are not similar., (b) Let α and 6α be roots of equation., 14, , Sum of roots : α + 6α =, p, 33., 34., , ⇒ 7α =, , 14, 2, ⇒p=, p, α, , Product of roots : (α) (6α) =, , 4, 8, ⇒p=, p, 3α 2

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Mathematics, , S-6, , 12., , x, , 3200 from (i),, Also (x + y) +  + y  =, 2, , , , , , , 2125 +, , The multiples of 3 from 18 to 38 are 18, 21, 24, 27,, 30, 33, 36., , x, x, 3200,, y 3200 − 2125, +y=, +=, 2, 2, , , , ⇒ x + 2y = 1075 ⇒ x + 2y = 2150, ....(ii), , Solving (i) and (ii), we get, , – y = – 25 or y = 25, , Putting the value of y = 25 in (i), , x + 25 = 2125, , x = 2125 – 25, x = 2100, full fare = ` 2100 and reservation charges = ` 25, 42, 6., (b) Diameter of each semi-circle =, = 14 cm, 3, Radius of each semi-circle = 7 cm, , (c) Initial number of workers = 120, , When 15 male workers are added, then the total number, of workers = 120 + 15 = 135, Number of female workers = 90, 90 2, , \ Probability of female workers =, =, 135 3, (a) When 2256 is divided by 17 then,, , 2256, 24 + 1, , =, , (24 + 1), , Here, f (a) = (24)64 and x = 24 and a = 1, ∴ Remainder = f (–1) = (–1)64 = 1, (c) As PQ is parallel to BC ⇒ DABC ~ DAPQ, , , ⇒, , 10., , (c), , Area of ∆ABC 2, =, Area of ∆APQ 1, , AB, Ratio of sides, = =, AP, , 2, 1, , \ AP : AB = 1 :, , 11. (b) Suppose the required ratio is m1 : m2, Then, using the section formula, we get, –2 =, , m1 ( 4 ) + m 2 ( –3), m1 + m 2, , ⇒ ––2m1 – 2m2 = 4m1 – 3m2, ⇒ m2 = 6m1 ⇒ m1 : m2 = 1 : 6, , 7 1, =, 21 3, , 13., , (b), , 14., , (b) 196 = 22 ⋅ 72, sum of exponents = 2 + 2 = 4, , 15. (b) We have,, Area of square metal plate = 40 × 40 = 1600 cm2, , , 2, , 22  1 , 11, Area of each hole =, πr 2 = ×   = cm 2, 7 2, 14, 11, = 346.5cm 2, 14, Hence, area of the remaining square plate, = (1600 – 346.5) = 1253.5 cm2, , 16., , (c), , 17., , (a) Given, AB = 2DE and ∆ABC ~ ∆DEF, , Hence,, or, , , 2, , area(∆ABC) AB2, =, area(∆DEF) DE 2, , 56, 4DE 2, = =, 4 [Q AB = 2DE], area(∆DEF) DE 2, , area (∆DEF), =, , (24 )64, , By remainder theorem when f (x) is divided by x + a the, remainder = f (– a), , 9., , , ∴ Required probability =, , , , , Area of the coloured portion = 588 – 462 = 126 cm2, , 8., , These are 7 in numbers, , , ∴ Area of 441 holes = 441×, , πr 2, Area of 6 semi-circle = 6 ×, = 3pr2, 2, 22, = 3×, × 7× 7 = 462 cm2, 7, Area of cloth piece = 42 × 14 = 588 cm2, 7., , (d) Total number of marbles = 38 – 18 + 1 = 21, , 56, = 14sq.cm., 4, 91 × 126, 91 × 126, = = 13, L.C.M.(91, 126), 182, , 18., , (a) H.C.F. (91, 126) =, , 19., , (b) Total number of cards = 52, , , , Total number of diamond cards = 13, , , , I., , P(diamond cards) = 13/52 = 1/4, , , , II., , P(an ace of heart) = 1/52, , , , III., , P(not a heart) = 1 −, , , , IV. P(king or queen) =, , 20., , 1 3, =, 4 4, 4, 4, 8, 2, +, =, =, 52 52 52 13, , (b) Let the required ratio be K : 1, ∴ The coordinates of the required point on the y-axis is, x=, , K(−4) + 3(1), K(2) + 5(1), ; y=, K +1, K +1, , Since, it lies on y-axis, ∴ Its x-cordinates = 0

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S-7, , Solutions, ∴, , ⇒K =, , i.e. (x – 5)2 + (0 – 4)2, = (x + 2)2 + (0 – 3)2, , 3, 4, , ⇒x=2, , ⇒ Required ratio =, ∴ ratio = 3 : 4, 21., , AP2 = BP2, , −4 K + 3, 0 ⇒ −4 K + 3 =0, =, K +1, , (d) We have,, , 3, :1, 4, , 25., , Let P be the point trisection of the line AB, then P divides, AB in the ratio 1 : 2, So coordinate of P is, æ1(5) + 2 (-2) 1(4) + 2 (-19)ö÷, çç, ÷÷, ,, ç, 1+ 2, 1+ 2, èç, ø÷, , cos θ − sin θ 1 − 3, =, cos θ + sin θ 1 + 3, , (cos θ − sin θ) + (cos θ + sin θ), ⇒, (cos θ − sin θ) − (cos θ + sin θ), , (1 – 3) + (1 + 3), =, (1 − 3) − (1 + 3), , , [Applying componendo and dividendo], 2 cos θ, 2, ⇒ =, ⇒ cot, =, θ, −2sin θ −2 3, , æ 5 - 4 4 - 38 ö÷ æ 1 -34 ö÷, = çç, ,, ÷ = çç ,, ÷, èç 3, 3 ø÷ èç 3 3 ø÷, 26., , 1, 3, , y, = cosec θ – cot θ , b, , ...(ii), , x y, ⇒ × = (cosec θ + cot θ) (cosec θ – cot θ), a b, xy, ⇒, = (cosec2θ – cot2 θ), ab, 23., , (b) Since (x, y) is midpoint of (3, 4) and (k, 7), ∴, , ⇒ tan θ = 3 ⇒ tan θ = tan 60° ⇒ θ = 60°, 22. (c) We have, x = a (cosec θ + cot θ), x, ⇒, = (cosec θ + cot θ) ...(i), a, y, 1, cos θ,  1 − cos θ , −, and y = b ,  ⇒=, sin, θ, b, sin, θ, sin θ, , , ⇒, , (c) Given points are A(–2, –19) and B(5, 4), , ∴ xy = ab, , (d) All the statements given in option (a, b, c) are correct., , 24., , (a) Since, the required point (say P) is on the x-axis, its, ordinate will be zero. Let the abscissa of the point be x., Therefore, coordinates of the point P are (x, 0)., , 3+ k, 4+7, and y =, 2, 2, , Also 2x + 2y + 1 = 0 putting values we get, 3+k+4+7+1=0, ⇒ k + 15 = 0 ⇒ k = – 15, 27., , (a) If the lines are parallel, then, , a1 b1 c1, a= b ≠ c, 2, 2, 2, , Here, a1 = 3, b1 = – 1, c1 = –5,, a2 = 6, b2 = – 2, c2 = – p, , ⇒, , , , , 3 −1 −5, =, ≠, , 6 −2 − p, , ... (i), , Taking II and III part of equation (i), we get, , , ⇒, , 28., , P (x, 0), , x=, , 1 −5, ≠, 2 −p, , ⇒ − p ≠ −10, , ⇒ p ≠ 10, , So, option (a) is correct., , (d) All equilateral triangles are similar, , , ∴ ∆ ABC~ ∆EBD, Area of ABC BC2, , Area of BDE BD 2, D is mid-point of BC, , A, , , ⇒, , , , \ BC =, = 2BD, , A (5, 4), , B (–2, 3), , Let A and B denote the points (5, 4) and (–2, 3), respectively., Given that AP = BP, we have, , , ⇒, 29., , ( 2BD) 2, , E, B, , 4, =, 2, 1, BD, , D, , Area (∆ABC) : Area (∆BDE) = 4 : 1, , (b) Coordinates of mid-point are given by,  x1 + x 2 y1 + y 2 , ,, , , 2 ,  2, , C

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Mathematics, , S-8, , a , Here, coordinates of mid-point are  , 4 , 3 , a −6 − 2, So, =, 3, 2, ∴ a = – 12, 30., , (b) [Hint. The outcomes are 1, 2, 3, 4, 5, 6. Out of these,, 4 is the only composite number which is less than 5]., 31. (a) In DABC, AB = AC, , Draw AL ⊥ BC,, then L is the mid-point of BC, Using Pythagoras theorem in ∆ABL, we get, AL = 8cm, Also, ∆BPS ≅ ∆CQR,, , \ BS = RC, SL = LR = x cm, , \ BS = CR = 6 – x, , In DABL, PS || AL, , \, , PS BS, y 6− x, =, ⇒ =, AL BL, 8, 6, , or x= 6 −, , 34., , 35., , 36., , (a) Required number = H.C.F. {(70 – 5), (125 – 8)}, , , 37., , m, 10, c, , cm, , 32., , y, , 2x, L, , y, , S, , R, , C, , (a) Since zeroes are reciprocal of each other, so product, k+2, of the roots will be 1, so, =1,, k2, , , k2 – k – 2 = 0 ⇒ (k – 2)(k + 1) = 0, , k = 2, k = –1, Since k > 0 ∴ k = 2, 33., , (a) Area of the shaded region, 40° 22, 40° 22, =, × × (7)2 −, × × (3.5) 2, 360° 7, 360° 7, 1 22, 1 22, 49, = × × (7 2 − 3.52 ) = × ×  49 − , 9 7, 9 7 , 4 , 1 22 49, 77 2, = × × × 3 = cm, 9 7 4, 6, , (a) In DAFD & DFEB,, , , ∠3 = ∠4 (Alternate angle), , \ ∆ FBE ~ ∆ FDA, EF FB, , FA DF, PQ = 13 ⇒ PQ2 = 169, (x – 2)2 + (–7 – 5)2 = 169, x2 – 4x + 4 + 144 = 169, x2 – 4x – 21 = 0, x2 – 7x + 3x – 21 = 0, (x – 7) (x + 3) = 0, x = 7, –3, , So,, , 10, , B, , Q, , = H.C.F. (65, 117) = 13., , , ∠1 = ∠2 (V.O.A), , A, , 2x, , 2, , 225, 2(1 − sin 2 θ) 2cos 2 θ,  15 , 2, =, = cot =, θ   =, 2, 2, 64, 8, 2(1 − cos θ) 2sin θ, (c) [Hint. The English alphabet has 26 letters in all. The, word ‘DELHI’ has 5 letter, so the number of favourable, outcomes = 5.], , 38., , P, , (2 + 2sin θ) (1 − sin θ), 2(1 + sin θ) (1 − sin θ), =, (1 + cos θ) (2 − 2 cos θ), (1 + cos θ) (2) (1 − cos θ), , =, , , 3, y, 4, , P, , (b), , (c), ⇒, ⇒, ⇒, ⇒, ⇒, ⇒, , 39. (b) Required number = H.C.F.{(245 – 5), (1029 – 5)}, = H.C.F. (240, 1024) = 16., 40. (c), AB, AC, 5, 3, =, ⇒ =, 41. (c), A ' B ' A ' C ' 15 A ' C ', 42., , ⇒ A′C′ = 9 cm, AB, BC, 5 BC, =, ⇒ =, (a), A ' B ' B ' C ' 15 12, , ⇒, 43. (b), 44. (a), 45. (b), , , BC = 4 cm, Q ∠A = ∠A′ = 80°, Q ∠B = ∠B′ = 60°, Q ∠A + ∠B + ∠C = 180°, 80° + 60° + ∠C = 180°, , ∠C = 40°, 46., , (a) , , 47., , (a), , 49., , (b) , , 50., , (c), , 48., , (c)

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Sample Paper, , 3, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., , (b), (b), (d), (d), (c), , 2, 12, 22, 32, 42, , (b), (a), (d), (a), (d), , 3, 13, 23, 33, 43, , (b), (d), (b), (b), (a), , 4, 14, 24, 34, 44, , (a), (d), (b), (d), (b), , (b) Here, x – y = 3 , , 5, 15, 25, 35, 45, , (a), (a), (b), (d), (b), , ...(i), , and xy = 54, \ (x +, , y)2, , = (x –, , y)2, , 6, 16, 26, 36, 46, , = (3)2 + 4(54) = 225, ⇒ ( x + y ) =225 =, ±15 ...(ii), Case I :, , (a), (c), (d), (a), (a), , 8, 18, 28, 38, 48, , (b), (c), (b), (b), (b), , 9, 19, 29, 39, 49, , 4., , On adding the above two equations, 2x = 18 ⇒ x = 9, , , ⇒ k = 2 and k = – 1, , \ x + y = 15 ⇒ 9 + y = 15 ⇒ y = 6, , But k > 0 \ k = 2, , Case II, , 5., , If x + y = – 15 and x – y = 3, , , p(– 2) = k (– 2 + 1)2 = 2, , On adding the above two equations, , , k=2, , 2x = – 12, , , p(x) = 2 (x + 1)2, , x=–6, , , p(2) = 2(2 + 1)2 = 2 × 3 × 3 = 18, , \ x + y = – 15 ⇒ – 6 + y = – 15, , 6., , ⇒ y = – 15 + 6 ⇒ y = – 9, , , , 2, , 2, , 2, , 16  AB , 16  12 , =, ⇒, , , ,  =,  and, 9  18 , 9  QR , 4 AB, 4 12, =, ⇒, =, and, , 3 18, 3 QR, , 10, 20, 30, 40, 50, , (d), (b), (a), (a), (d), , 1, = cos θ and maximum value of cos θ is 1, sec θ, 1, ⇒ Maximum value of, is 1, sec θ, 1, (a) α, are the roots of k2x2 – 17x + (k +2), α, , (b), , 1 k+2, =, α, k2, , ⇒ k2 = k + 2 ⇒ k2 – k – 2 = 0, , 16  AB ,  BC , (b), = =, , , , 9  PQ ,  QR , , (d), (c), (b), (d), (c), , α ×, , If x + y = 15 and x – y = 3, , 2., , 7, 17, 27, 37, 47, , , ⇒ AB = 24 cm QR = 9 cm, 3., , + 4xy, , (c), (b), (c), (d), (d), , (a) Quadratic polynomial p(x) = k (x + 1)2, , (c) P(raining on both day) = 0.2 × 0.3 = 0.06, (Because both independent event), , 7., , (a) Statement given in option (a) is false., , 8., , (b) 2 πr1 =, 503 and 2 πr2 =, 437, , 2, , ∴, , r1 =, , 503, 437, and r2 =, 2π, 2π, , Area of ring = π (r1 + r2 ) (r1 − r2 )

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Mathematics, , S-10, , 18., ,  503 + 437   503 − 437 , = π, , , 2π, 2π, , , , =, 9., 10., , 11., , ∴, , 940  66 , 66,  = 235 × × 7= 235 × 21= 4935 sq. cm., 2  2π , 22, , (d), (d) L.C.M × H.C.F = First number × second number, 36 × 2, Hence, required number =, =4., 18, (b), , 12., 13., , (a), (d) Sum is 888 ⇒ unit’s digit should add up to 8. This is, possible only for option (d) as “3” + “5” = “8”., , 14., , x, (d) Let the fraction be, y, x +1, = 4 , y +1, , ... (i), , x −1, = 7 , y −1, , ... (ii), , 15., , Solving (i) and (ii), we have x = 15, y = 3, i.e. numbers = 15, (a) Let the radii of the two circles be r1 and r2, then, r1 + r2 = 15       (given), ..... (i), 2, and πr12 + πr2=, 153π   , , ⇒, , (c) Let the ages of father and son be 7x, 3x, After 10 years,, \ (7x + 10) : (3x + 10) = 2 : 1 or x = 10, \ Age of the father is 7x i.e. 70 years., , 20., , (b) 24 out of the 90 two digit numbers are divisible by ‘3’, and not by ‘5’., The required probability is therefore,, , 21., , 22., , (given), , r12 + r22 =, 153 , , AD AE, 1.5, 1, =, ⇒, =, ⇒ EC = 2 cm, DB EC, 3, EC, , 19., , According to given conditions,, , and, , (c) Since, DE || BC ∴ ∆ADE ~ ∆ABC, , ..... (ii), , (d) Let x2 = u, y2 = v, u v, 5, ⇒ 2u + 3v = 35 and + =, 2 3, ⇒ 2u + 3v = 35, ...(i), ⇒ 3u + 2v = 30, ...(ii), Multiply (i) by 3 and (ii) by 2 and subtracting (ii) from (i),, we have, 6u + 9v – 6u + 4v = 60, ⇒ 6u – 6u + 9v – 4v = 105 – 60, ⇒ 5v = 45 ⇒ v = 9, substituting v = 9 in (1), we get 2u + 27 = 35, ⇒ 2u = 8 ⇒ u = 4 ⇒ x2 = 4, y2 = 9, \ x=, ± 2, y =, ± 3 is the required solution., (d) Let Area of ∆BEF = x, ∴ Area of ∆AFE = 3x, Let Area of ∆ABF = 3y, ∴ Area of ∆CAF = 2y, A, , On solving, we get, r1 = 12, r2 = 3, Required ratio = 12 : 3 = 4 : 1, 16., , E, 1, , (b) x2 – (m +3)x + mx – m(m + 3) = 0, , , ⇒, , x[x – (m + 3)] + m[x – (m + 3)] = 0, , , ⇒, , (x + m) [x – (m + 3)] = 0, , , \, , x + m = 0  , , x – (m + 3) = 0, , x = –m, 17., , x=m+3, , (c) We have, sum of zeroes, , B, C, F 2, 3, , Area ∆ABF = Area ∆BEF + Area ∆AEF, 3y = x + 3x, 3y = 4x, 3 x, , 4 y, Area of ∆ABC = Area of ∆ABF + Area of ∆CAF, = 3y + 2y = 5y, Area BEF, x 1 3 3, ,   , Area ABC 5y 5 4 20, , ( −4), , 2, = a + b =, −, =, 2, , , Product of zeroes = ab =, , , ∴, , a2b, , +, , ab2, , 24, 4, =, ., 90 15, , 3, 2, , 3, = ab (a + b) = × 2 = 3, 2, , 23., , (b) sin θ + 2 cos θ = 1 ⇒ (sin θ + 2 cos θ)2 = 1, ⇒ sin2 θ + 4 cos2 θ + 4 sin θ cos θ = 1, ⇒ 1 – cos2 θ + 4, (1 – sin2 θ + 4 sin θ cos θ =1

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S-11, , Solutions, ⇒ 4 sin2 θ + cos2 θ – 4 sin θ cos θ = 4, ⇒ (2 sin θ – cos θ)2 = 4, , ⇒ 2 sin θ – cos θ = 2, , [ 2 sin θ – cos θ ≠ –2], 24. (b) α + β = 5 , ...(i), αβ = k , ...(ii), α – β = 1 , ...(iii), , Solving (i) and (iii), we get α = 3 and β = 2., , Putting the value of α and β in (ii), we get, 25. (b) x + y = 1 & x3 + y3 + 3xy, , = (x + y)3 – 3xy(x + y) + 3xy = 1, , , 26., ,  2 A − 3A, ,  × 100, 2 , ∴ Required percentage increase =, 3A, 2, 1, 1, = × 100 ⇒ 33 %, 3, 3, 30., , ∴ p(x) = (x –15) (x + 5), , ⇒ 2r +, , So, p(x) = 0 when x = 15 or x = –5. Therefore required, zeroes are 15 and –5., (d) Let cosec x – cot x =, , ⇒, , θ, × 2pr = 25, 360°, , 90°, 22, ⇒ 2r + 360° × 2 ×, × r = 25, 7, , = x (x – 15) + 5 (x –15) = (x – 15) (x + 5), , 27., , (a) Perimeter of sector = 25 cm, ⇒ 2r +, , (c) We have, p(x) = x2 –10x –75 = x2 – 15x + 5x – 75, , [from (i) & (ii) eqns.], , 11, 25, r = 25 ⇒, r = 25 ⇒ r = 7, 7, 7, ,  πθ sinθ  2, −, Area of minor segment = , r,  360°, 2 , , 1, 3, ,  22 90° sin 90°  2, −, = ×,  (7), 2 ,  7 360°, , 1, cos x 1, –, =, sin x sin x 3, , x, 2sin 2, 1 – cos x 1, 1, 2, ⇒, =, ⇒, =, x, x 3, sin x, 3, 2 sin cos, 2, 2, x 1, =, 2 3, Consider, ,  11 1 , 4, =  −  × 49 =, × 49 = 14 cm2., 14,  14 2 , , 31., , (d) ∵, , Perimeter of ∆ABC AB BC AC, = = =, Perimeter of ∆PQR PQ QR PR, , (a) The number divisible by 15, 25 and 35 = L.C.M. (15,, 25, 35) = 525, , Since, the number is short by 10 for complete division, by 15, 25 and 35., Hence, the required least number = 525 – 10 = 515., , ⇒ tan, , 32., , 2, x, 2 tan, 3, 3, 2, = =, tan x =, 1 4, 2 x, 1 – tan, 1–, 2, 9, , 33., , (b) [Hint. One digit prime numbers are 2, 3, 5, 7. Out of, these numbers, only the number 2 is even.], , 34., , (d) Work ratio of A : B = 100 : 160 or 5 : 8, , 3, 4, =, , cos x, 5, 5, 16, 9, 7, –, ∴ cos2 x – sin2x =, =, 25 25 25, Thus, =, sin x, , 28., , (b) The point satisfy the line 4y = x + 1, , 29., , (b) Let salary of Y be = A and of X is =, , ∴ time ratio = 8 : 5 or 24 : 15, , A, 2, , 3A, , ... (i), 2, Let X’ and Y ’ be the new salary after increment, then we, get, ∴ Total salary of X and Y =, , X' =, , 5A, 3A, and Y ' =, ⇒ X '+ Y' =, 2 A ... (ii), 4, 4, , If A takes 24 days, B takes 15 days. Hence, B takes, 30 days to do double the work., 35. (d) Out of n and n + 2, one is divisible by 2 and the, other by 4, hence n (n + 2) is divisible by 8. Also n, n + 1,, n + 2 are three consecutive numbers, hence one of them is, divisible by 3. Hence, n (n + 1) (n + 2) must be divisible by, 24. This will be true for any even number n., 36. (d) The L.C.M. of 16, 20 and 24 is 240. The least multiple, of 240 that is a perfect square is 3600 and also we can, easily eliminate choices (a) and (c) since they are not, perfect number. Hence, the required least number which

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Mathematics, , S-12, , is also a perfect square is 3600 which is divisible by each, of 16, 20 and 24., 37., , (a) Since, ∆ABC ~ ∆PQR, , , \, 38., , 2, , (b) Area of rectangle = 28 × 23 = 644 cm2, Radius of semi-circle = 28 ÷ 2 = 14 cm, Radius of quadrant = 23 – 16 = 7 cm, Area of unshaded region, 1 22, , , 2,  2 × 4 × 7 × 7 × 7  = 385 cm, , , , \ Shaded area = (644 – 385) = 259, 39., , (d), , 1 2 −3, = ≠, , 5 k, 7, , , ⇒, , k = 10, , (a) Required probability =, , cm2, , ∵ For inconsistent , , a1 b1 c1 , , =, ≠, , a2 b2 c2 , , 4 2, = ., 6 3, , (c) For getting least number of books, taking LCM of 64, 72, 8, 64, 72, 8, 9, , , ⇒ 8 × 8 × 9 = 576, 42. (d), 43. (a) 72 is expressed as prime, 72 = 2 × 2 × 2 × 3 × 3 = 23 × 32, 44. (b) 5 × 13 × 17 × 19 + 19, ⇒ 19 × (5 × 13 × 17 + 1), so given no. is a composite number., 45. (b), 41., , 2, , ar(∆PQR) PR, QR, 9  QR 3 , =, =, =, ∵, =, =9, ar(∆ABC ) AC 2 BC 2 1  BC 1 , ,  1 22, , =  × × 14 × 14  +, 2 7, , , 40., , 46., , (d) parabola, , 47., , (a) 2, , 48., , (b) –1, 3, , 49., , (c) x2 – 2x – 3, , 50., , (d) 0

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Sample Paper, , 4, , ANSWERKEY, 1, 11, 21, 31, 41, , (b), (a), (a), (d), (a), , 2, 12, 22, 32, 42, , (a), (c), (d), (b), (d), , 3, 13, 23, 33, 43, , (b), (c), (d), (b), (a), , 4, 14, 24, 34, 44, , (b), (c), (a), (b), (b), , 5, 15, 25, 35, 45, , (d), (c), (b), (d), (c), , 1., (b) x + y = 1 & x3 + y3 + 3xy, , = (x + y)3 – 3xy(x + y) + 3xy = 1, 2. (a) Since, the required point (say P) is on the x-axis, its, ordinate will be zero. Let the abscissa of the point be, x., , Therefore, coordinates of the point P are (x, 0)., P (x, 0), , A (5, 4), , , B (–2, 3), , Let A and B denote the points (5, 4) and, (–2, 3) respectively., , Given that AP = BP, we have, AP2 = BP2, i.e. (x – 5)2 + (0 – 4)2, , = (x + 2)2 + (0 – 3)2, ⇒x=2, 3., (b) Here, x – y = 3 , ...(i), , and xy = 54, ∴ (x + y)2 = (x – y)2 + 4xy, = (3)2 + 4(54) = 225, ±15, ⇒ ( x + y ) =225 =, ...(ii), Case I :, , If x + y = 15 and x – y = 3, , On adding the above two equations, , 2x = 18 ⇒ x = 9, , 6, 16, 26, 36, 46, , (d), (c), (d), (c), (a), , 7, 17, 27, 37, 47, , (c), (b), (a), (b), (c), , 8, 18, 28, 38, 48, , (a), (b), (b), (c), (b), , 9, 19, 29, 39, 49, , (c), (c), (a), (a), (d), , 10, 20, 30, 40, 50, , (b), (a), (c), (b), (a), , ∴ x + y = 15 ⇒ 9 + y = 15 ⇒ y = 6, , Case II, , If x + y = – 15 and x – y = 3, , On adding the above two equations, 2x = – 12, x = – 6, ∴ x + y = – 15 ⇒ – 6 + y = – 15, ⇒ y = – 15 + 6 ⇒, y=–9, 4., (b) Let full fare = ` x, , and reservation charges = ` y, ∴ x + y = 2125 , ....(i), x, , 3200 from (i), , , Also (x + y) +  + y  =, 2, , x, x, 2125 + + y =, 3200,, y 3200 − 2125, +=, 2, 2, , ⇒ x + 2y = 1075 ⇒ x + 2y = 2150 ....(ii), , Solving (i) and (ii), we get, , – y = – 25 or y = 25, , Putting the value of y = 25 in (i), , x + 25 = 2125, , x = 2125 – 25, x = 2100, full fare = ` 2100 and reservation charges = ` 25, 5., , (d) We have,, , tan θ − cot θ, sin θ cos θ, , =, , tan θ, cot θ, −, sin θ cos θ sin θ cos θ, , =, , sin θ, cos θ, −, cos θ sin θ cos θ sin θ cos θ cos θ, , =, , 1, 2, , cos θ, , −, , 1, = sec2 θ − cosec2θ, 2, sin θ, , 2, = 1 + tan 2 θ − 1 − cot=, θ tan 2 θ − cot 2 θ

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Mathematics, , S-14, , 6., 7., , (d) L.C.M × H.C.F = First number × second number, 36 × 2, Hence, required number =, =4., 18, (c) Let BD = x cm, Since AC = BC, therefore DABC is an isoscele triangle., ⇒ ∠B = ∠CAB = 72°, Since AD bisects ∠A, \ ∠DAB = 36° so, In DADB, ∠ADB = 72°, ⇒ DADB is an isoscele triangle, \ AB = AD = 1cm, ⇒ AB = 1 cm, Similarly, DADC is also an isoscele triangle., \ AD = CD ⇒ AD = 1 cm, C, , = p(5.7 – 4.3) (5.7 + 4.3) = p × 1.4 × 10 sq. cm, = 3.1416 × 14sq. cm. = 43.98 sq. cms., 12., , (c) Given, area of two similar triangles,, A1 = 81cm2 , A2 = 49 cm2, , 13., , (c) We have,, , 1, , x, , 14., , 1+, , D, , 2 cos θ, 2, , 72°, x, 36°, , Also, y, =, , AC CD, =, Now  , AB BD, , 8., , 15., , 3, :1 or 3 : 2, 2, , 3(3) + 2(3), = 3, 3+ 2, , (c) Let unit’s digit : x, tens digit : y, then x = 2y, number = 10y + x, Also 10y + x + 36 = 10x + y, , – 1 ± (1) 2 – 4(1)(–1) –1 ± 5, =, 2, 2, , ⇒ x, =, BD =, , ⇒ x + x2 – 1 = 0, , 1, ⇒ θ = 60°, 2, , 6k − 4(1), 3, =, or k, k +1, 2, , ∴ The required ratio is, B, , 1+ x 1, =, 1, x, , = 4 ⇒ cos θ =, , 72°, , A, , ⇒, , cos θ, cos θ, 4, +, =, 1 − sin θ 1 + sin θ, , cos θ, (c) Let the required ratio be k : 1, , =, Then, 2, , 36°, , 81 9, =, 49 7, ,  1 + sin θ + 1 − sin θ , 4, ⇒ cos θ ,  =, , 1 − sin 2 θ, ⇒, , 36°, , A1, =, A2, , Ratio of corresponding medians =, , ∴ 9x – 9y = 36 or x – y = 4, Solve, x = 2y,, , 5 –1, 2, , x–y=4, , Substitute x = 2y in x – y = 4, we get, 2y – y = 4 ⇒ y = 4, , (a) Since, C (y, – 1) is the mid-point of P (4, x) and Q (–2, 4)., , and x = 8, So, the number = 10y + x = 48, 16., , We have,, , 9., 10., , (b) Value of n = 2., , 11., , (a) Let the radii of the outer and inner circles be r1 and r2, respectively; we have, pr12, , –, , pr22, , =, , p(r12, , –, , r22), , = p(r1 – r2) (r1 + r2), , 3, ., 4, (b) Given an equilateral triangle ABC in which, P(at most one head) =, , 17., , 5, 1, (c) Required probability =, = ., 25, 5, , Area =, , Favourable outcomes = HT, TH, TT, , 4−2, 4+ x, = y and, = −1, 2, 2, , ∴ y = 1 and x = – 6, , (c) Total outcomes = HH, HT, TH, TT, , AB = BC = CA = 2p, , A, , and AD ⊥ BC., , 2p, , ∴ In ∆ADB,, AB2 = AD2 + BD2, , B, , (By Pythagoras theorem), ⇒ (2p)2 = AD2 + p2 ⇒ AD2 =, , 3 p., , 2p, D, , C

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S-15, , Solutions, 18., , 19., , 20., , (b) x2 + 4x + 2 = (x2 + 4x + 2) – 2 = (x + 2)2 – 2, , Alternate method:, , Lowest value = – 2 when x + 2 = 0, , 2 tan A , ,  Using identity,sin 2 A =, , 1 + tan 2 A , , , (c) −, , 3(1) + 4(2) − 7, 4 4, =, = −, 3(−2) + 4(1) − 7, −9 9, ,  2 1 22 2 , (a) Required area =  7 – × × 7  cm2, 4 7, , , , = (49 – 38.5) cm2 = 10.5 cm2, 21. (a) Quadratic polynomial p(x) = k (x + 1)2, , p(– 2) = k (– 2 + 1)2 = 2, , k=2, , p(x) = 2 (x + 1)2, , p(2) = 2(2 + 1)2 = 2 × 3 × 3 = 18, 22. (d) All the statements given in option (a, b, c) are correct., 23. (d) We have, sin 5θ = cos 4θ, ⇒ 5θ + 4θ = 90°, , [ sin α = cos β, than α + β = 90°], ⇒ 9θ = 90° ⇒ θ = 10°, , Now, 2 sin 3θ – 3 tan 3θ, = 2sin 30° – 3 tan 30°, 1, 1, = 2× − 3 ×, = 1−1 = 0, 2, 3, , , 28., , 7 −1 5, 1 5, ∴ = =, ⇒ = ⇒ k = 15, 3 k, 21 −3 k, , Hence, for k = 15, the system becomes consistent., 25. (b) α + β = 5 , ...(i), αβ = k , ...(ii), α – β = 1 , ...(iii), , Solving (i) and (iii), we get α = 3 and β = 2., , Putting the value of α and β in (ii), we get, 26., , (d) We have,, , (b) Let P (x, 0) be a point on X-axis such that AP = BP, , ⇒ x2 + 4x + 4 + 9 = x2 – 10x + 25 + 16, ⇒ 14x = 28 ⇒ x = 2, Hence, required point is (2, 0)., 29., , (a) Let side of a square = x cm, ∴ By Pythagoras theorem, x2 + x2 = (16)2 = 256, ⇒ 2x2 = 256 ⇒ x2 = 128 ⇒ x = 8 2 cm., , 30., , (c), , Hence, 12x + 16y = 9x + 18y or 3x = 2y, \ x=, , 2, y., 3, , 2, y in the required expression., 3, i.e. 3 x + 5 y : 3 x − y, Substitute x =, , 2 , 2 , = 3  y + 5 y : 3  y − y, 3 , 3 , = 2y + 5y : 2y – y = 7y : y = 7 : 1, 31., , (d), , 32., , (b) 10 x = 7.7, , or x = 0.7, , Subtracting, 9x = 7 \, 2=, x, , 3, 2, , 3x + 4 y 9, =, x + 2y 4, , ⇒ 4(3x + 4y) = 9(x + 2y), , 1 + tan 2 30°, 2×3, =, 3×4, , =, , ⇒ (x + 2)2 + (0 – 3)2 = (x – 5)2 + (0 + 4)2, , 2 tan 30°, , 1, 2, 3, 3, =, =, =, 2, 1,  1 , 1+, 1+  , 3,  3, 2×, , 2, , ⇒ AP2 = BP2, , , , 24. (a) Given equations are :, 7x – y = 5 and 21x – 3y = k, Here a1 = 7, b1 = –1, c1 = 5, a2 = 21, b2 = –3, c2 = k, , We know that the equations are consistent with unique, solution, a, b, if 1 ≠ 1, a2 b2, , Also, the equations are consistent with many solutions, a1 b1 c1, if =, =, a2 b2 c2, , 2 tan 30°, , 3, 2, 1 + tan 30°, 27. (a) The largest number of four digits is 9999. Least, number divisible by 12, 15, 18, 27 is 540., On dividing 9999 by 540, we get 279 as remainder., Required number = (9999 – 279) = 9720., sin 60° =, , x=, , 7, 9, , 14, = 1.555........, = 1.5, 9, , 33., , (b), , 34., , (b) The numbers that can be formed are xy and yx. Hence, (10x + y) + (10y + x) = 11(x + y). If this is a perfect, square then x + y = 11.

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Mathematics, , S-16, , 35., , (d) Since, ∆ABC ~ ∆PQR, ar(DABC ) BC 2, 9 (4.5) 2, ∴, =, ⇒, =, ar(DPQR) QR 2, 16 QR 2, 16 × (4.5) 2, ⇒=, QR, ⇒, =, QR 6 cm, 9, 2, , 36., , (c), , 37., , x + x + x y + y + y3 , (b) Centroid is  1 2 3 , 1 2, , 3, 3, , , 3 + (–8) + 5 –7 + 6 + 10   0 9 , i.e. , ,,  =  3 , 3  = (0, 3), 3, 3, , ,  , 1+ 2 +1 4, = ., 11, 11, , 38., , (c) Required probability =, , 39., , (a) Given: The natural number, when divided by 13 leaves, remainder 3, The natural number, when divided by 21 leaves remainder 11, So, 13 – 3 = 21 – 11 = 10 = k, Now, LCM (13, 21) = 273, But the number lies between 500 and 600, \, , 2 LCM (13, 21) – k = 546 – 10 = 536, , 536 = 19 × 8 + 4 \ remainder = 4, 40., , (b) Since, ∆ABC ~ ∆APQ, ∴, , 41., , ar( DABC ) BC 2, =, ar( DAPQ) PQ 2, 2, , ⇒, , ar( DABC ), BC 2, 1,  BC , = 2 ⇒, =, , 4 ⋅ ar( DABC ) PQ,  PQ , 4, , ⇒, , BC 1, =, PQ 2, , (a) Area of DABC =, , , , 17320.5 =, , 3 2, a, 4, , 17320.5 × 4, = 40000, 1.73205, a = 200 cm, , a2 =, , , 3 2, a, 4, , 42., , (d) Radius of circle =, , , 43., , = 100 cm, , 200, 2, , (a) Area of each sector, , =, , 60, × pr2, 360, , 1, × 3.14 × 10000 = 5233.3 cm2, 6, (b) Area of the shaded region, , =, 44., , = Area of DABC – 3 × Area of each sector, = 17320.5 – 3 ×, 45., 46., , 31400, = 1620.5 cm2, 6, , (c) Perimeter of DABC = 3 × 200 = 600 cm, (a) Radius of inner semicircular end, , 60, = = 30 m, 2, 47. (c) Radius of outer semicircular end, , = 30 + 10 = 40 m, 48. (b) The distance arounded the track along its inner edge, = 106 × 2 + 2 × πr, 22, = 212 + 2 ×, × 30= 212 + 188.57, 7, , , = 400.57 m, 49. (d) The distance arounded the track along its outer edge, , = 106 × 2 + 2 × πr, 22, = 212 + 2 ×, × 40= 212 + 251.43, 7, , , = 463.43 m, 50. (a) The area of the track, , = 2 × Area of ractangle + 2 × Area of, semicircular ring., 1 22, ×, × (402 – 302), 2 7, , , , = 2(10 × 106) + 2 ×, , , , = 2120 + 2200 = 4320 m2

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Sample Paper, , 5, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., , (c), (c), (b), (a), (a), , (c), , 2, 12, 22, 32, 42, , (d), (a), (c), (c), (d), , 3, 13, 23, 33, 43, , (c), (b), (a), (d), (c), , (b), (a), (b), (c), (d), , (, , 6, 16, 26, 36, 46, , (d), (b), (c), (d), (b), , 7, 17, 27, 37, 47, , = (3 –, , 3)2, , +, , (, , 3 -λ, , (, , ⇒, , (λ - 3, , ) (, ), 3 )(λ + 3 ) = 0, , ⇒ λ = 3 3 or λ = - 3 ., 2, (d)  2, k  14, 2,  1 (Given), ⇒, 2, k  14, ⇒ k=±4, (c) We have ABCD is square,, , 1, BC,, 3, Area ∆FBE = 108 sq cm., AF = BF, BE =, , , , Let AB = x ⇒ BF =, , 8, 18, 28, 38, 48, , (c), (c), (a), (b), (c), , 9, 19, 29, 39, 49, , x, x, and BE =, 2, 3, , (b), (c), (b), (a), (a), , 10, 20, 30, 40, 50, , (b), (c), (d), (d), (d), , C, , E, x/3, A, x, , 2, , ), , , , (d), (a), (c), (a), (a), , D, , ), , ⇒ λ λ -3 3 + 3 λ -3 3 = 0, , 3., , (c), (c), (a), (a), (b), , 2, , ⇒ 9 + 3 = 0 + 3 – 2λ 3 + λ 2, ⇒ λ 2 - 2 3λ - 9 = 0, ⇒ λ 2 - 3 3λ + 3λ - 9 = 0, , 2., , 5, 15, 25, 35, 45, , Let the three points be A(0, 0), B(3, 3 ) and, C (3, λ)., ∴ AB = BC = CA, [Q ∆ABC an equilateral ∆], 2, 2, 2, ⇒ AB = BC = CA Now AB2 = BC2, ⇒ (0 – 3)2 + 0 - 3, , , , 4, 14, 24, 34, 44, , , , Area of DFBE =, , F, , x/2, , B, , 1, BF × BE, 2, , 1 x x x2,   , 2 2 3 12, = 108 sq. cm. (Given), , =, , , ⇒ x2 = 12 × 108 ⇒ x2 = 12 × 12 × 3 × 3, ⇒ x = 12 × 3 = 36 cm, , , In rt. ∆ABC, AC =, , AB2  BC2, (By Pythagoras Theorem), , = 362 + 362 = 36 2 cm, , 4., (b) Let x = 0.31783178...... , ...(i), , Multiply by 10000, , 10000 x = 3178.31783178........, ...(ii), , Subtracting (i) from (ii), , 10000 x = 3178.3178........, x = 0.3178........, –, –, 999 x = 3178, 3178, , x=, 999

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Mathematics, , S-18, , 5., , (c) Either a rational number or an irrational number., , 9., , 6., , (d), , 10., , (b) Let the number of boys and girls in classroom is x, and y., According to question, , A, b, , x −x/5 2, 4x 2, , = ⇒, = ⇒, y, 3, 5y 3, , b, , Q, , P, 2b, , a, , 2b, a, , Also,, , a, C, , B, In DABC, , , , AB = AC, , By angle sum properly in DABC,, , , b + a + a = 180, , 11., , , ⇒  b + 2a = 180° , , , ...(i), , In  DQPB, , , ⇒  ∠QPB = 180 – 4b, , , Since ‘APC’ is a straight line, , (c) Perimeter =, , \, ...(ii), , From equations (i) & (ii), , , b + 2(3b) = 180 ⇒ b =, , 180, 7, ,  180  5, , ∠AQP = 180° – 2 , = p,  7  7, 7., (d) 2πr = 4π ⇒ r = 2, Area = π(2)2 = 4π, When, 2πr = 8π, ⇒r=4, Area = 16π, 8., , (c) (1 + tan θ + sec θ)(1 + cot θ – cosec θ), , 1   cos θ, 1 ,  sin θ, –, = 1 +, +,  × 1 +, ,  cos θ cos θ   sin θ sin θ , {(cos θ + sin θ) + 1} × {(cos θ + sin θ) –1}, =, cos θ × sin θ, 2, 2, a 2 – b2 }, = (cos θ + sin θ) – (1) {∵ (a + b)(a – b) =, cos θ × sin θ, 1 + 2 cos θ sin θ –1, = 2., =, cos θ × sin θ, , 1, × 2πr + 2r, 4, , (a) DABC ~ DANM, , , ⇒  180 – 4b + a + b = 180, , , x − x/5 5, 4x, 5, =, ⇒, =, y − 44, 2, 5 (y − 44) 2, ,  1 22, , =  × × 7 + 2 × 7  cm = 25 cm, 2 7, , 12., , , ⇒  a = 3b , , x 5, = ...(i), y 6, , , ⇒ 8x = 25y – 1100, ...(ii), From Eqs. (i) and (ii), we get, x = 50, y = 60, Let n number of boy leaves the class so number of, boys and number of girls become equal., , \ 50 – 10 – n = 60 – 44, , n = 40 – 16 = 24, , , ⇒  ∠C = ∠B ⇒  ∠B = ∠C = a, , , (b), , Area of ∆ABC, AC 2, ...(i), =, Area of ∆ANM AM 2, , , DABC ~ DMPC, \, , , Area of ∆ABC, AC 2 ...(ii), =, Area of ∆MPC MC 2, From Eqs. (i) and (ii,) we get, , Area of ∆ANM AM 2, , =, Area of ∆MPC MC 2, Area of ∆ANM + Area of ∆MPC, AM 2 + MC 2, , =, Area of ∆MPC, MC 2, , , Now, Area of DANM + Area of DMPC, , , , = Area of DABC – Area of BNMP, , , Using Area of BNMP =, , 5, of area of DABC, 18, , 2, 2, \ 13 (Area of ∆ABC ) = AM + MC ...(iii), 18 (Area of ∆MPC ), MC 2, , , , From Eq. (iii),, , 13  AC 2  AM 2 + MC 2, =, 18  MC 2 , MC 2, , ⇒ 13 (AM + MC)2 = 18 (AM2 + MC2), ⇒, , 1, AM, = 5, . Hence, option (a) is correct., MC, 5

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S-19, , Solutions, 13., , (b), , 14., , (a) Condition for infinite many solutions., , , As QAR are collinear, , p 3, = =, 12 p, , , , p2 = 36 ; p =, , p2 – 3p = 3p, , p=6, , \ p=6, 15., , \ ∠QAR = 180°, , p – 3  a1 b1 c1 ,  = =, , p  a2 b2 c2 , , Q is reflection of P on AB, \ ∠QAB = ∠BAP, , {From I and II}, {From II and III}, , R is reflection of P on AC, \ ∠RAC = ∠CAP, , ∠QAR = 180°, , (c) Radius of outer concentric circle = (35 + 7) m = 42 m., 22, Area of path = π (422 – 352) m2 =, (422 – 352) m2, 7, , 16. (b) 9sec2 A – 9 tan 2 A = 9(sec2 A – tan 2 A), = 9 × 1 = 9., 17., , , , Now, numbers divisible by 5 are :, , , , 2 × 3 × 1 + 2 × 2 × 1 = 10, , , , So, probability that the slip bears a number divisible, , 18., , , ∠BAP + ∠CAP = 90° ⇒ ∠BAC = 90°, 21. (b) α + β = –a and αβ = –b, 1 1 α +β −a a, + =, = =, α, β αβ −b b, , 1 1 1, 1, × =, =, α β αβ –β, , (a) Total three digit number are : 3 × 3 × 2 = 18, , by 5 =, , \ 2∠BAP + 2∠CAP = 180°, , ∴, , 10 5, =, 18 9, , (c) Let x = 0.235, , ...(i), , 1000 x = 235.235, , ...(ii), , 235, Subtract (i) from (ii), 999 x= 235 ⇒ x=, 999, 19. (c) Joining B to O and C to O, , The required polynomial is, , a, 1, x2  x  ., , b, b, , 22. (c) H.C.F. of 20 and 15 = 5, , So the 5 students are in each group so, 20 + 15, 35, n =, =, =7, 5, 5, , Hence, x = 4, y = 3 and n = 7, 23. (a) Let AB = BC = x units., , Then Hyp. AC = side 2 = x 2 units, , Let the radius of the outer cirlce be r, , 3, , 2, , x, , x 2, , ∴ perimeter = 2πr, But OQ = BC = r, , [diagonals of the square BQCO], , D, , 4, , 2πr π, =, 4r, 2, (c) Here, ABC is a triangle & P be interior point of a, DABC, Q and R be the reflections of P in AB and AC,, respectively., Hence, ratio =, , R, A, , Q, , , , θ, φ, , x, C, B, x, In ∆ABD and ∆CAE, ∠1 = ∠3 [In equilateral ∆ each angle = 60°], ∠2 = ∠4, ∴ ∆ABD ~ ∆CAE (By AA rule), area ABD AB2, , ∴, area CAE CA 2, (By the theorem), =, , P, , B, , x 2, , x, , 1, , ∴ Perimeter of ABCD = 4r., , 20., , E, , x 2, A, , C, , x2, , x 2 , , 2, , , , x2, 2x, , 2, , , , 1, 2

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Mathematics, , S-20, , ∴, , area ABD 1, , area CAE 2, , , ⇒, , 1, ⇒ area (∆ABD) =, area (∆CAE), 2, 24., , (b) Degree of quotient = degree of dividend – degree of, divisor, , Degree of quotient = 7 – 4 = 3., 25. (a), , , Since, the required point (say P) is on the x-axis, its, ordinate will be zero. Let the abscissa of the point be x., Therefore, coordinates of the point P are (x, 0)., P (x, 0), , 2(x + y) = 10 + y ⇒ x + y = 5 +, , y, 2, , Now, (x + y)max when y is maximum & maximum, value of y will be 10. (Q y = 10 – 2x), , , So (x + y)max = 5 + 5 = 10 & (x + y)min when y = 0, , , \, , 31., , minimum value of x + y = 5, , So, sum of (x + y)max & (x + y)min = 15, , (a) Here, when A = 0°, , , , , LHS = sin 2 A = sin 0° = 0, and RHS = 2 sin A= 2 sin 0° = 2 × 0 = 0, In the other options, we will find that, , , LHS ≠ RHS, 32., , (c), , 1, = 0.142857, 7, , The second positive integer whose reciprocal have six, different repeating decimals is, , B (–2, 3), , A (5, 4), , , Let A and B denote the points (5, 4) and, (–2, 3) respectively., , , , Given that AP = BP, we have, , AP2 = BP2, i.e. (x – 5)2 + (0 – 4)2, , 27., , (c), , (c), , 2 tan 30°, 1 – tan 2 30°, , =, , 34., ,  1 , 2 ,  3,  1 , 1–  ,  3, , So, quadratic polynomial p(x) = k(x + 1)2, , p(–2), , = k(–2 + 1)2 = 2 ⇒ k = 2 ∴ p(x) = 2(x + 1)2, , Also, p(2) = 2(2 +, , 1)2, , (c) Let distance = d,, d, d, =, 15 − 5 10, , , , Time taken upstream =, , , , Time taken downstream =, , , , Hence, average speed, , 2, , 2, 2 3, 3, =, =, × = 3 = tan 60°., 1, 3 2, 1–, 3, 28. (a) x = – 1 is the root of the quadratic polynomial p(x), , , Therefore, the values of x are 7, 13, 21, , Hence, the required sum is = 7 + 13 + 21 = 41, 33. (d) , , ⇒x=2, 26., , And the third positive integer whose reciprocal have six, different repeating decimals is, 1, = 0.047619, 21, , , , = (x + 2)2 + (0 – 3)2, , , , 1, = 0.076923, 13, , , d, d, =, 15 + 5 20, , 2d, 2d × 20 40, km/hr, = = =, d, d, 3d, 3, +, 10 20, , , Ratio =, , 40, :15 = 40 : 45 = 8 : 9, 3, , 35., , (a) P (E) + P ( ) = 1, , = 2 × 3 × 3 = 18, , 1 – tan 2 45° 1 – (1)2, = = 0., 1 + tan 2 45° 1 + (1)2, , 29., , (b), , 36., , (d), , 30., , (d) Given 2x + y = 10, , 37., , (a) Let the required number be 33a and 33b., Then 33a + 33b = 528 ⇒ a + b =16., , , , on adding y both sides, we get, 2x + y + y = 10 + y

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S-21, , Solutions, Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11), and (7, 9)., \ Required numbers are (33 × 1, 33 × 15),, (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9)., 38., , The number of such pairs is 4., (b) Upstream speed = 4 km/hr and time = x hrs., , , , Downstream speed = 8 km/hr and, , , , time taken = x/2 hrs., , , , Hence average speed =, , 39., , (a), , 2 tan 30°, 1 + tan 2 30°, , =, , 4 x + 8 × x / 2 16, = km/hr., x+ x/2, 3, ,  1 , 2 ,  3,  1 , 1+  ,  3, , 2, , 2, 3, , =, =, 1, 1+, 3, 40., , 2, , 3, × =, 3 4, , 3, = sin 60°., 2, , (d) For given numbers,, , (55)725, unit digit = 5; (73)5810, unit digit = 9, (22)853, unit digit = 2, Unit digit in the expression, 55725 + 735810 + 22853 is 6, 41., , (a) , , 42., , (d), , 43., , (c), , 44., , (d) , , 45., , (b), , 46., , (b), , 47., , (a) , , 48., , (c), , 49., , (a), , 50., , (d)

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Sample Paper, , 6, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., , (a), (a), (a), (b), (c), , 2, 12, 22, 32, 42, , (a), (a), (b), (c), (a), , 3, 13, 23, 33, 43, , (a), (a), (b), (d), (a), , 4, 14, 24, 34, 44, , (d), (c), (b), (d), (a), , 5, 15, 25, 35, 45, , (a), (d), (b), (b), (b), , 6, 16, 26, 36, 46, , (b), (b), (d), (b), (a), , (a) P(x) is a polynomial of degree 3., , D, , 1, and P(n) =, ⇒ n P(n) – 1 = 0, n, , a, , n(P(n)) is a polynomial of degree 4, , 7, 17, 27, 37, 47, , (b), (d), (a), (c), (b), , a, , O, , 8, 18, 28, 38, 48, , (a), (c), (a), (a), (c), , 9, 19, 29, 39, 49, , (b), (a), (d), (d), (b), , 10, 20, 30, 40, 50, , (c), (a), (b), (a), (a), , C, , a, , , , 1, ,  × d1 × d 2 , 2, , , \ n P(n) – 1 = k(n – 1)(n – 2)(n – 3)(n – 4), For n = 0; –1 = 24 k ⇒ k =, For n = 5; 5 × P(5) – 1 =, , 2., , −1, 24, , A, , −1, (4)(3)(2)(1), 24, , ⇒ 5 ⋅ P(5) – 1 = – 1 ⇒ P(5) = 0, (a) Area of square = 2 cm2, Side of square = 2 cm, 2, cm, OQ = x cm, 2, , OP =, , Q, , P, x, , ⇒, , x2, , =, , ( 2), , ⇒ x 2= 2 +, 2, ⇒ x =, , 2, 4, ,  2, + , ,  2 , , 5, ⇒x=, 2, , 5, cm, AC = 2, 2, , (AC = Diameter), , Area of square =, , 1, × d1 × d 2, 2, , 1, 5, 5, ×2, ×2, = 5 cm2, 2, 2, 2, , (a), (d) Let x & y be the unit and tenth digits respectively of, a two digit number. Then,, x + y = 9 (Q Given) , ... (i), and according to given condition,, 10x + y = 10 y + x + 27, ⇒ 9x – 9y = 27, ⇒ x – y = 3 , ... (ii), On adding (i) & (ii), 2x = 12 ⇒ x = 6, Hence, from equation (i),, 6+y=9⇒y=3, So number will be 10 × 3 + 6 = 36, , 5., , (a) The largest number of four digits is 9999. Least number, divisible by 12, 15, 18, 27 is 540., On dividing 9999 by 540, we get 279 as remainder., Required number = (9999 – 279) = 9720., , 2, , 5, cm., 2, , 1, × AC × BD, 2, , 3., 4., , O, 2, , B, , Area of square =, , =, , x, , 2, , a

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S-23, , Solutions, (b), , A (–2, 5), , 12., , (a), , 13., , (a), , A, , 53, , O, (x, y), , s, =, , Q OA = OB = OC, =, , 2, , 2, , ( x  2)  ( y  5)  ( x  2)  ( y  3), , ⇒, x2, , ⇒, , ⇒ r2 =, , + 4 + 4x + y2 + 25 – 10y = x2 + 4 + 4x + y2 + 9 + 6x, , 2 × 924 + 7, ⇒ r2 = 588, 22, , r, , ⇒ r = 14 3 m, , Again: OB2 = OC2, ( x  2) 2  ( y  3) 2  ( x  2) 2  ( y  3) 2, , 14., , a1 b1 c1 5, (c) a= b= c= 3, 2, 2, 2, , 15., , tan 30°, (d), =, cot 60°, , 16., , (b), , 17., , (d), , 18., , (c) Total number of cards = 25, , ⇒ x2 + 4 + 4x + (y + 3)2 = x2 + 4 – 4x + (y + 3)2, ⇒ 8x = 0 ⇒ x = 0, \ centre of the circle is (0, 1)., 1, , 1, + =, 2, 2, , 2, =, 2, , 7., , °, (b) sin 45° + cos 45=, , 8., , (a), , 9., , (b) Since, H.C.F. of co-prime number is 1., , 2, , ∴ Product of two co-prime numbers is equal to their, L.C.M. So, L.C.M. = 117, 10., , 77(11)(24)(42) = 924, , πr2 = 2(924), , 2, , ⇒ 16y = 16 ⇒ y = 1, , ⇒, , s(s − a)(s − b)(s − c), , So, area of ∆ =, , OB2, 2, , a + b + c 66 + 53 + 35, =, = 77m, 2, 2, , Area of ∆ =, , Join OA, OB & OC., \, , C, , Here, a = 66 m, b = 53 m & c = 35 m, , C, (2, –3), , Let O(x, y) is the centre of the given circle., , OA2, , 66 m, , B, , B, (–2, 3), , m, , 35, , m, , 6., , 1, 3, = 1, 1, 3, , Prime number are 3, 5, 7, 11, 13, 17, 19, 23,, , (c) (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2...(i), , ∴ Probability of prime number card =, , Also, (x – 3)2 + (y – 3)2 = (x – 3)2 + (y + 7)2, y2 – 6y + 9 = y2 + 14y + 49, , A, , – 20y = 40 ⇒ y = – 2, Putting y = – 2 in equation (i), we have, (x – 6)2 + (4)2 = (x – 3)2 + (5)2, , 19., , (a), , q, , 90 – q, O, , x2 – 12x + 36 + 16 = x2 – 6x + 9 + 25, , y, , –6x = – 18 ⇒ x = 3, 11., , (a) Since a, b are co-prime, , x, , ⇒ g. c.d of a, b = 1 ⇒ g. c. d. of a2, b2 = 1, ⇒ a2, b2 are co-prime., , C, , (b) does not hold. (c) does not hold, (d) does not hold, , ∠AOB = q, , Q If a = 2, b = 3, then a2 = 4, b2 = 9, , Q CO ^ OA, , \, , a2, , is even,, , b2, , is odd., , B, , 90 – q, , \ ∠BOC = (90° – q), , 8, 25

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Mathematics, , S-24, , P(h, 5h + 3), , 3, 4 , 2 , sin θ =, cos θ =, ∵ cos θ = 1 −sin θ , 5, 5, , ;, sin q = x =, cos q = y =, , 3, 5, , M, 4, 5, , Q (3, –2), Since, M is the mid-point of PQ, therefore by mid-point, ,  3 4, \ point on fourth quadrant is  , − ,  5 5, 20., , y = 5x + 3, ,  h + 3 5h + 3 − 2 , ,, formula, we have M = ,  .,  2, 2, , (a) Required area = π (r + d )2 − r 2 , , , , Clearly by observing the options, we can say that M must, lie on the line, , d, , , y = 5x – 7, 25., , r, , (b) Perimeter =, , 2πr, + 2r, 2, , = πr + 2r, ⇒ (π + 2) r = 36, , = π[r2 + d2 + 2rd – r2], =, 21., , π[d2, ,  36 , ⇒   – r = 36,  7 , , + 2rd] = πd[d + 2r], , (a) We know that height of an, , , , Hence, diameter = 7 × 2 = 14 cm., , A, , equilateral triangle, , 26., , 30° 30°, , (d) f(x) = (x – 1)2 + (x – 2)2 + (x – 3)2 + (x – 4)2, 2, , 5, , = 4  x −  + 5, 2, , , equilateral triangle, , 3, 3, \ AD 2 = a 2 = BC 2, 4, 4, , , , r, , ⇒ r = 7 cm, , 3, a,, 2, , where a is the side of, , , r, , 60°, B, , 60°, D, , C, , 22. (b) Let speed of boat in still water be x km/hr, and speed of stream be y km/hr, , , 30, =3, x+y, , ⇒ x + y = 10, , …(i), , , , 30, =5, x−y, , ⇒ x – y = 6, , …(ii), , From solving equations (i) and (ii), , –x + y = 10, –x – y = 6,   +  –, 2y = 4   y = 2 km/hr. and, x = 8 km/hr, No. of favourable outcomes 1, =, 23. (b) Probability, =, Total number of outcomes 5, 24. (b) Let coordinate of point p be (h, 5h + 3), , 5, = 2.5, 2, 27. (a) Let one woman can paint a large mural in W hours, and one girl can paint it in G hours, According to question,, 8 12 1, 2 3, 1, ⇒ + =, + =, ...(i), W G 10 W G 40, , f(x) is minimum at x =, , Also,, , 6 8 1, 3 4, 1, + =, ⇒ + =, ...(ii), W G 14 W G 28, , On solving equation (i) and (ii), we get, , W = 140 and G = 280, Now,, , 7, 14, 1, 1, ,  (say), 140 280 Time taken t, , 1 1, 1, , ⇒, , ⇒ t = 10 hours, t 20 20

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S-25, , Solutions, 28. (a) Here, BAC is a right angle triangle, , , ,  81 , πr , then, area = π  r −  =, , 10, , ,  100 , , B, , AB = 15 & BC = 25, \ AC=, , BC 2 − AB 2= 20, , F, , 1, Area of ∆ABC =, BC. AD, 2, 1, = AB. AC, 2, AB. AC, ⇒ BC. AD =, , A, , , , ⇒ 25(AD) =, 15(20), , D, , E, , 81 , , Thus, area is diminished by 1 −,  % = 19%,  100 , C, , 37., , (c) Q ∠BAC = ∠ADC(given), ∠C = ∠C , A, , ⇒ AD = 12, , Q AEDF is rectangle then, AD = EF = 12, (d) As (a, 0), (0, b) and (1, 1) are collinear, \ a(b – 1) + 0(1 – 0) + 1(0 – b) = 0, , ab – a – b = 0, , ab = a + b, 29., , , 30., , (common), , B, , D, , \ DABC ~ DDAC , ⇒, , C, , (by AA similarity criterion), , 2, BC AC, ⇒ BC × DC = AC, , AC DC, 2, , ⇒ BC × DC = (21) = area of rectangle with sides BC & DC, Now, Area of equilateral triangle = area of rectangle, , 1 1, 1= +, a b, , (b) (sin 30° + cos 30°) – (sin 60° + cos 60°), , ⇒, , 3 (side)2 = (21)2 ⇒ Side = 14 × 33/4, 4, , 1, 3  1, 3, =  2 + 2  −  2 + 2  = 0, ,  , , , 38., , 31. (b) Since, 2 is the zero of x2 + 3x + k,, , \ (2)2 + 3(2) + k = 0 ⇒ k + 10 = 0 ⇒ k = – 10, , , \, , (k – 1) (–3)2 + k(–3) + 1 = 0, , , ⇒, , 9k – 9 – 3k + 1 = 0 ⇒ 6k – 8 = 0 ⇒ k =, , 32., , (c) Possible products are 1, 4, 9, 16, 2, 8, 18, 32, 3, 12,, 27, 48, 4, 16, 36, 64, So, required probability of getting the product of the two, 6 3, numbers so obtained is, =, 16 8, , 33., , (d), , 4, , (d), , 1 + tan 2 A, 1 + cot 2 A, , =, , (sec2 A – tan 2 A) + tan 2 A, (cosec2 A – cot 2 A) + cot 2 A, 2, , sec2 A, sin 2 A  sin A , tan 2 A., =  =, = =, , cosec2 A, cos 2 A  cos A , (b) (I) Statement I is false. Consistent Linear equations, may have unique or infinite solutions., (II) Statement or is also false, , Q 132 + 142 = 365, 35., , 36., , (b) Let r be the radius of circle, then area = πr2, When r is diminished by 10%, , 4, 3, , (d) (sec A + tan A) (1 – sin A), , sin A ,  1, = , +, × (1 – sin A),  cos A cos A , = (1 + sin A)(1 – sin A), cos A, 1 – sin 2 A cos 2 A, , , = =, cos A, cos A, = cosA., , P (4, 5), 5, , 34., , 39., , (a) Since –3 is the zero of (k –1) x2 + kx + 1,, , (∵ cos 2 A = 1 – sin 2 A), , 1, 1, ⇒ a = 10 and y = ⇒ b = 5, 10, 5, , 40., , (a) x =, , 41., , (c) (0, 0), , 42., , (a) (4, 6), , 43., , (a) (6, 5), , 44., , (a) (16, 0), , 45., 46., , (b) (–12, 6), (a) , 47. (b) , , 49., , (b) , , 50. (a), , 48. (c)

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Sample Paper, , 7, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., 2., , (b), (a), (c), (a), (c), , 2, 12, 22, 32, 42, , (c), (b), (d), (a), (b), , 3, 13, 23, 33, 43, , (a), (b), (c), (d), (a), , 4, 14, 24, 34, 44, , (b), (a), (c), (b), (d), , 5, 15, 25, 35, 45, , (d), (d), (b), (b), (a), , (b), (c), , 2, , 2, , 1, – 3x 2, , (b) 2 x – =, 3 x + 1 2x, + 1 is not a polynomial,, since the exponent of variable in 2nd term is a, rational number., (c) x3 – 3x + 1 is a polynomial., 3, 2x 2, , – 5 x is also not a polynomial, since the, exponents of variable in 1st term is a rational, number, Hence, (a), (b) and (d) is not a polynomial., 3., (a) In ∆ABC, we have DE || BC, AD AE, ∴, =, [By Thale’s Theorem], DB EC, ⇒, , x, x+2, =, x − 2 x −1, , x(x –1) = (x – 2)(x + 2), x2 – x = x2 – 4 ⇒ x = 4, No. of sample space = 6 × 6 = 36, Sum total of 9 = (3, 6), (4, 5), (5, 4), (6, 3), 4 1, , ∴ P=, =, 36 9, 5., (d), 6., (c), 12, 7., (d) Given : 13 tan q = 12 ⇒ tan θ=, 13, 2sin θ.cos θ, Now given expression is,, cos 2 θ − sin 2 θ, ⇒, ⇒, 4., (b), , , (c), (a), (b), (c), (d), , 7, 17, 27, 37, 47, , (d), (a), (d), (c), (c), , 8, 18, 28, 38, 48, , (b), (a), (d), (b), (d), , 9, 19, 29, 39, 49, , (a), (d), (c), (a), (a), , 10, 20, 30, 40, 50, , (c), (d), (b), (d), (b), , Dividing numerator and denominator by cos2q,, 2sin θ cos θ, 12, 2×, 2, 2 tan θ, cos=, 13, θ, =, 2, cos 2 θ sin 2 θ 1– tan 2 θ,  12 , –, 1–,  , 2, 2, 13, cos θ cos θ, 12 , , ∵ tan θ = , 13, , 8., (b) n(S) = [1, 2, 3, ..., 100] = 100, 1, , Q x+, >2, x, , , 1, 2, –1, (a) x2 + = x + x, is not a polynomial since the, x, exponent of variable in 2nd term is negative, , (d), , 6, 16, 26, 36, 46, , , \ x2 + 1 > 2x, , ⇒ x2 – 2x + 1 > 0, , ⇒ (x – 1)2 > 0, , x = [2, 3, ... ,100], , n(E) = [2, 3, 4, ..., 100] = 99, 99, P(E) =, = 0.99, 100, 9., , (a) Let the third side be x cm. Then, by Pythagoras, theorem, we have, p2 = q2 + x2, , ⇒ x 2 = p 2 − q 2 =( p − q )( p + q ) = p + q [∵ p – q = 1], ⇒ x =, p + q = 2q + 1, [∵ p – q = 1 ∴ p = q + 1], , Hence, the length of the third side is, 2q + 1 cm., , 10., , (c) Given,, Two circle each of radius is 2 and difference between their, centre is 2 3, 1, AB = 2 3 ⇒ AC = AB, 2

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S-27, , Solutions, AC =, , 3 = CB, P, 2, , 2, A, , , 3, , C, 3, , B, , Q, , In ∆APC, cos θ =, , , , , AC, 3, =, (∠C = 90°), AP, 2, , ⇒ θ = 30°, We know,, Area of common region, = 2 (Area of sector – Area of ∆APQ), , 1,  60°, , × π(2) 2 − × (2) 2 × sin 60°, = 2 , , 360°, 2,  4π 4 3 , 2, , = 2 , −,  = 2  (3.14) − (1.73),  6, 3, 4, = 2 (2.09 – 1.73) = 2 (0.36) = 0.72., ∴ Area of region lie between 0.7 and 0.75., 11. (a), 12. (b) (a) is not true [By def.], (b) holds [Q degree of a zero polynomial is not defined], (c) is not true [Q degree of a constant polynomial is ‘0’], (d) is not true, [Q a polynomial of degree n has at most n zeroes]., (2 + 2sin θ) (1 − sin θ), 2(1 + sin θ) (1 − sin θ), 13. (b), =, (1 + cos θ) (2 − 2 cos θ), (1 + cos θ) (2) (1 − cos θ), 2, , 225, 2(1 − sin 2 θ) 2cos 2 θ,  15 , 2, =, =, = cot =, θ   =, , 2, 2, 64, 8, 2(1 − cos θ) 2sin θ, 14. (a), 15. (d) , All the statements given in option ‘a’, ‘b’ and ‘c’ are, correct., 16. (a) A, , , , r, O, , r, r, B, , 17. (a) S and T trisect the side QR., Let QS = ST = TR = x units, Let PQ = y units, , In right ∆PQS, PS2 = PQ2 + QS2 , (By Pythagoras Theorem), , = y2 + x2 , ...(i), , In right ∆PQT, PT2 = PQ2 + QT2, (By Pythagoras Theorem), = y2 + (2x)2 = y2 + 4x2, ...(ii), In right ∆PQT, PR2 = PQ2 + QR2, (By Pythagoras Theorem), = y2 + (3x)2 = y2 + 9x2, ...(iii), , R.H.S. = 3PR2 + 5PS2, , = 3(y2 + 9x2) + 5(y2 + x2), [From (i) and (iii)], , = 3y2 + 27x2 + 5y2 + 5x2 = 8y2 + 32x2, = 8(y2 + 4x2) = 8PT2 = L.H.S. [From (ii)], , Thus 8PT2 = 3PR2 + 5PS2, 18. (a) Unit digit in (795) = Unit digit in [(74)23 × 73], = Unit digit in 73 (as unit digit in 74 = 1), = Unit digit in 343, Unit digit in 358 = Unit digit in (34)4 × 32, , [as unit digit 34 = 1], = Unit digit is 9, So, unit digit in (795 – 358), = Unit digit in (343 – 9) = Unit digit in 334 = 4, Unit digit in (795 + 358) = Unit digit in (343 + 9), = Unit digit in 352 = 2, So, the product is 4 × 2 = 8, 19. (d) In (a) power of x is –1 i.e. negative, , \ (a) is not true., 1, In (b) power of x = , not an integer., 2, , \ (b) is not true, In (c) Here also power of x is not an integer, , \ (c) is not true, (d) holds [Q all the powers of x are non-negative, integers.], 20. (d) We have, sin 5θ = cos 4θ, ⇒ 5θ + 4θ = 90° [Q sin α = cos β, than α + β = 90°], ⇒ 9θ = 90° ⇒ θ = 10°, Now, 2 sin 3θ – 3 tan 3θ, , C, , circumference of circle = 2pr, ...(i), Area of DABC = [ar(DAOB) + ar(DBOC) + ar(DAOC)], 1, 1, = AB × r + × BC × r + AC × r, 2, 2, 1, 1, = r [AB + BC + AC] = r × 7p...(ii), 2, 2, From (i) and (ii),, Circmference of circle, 2r, 4,  , 1, Area of triangle, r  7 7, 2, , = 2sin 30° –, 1, 1, = 2× − 3×, = 1−1 = 0, 2, 3, , 21. (c) , 22. (d), 23. (c) In ∆PAC and ∆QBC, We have, ∠PAC = ∠QBC, [Each = 90°], ∠PCA = ∠QCB, [Common], \ ∆PAC ~ ∆QBC, x, AC, y BC, \, =, i.e. =, ...(i), y, BC, x AC, z, AC, y, AB, Similarly, =, i.e. =, ...(ii), y, AB, z, AC, , , Adding (i) and (ii), we get, , , ,  1 1, BC + AB, y y, = + = y  + , x z, x z, AC

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Mathematics, , S-28, , , ,  1 1,  1 1, AC, = y  +  ⇒ 1 = y  + , x, z, x z, AC, , 1, 1 1, = +, y, x z, 24. (c) On adding both the equations, we get x = 3, y = 1, 25. (b) A(2 – 2), B( – 1, x), AB = 5, ⇒ AB2 = 25, ⇒ ( – 1–2)2 + (x + 2 )2 = 25, ⇒ 9 + x2 + 4x + 4 = 25, ⇒ x2 + 4x – 12 = 0, ⇒ x2 + 6x – 2x – 12 = 0, ⇒ x (x + 6) – 2(x + 6) = 0, ⇒ (x – 2) (x + 6) = 0, ⇒ x = 2, –6, 180°, 26. (b) As 1 radian = 1 degree ×, π,  2π 180° , 2π, ×, , \, radian = , ,  3, π , 3, ⇒, , , \ Time =, , 120, = 20 min., 6, , −c −1 −2, (d) For solution to be infinite, = =, must satisfy., 6, 2 −3, −1 2, but, ≠ , so, infinite solution don’t exist, for given, 2 3, equations., 28. (d) All the statements given in option (a, b, c) are correct., 29. (c) Let the coordinate of other end be B(10, y) Given, point is A(2, –3), , AB = 10 ⇒ AB2 = 100, ⇒ (10 – 2)2 + (y + 3)2 = 100, ⇒ y2 + 6y – 27 = 0, ⇒ (y + 9) (y – 3) = 0, ⇒ y = –9, 3, 30. (b) The probability of an event can never be negative., 31. (a) Given, sin A + sin2A = 1, , ⇒ sin A = 1 – sin2A = cos2 A, , Consider, cos2A + cos4A = sinA + (sin A)2 = 1, 32. (a), 2, , π ( 49),  7 , (d) Area of the circle = π , = 49 cm2.,  =, π,  π, , 154 154 × 7, =, = 49 cm2, π, 22, 34. (b) Coefficient of all the terms are positive. So, both, roots will be negative., 35. (b) Let (x, y) be the point which will be collinear with, the points (–3, 4) and (2, –5), ∴ x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) = 0, ⇒ x (4 + 5) – 3(–5 – y) + 2 (y – 4) = 0, ⇒ 9x + 15 + 3y + 2y – 8 = 0, ⇒ 9x + 5y = –7, By plotting the points given in the options we find, that (7, –14) satisfies it., Now, consider, , 36., , (c) cos θ=, , 1 − sin 2 θ=, , 1−, , (c), (b) A die is thrown once therefore, total number of, outcomes are {1, 2, 3, 4, 5, 6}, (a) P(odd number) = 3/6 = 1/2, (b) P(multiple of 3) = 2/6 = 1/3, (c) P(prime number) = 3/6 = 1/2, (d) P(greater than 5) = 1/6, 39. (a) (By definition of similar triangles)., 40. (d) Radius of the circle is 13/4,  3 , Distance between (0, 0) and  − , 1 is,  4 , , 2, 3, 9, , 2, , +, +1, 0, ,  + (0 − 1) =, 4, 16, , , 25 5 13, =, <, 16 4 4,  7, Distance between (0, 0) and  2,  is,  3, =, , 2, , 27., , 33., , 37., 38., , a2, b, , 2, , =, , b2 − a 2, b, , 49, 85, 7, , =, (2 − 0)2 +  − 0  = 4 +, 9, 9, 3, , , 13, , = 3.073 <, 4,  −1 , Distance between (0, 0) and  3,  is,,  2 , 2, , 1,  −1 , (3 − 0)2 +  − 0  = 9 +, 4,  2, , 13, , = 3.041 <, 4, 5, , Distance between points (0, 0) and  −6,  is, 2, , 2, , 5, , (−6 − 0)2 +  − 0 =, 2, , =, , 25, 36 + =, 4, , 169, 4, , 13, 13, = 6.5 >, 2, 4 , , 41. (c) AB =, , (2.4)2 + (1.8)2 = 3m., , 42. (b) CD = 3.6 – 2.4 = 1.2 m, 43. (a) Q DABC ~ DAEF, AE, AC, , \, =, AF, AB, 1.8, 0.9, ⇒, =, ⇒ AF = 1.5 m, 3, AF, 44. (d), D, 300, 45. (a) Time =, =, = 60 sec = 1 min., S, 5, 46., (d), 47., (c), 48., (d), 49., (a), 50., (b)

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Sample Paper, , 8, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., , (c), (d), (b), (a), (d), , 2, 12, 22, 32, 42, , (d), (d), (a), (d), (a), , 3, 13, 23, 33, 43, , (d), (c), (b), (b), (d), , 4, 14, 24, 34, 44, , (d), (a), (d), (d), (c), , 5, 15, 25, 35, 45, , (b), (a), (a), (a), (b), , (c) Let the speed of the boat in still water be x km/hr and, the speed of the stream be y km/hr then speed of boat, in downstream is (x + y) km/hr and the speed of boat, upstream is (x – y) km/hr., Ist case : Distance covered upstream = 12 km, 12, hr, \ time =, x− y, , 6, 16, 26, 36, 46, , 12, 40, +, =, 8, x− y x+ y, , 3., 4., , 5., ...(ii), , AB =, , (, , 6., , ), , 2, , ) (, , 3 -1 – 3 -1 +, , 9, 19, 29, 39, 49, , (c), (c), (b), (d), (c), , 10, 20, 30, 40, 50, , 2, , ), , 2 + 1- 2 + 1, , (c), (d), (a), (b), (a), , (d) isosceles and similar, (d) Let us first find the H.C.F. of 210 and 55., Applying Euclid’s division lemma on 210 and 55, we, get, 210 = 55 × 3 + 45, ..... (i), Since, the remainder 45 ≠ 0. So, we now apply division, lemma on the divisor 55 and the remainder 45 to get, 55 = 45 × 1 + 10, ..... (ii), We consider the divisor 45 and the remainder 10 and, apply division lemma to get, 45 = 4 × 10 + 5, ..... (iii), We consider the divisor 10 and the remainder 5 and, apply division lemma to get, 10 = 5 × 2 + 0, ..... (iv), We observe that the remainder at this stage is zero., So, the last divisor i.e., 5 is the H.C.F of 210 and 55., −1045, = −19, \ 5 = 210 × 5 + 55y ⇒ y =, 55, (b) There are a total of six digits (1, 2, 2, 3, 4, 6), out of which four are even (2, 2, 4, 6), So, required probility =, , Solving (i) and (ii), we get,, x = speed of boat in still water = 6 km/hr,, y = speed of stream = 2 km/hr, (d) A 3 + 1, 2 -1 , B 3 -1, 2 + 1, , ) (, , (a), (b), (b), (c), (a), , 4+4 = 8 = 2 2, , Distance covered downstream, 32, = 32 km \ time =, hr, x+ y, , (, , 8, 18, 28, 38, 48, , =, , IInd case :, Distance covered upstream = 16 km, 16, \ time =, hr, x− y, , 2., , (c), (b), (d), (b), (b), , (-2)2 + (2)2, , ...(i), , Total time taken = 8 hr, 16, 32, \, +, =, 8, x− y x+ y, , 7, 17, 27, 37, 47, , =, , Distance covered downstream = 40 km, 40, hr, \ time =, x+ y, Total time is 8 hr \, , (d), (b), (b), (a), (a), , (d), (a) It is quadratic polynomial, [∵ the graph meets the x-axis in two points], (b) It is a quadratic polynomial, [∵ the graph meets the x-axis in two points]

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Mathematics, , S-30, , (c) It is a quadratic polynomial, [∵ the graph meets the x-axis in two points], (d) It is a not quadratic polynomial, [∵ the graph meets the x-axis in one point], 7. (c) Let ABCD be a square and two opposite vertices of, it are A(–1, 2) and C(3, 2). ABCD is square., D, C(3, 2), , 12., , (d) Let the y-axis divides the line segment joining (4, 5), and (– 10, 2) in the ratio k : 1., x coordinate will be zero on y-axis., We know that the coordinates of the point dividing, the line segment joining (x1, y1) and (x2, y2) in the,  mx 2 + nx1 m y 2 + ny1 , ,, ratio m : n are given by , m + n ,  m+n, Here, x coordinate of the point, dividing the line segment joining (4, 5) and (– 10, 2), is equal to zero., k × ( −10 ) + 1× 4, 2, = 0 ⇒ – 10k = – 4 ⇒ k =, So,, k +1, 5, , A (–1, 2), , B(x, y), , ⇒ AB = BC, ⇒ AB2 = BC2, ⇒ (x + 1)2 + (y – 2)2 = (x – 3)2 + (y – 2)2, ⇒ x2 + 2x + 1 = x2 – 6x + 9, ⇒ 2x + 6x = 9 – 1 = 8, ⇒ 8x = 8 ⇒ x = 1, ABC is right ∆ at B, then, AC2 = AB2 + BC2 (Pythagoras theorem), ⇒ (3 + 1)2 + (2 – 2)2, = (x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2, ⇒ 16 = 2(y – 2)2 + (1 + 1)2 + (1– 3) 2, ⇒ 16 = 2(y – 2)2 + 4 + 4, ⇒ (y – 2)2 = 4 ⇒ y – 2 = ± 2, ⇒ y = 4 and 0, , i.e., when x = 1 then y = 4 and 0, Coordinates of the opposite vertices are :, , B(1, 0) or D(1, 4), 8., (a), 9., (c) In right angled triangle POR, PR2 = PO2 + OR2 = (6)2 + (8)2 = 36 + 64 = 100, ∴ PR = 10 cm, , Again in right angled triangle, , PQR, QR2 = (26)2 = 676, PQ2 + PR2 = (24)2 + (10)2 = 576 + 100 = 676, ∴ QR2 = PQ2 + PR2, ∴ ∆PQR is a right angled triangle with right angle, at P., i.e., ∠QPR = 90°, πr12, , r1 5, 2πr1 5 5 × 125 625, 25, ⇒, = ⇒, = =, =, r2 4 2πr2 4 4 × 125 500, 16, , 10., , (c), , 11., , (d) For any rational number, , πr22, , =, , p, , where prime factorization, q, , of q is of the form 2n.5m, where n and m are nonnegative integers, the decimal representation is, terminating., , 13., , 14., 15., , 16., , 17., 18., 19., 20., , 21., , Therefore, the line segment joining (4, 5) and (– 10, 2), is cut by the y-axis in the ratio 2 : 5., (c) There are 4 cards of king and 4 cards of Jack n(S) =, 52, n(E) = 4 + 4 = 8, n(E) 8, 2, P(E), =, = =, n(S) 52 13, (a) The graph of y = ax2 + bx + c is a parabola open upward, if a > 0. So, for y = x2 – 6x + 9, a = 1 > 0, the graph, is a parabola open upward., (a) If 1, 1 and 2 are sides of a right triangle then sum of, squares of any two sides is equal to square of third, side., Case 1 (1)2 + (1)2 = 2 ≠ (2)2, Case 2 (1)2 + (2)2 = 1 + 4 = 5 ≠ (1)2, Case 3 (2)2 + (1)2 = 5 ≠ (1)2, (b) n(S) = 6 × 6 = 36, E = {(1, 2), (2, 1), (2, 3), (3, 2), (3, 4), (4, 3), (4, 5),, (5, 4), (5, 6), (6, 5)}, n(E) = 10, n(E) 10 5, P(E), =, = =, n(S) 36 18, (a), , 2 is not a rational number. It can’t be expressed in, the fractional form., θ, 30°, 49π, 2, × πr 2 =, × π ( 7) =, (b) Area =, 360°, 360°, 12, (c) Inconsistent system, (d) S = {1, 2, 3, 4......, 25}, n(S) = 25, E = {2, 3, 5, 7, 11, 13, 17, 19, 23}, n(E) = 9, 9, \, P(E) =, 25, , (b), , 1 α3 + β3, +=, =, α3 β3, (αβ)3, 1, , ⇒, , 1, α3, , +, , 3abc − b3, a3, 3, c,  , a, , 3abc − b3, =, β3, c3, 1

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S-31, , Solutions, 22., , 23., , 24., , (a) S = {S, M, T, W, Th, F, Sa}, n(S) = 7, A non-leap year contains 365 days,, i.e., 52 weeks + 1 day., E = {Sa}, n(E) = 1, n(E) 1, = =, \ P(E), n(S) 7, , 5 × 12, = 6, 10, Substituting x = 6 in (i), 3 4, 4, 1 1, 1 ⇒= 1=, –, + =, ⇒, 6 y, y, 2 2, x, \=, , Hence, x = 6 and y = 8, , (b) Let the given points be A(4, 3) and B(x, 5), Since A and B lies on the circumference of a circle, with centre O(2, 3), we have, OA = OB, ⇒ OA2 = OB2, ⇒ (4 – 2)2 + (3 – 3)2 = (x –2)2 + (5 – 3)2, ⇒ 4 + 0 = (x – 2)2 + 4, ⇒ (x – 2)2 = 0 ⇒ x = 2, (d) Since, , 27., 28., , 29., , 13 13, 13 23, 104, = =, =, = 0.104, 125 53 (2)3 (5)3 1000, 25., , (a), , A, , 30., , B, , 26., , D, , C, , Given : A DABC in which ∠B = 90° and D is the, midpoint of BC., Join AD., In DABC, ∠B = 90°., \ AC2 = AB2 + BC2, ....(i), , [by Pythagoras’ theorem], In DABD, ∠B = 90°, \ AD2 = AB2 + BD2, ....(ii), , [by Pythagoras’ theorem], ⇒ AB2 = (AD2 – BD2)., \ AC2 = (AD2 – BD2) + BC2, [using (i)], ⇒ AC2 = AD2 – CD2 + (2CD)2, , [∵ BD = CD and BC = 2CD], ⇒ AC2 = AD2 + 3CD2, Hence, AC2 = AD2 + 3CD2, 4 2 11, 3 4, + =, 1 ...(i), (b) + =, ...(ii), x y 12, x y, Multiplying (ii) by 2, 8 4 22, ⇒, + =, x y 12, Subtracting (i) from (iii) ⇒, , ...(iii), 5 10, =, x 12, , \ y = 8., , 31., 32., , 33., , (d), (b) n(S) = 6× 6 = 36, E = {(1, 5), (2, 5), (3, 5), (4, 5), (5,, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}, n(E) = 12, n(E) 12 1, P(E), =, = =, n(S) 36 3, (b) Put x + 1 = 0 or x = – 1 and x + 2 = 0 or, x = – 2 in p (x), Then, p (–1) = 0 and p (–2) = 0, ⇒ p (–1) =, ⇒ −1 + 3 + 2α + β = 0 ⇒ β = −2α − 2 .... (i), 3, 2, p (−2) = (−2) + 3(−2) − 2α(−2) + β = 0, ⇒ −8 + 12 + 4α + β = 0 ⇒ β = − 4α − 4 .... (ii), By equalising both of the above equations, we get, −2α − 2 = − 4α − 4, ⇒ 2α = −2 ⇒ α = −1, put a in eq. (i), ⇒ β =−2 (−1) − 2 =2 − 2 =0, Hence, α = −1, β = 0, (a) Let D be the window at a height of 9m on one side of, the street and E be the another window at a height of, 12 m on the other side., In DADC, AC2 = 152 – 92 = 225 – 81, AC = 12 m, In DECB, CB2 = 152 – 122 = 225 – 144, CB = 9 m, Width of the street = (12 + 9)m = 21 m, (a), (d) All equilateral triangles are similar, \ D ABC~ D EBD, Area of ABC BC2, , ⇒, Area of BDE BD 2, A, 2BD 2 4, , , E, 1, BD 2, , ⇒ Area (DABC) : Area (DBDE), B, C, D, , =4:1, (b) As A lies on x-axis and B lies on y-axis, so their, coordinates are (x, 0) and (0, y), respectively. Then,, 0+y, x +0, = 4 and, = –3 ⇒ x = 8 and y = – 6, 2, 2, Hence, the points A and B are (8, 0) and (0, –6).

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Mathematics, , S-32, , 34., , (d) If 6x ends with 5, then 6x would contain the prime 5., But 6x = (2 × 3)x = 2x × 3x., ⇒ The only prime numbers in the factorization of 6x, are 2 and 3., \ By uniqueness of fundamental theorem, there are, no primes other than 2 & 3 in 6x. So, 6x will never end, with 5., , θ, 60° 22, 2 132, × π=, r2, ×, × (=, 6), cm2, 360°, 360° 7, 7, θ, × πr 2, (b) Area of minor sector =, 360°, 60° 22, × × 14 × 14 = 102.57 cm2, =, 360° 7, Area of major sector, = Area of circle – Area of minor sector, 22, (14) 2 –102.57, =, 7, = 615.44 – 102.57 = 512.87 cm2, , 35., , (a) (a) Area =, , (c), , 2π ( 5) 2, C, =, =, 2, 5, A, π ( 5), ,  θ  2  θ  πr, , ∴ Area of sector = ,  πr =, ,  ( 2r ),  360° ,  360°  2,  θ ,  r  22 × 6, = , = 66 cm2,  2πr   =, 2,  360° , 2, (a) Let AD = 5x cm and DB = 4x cm., Then,, AB = (AD + DB) = (5x + 4x) cm = 9x cm., In DADE and DABC, we have, ∠ADE = ∠ABC, (corres. ∠s), ∠AED = ∠ACB, (corres, ∠s), \ DADE ~ DABC, [by AA-similarity], DE AD 5x 5, =, =, =, ⇒, BC AB 9x 9, DE 5, =, BC 9, In DDFE and DCFB, we have, ∠EDF = ∠BCF, ∠DEF = ∠CBF, \ DDFE ~ DCBF, 2, ar (∆DFE ) DE 2 DE 2,  DE , =, =, =, ⇒, =, , , ar (∆CFB) CB2,  BC , BC2, , ...(i), , ⇒, , 37., , (alt. int. ∠s), (alt. int. ∠s), , ⇒ ar (DDFE) : ar (DCFB) = 25 : 81, (b) Let f (x) = 6x3 – 11x2 + kx – 20, 3, , Q (0, b), , R(2, –5), , O (0, 0), , 2, , 4, 4, 4, 4, 20 0, f =,  6   − 11   + k   − =, 3, 3, 3, 3, , 2, , 25, 5, =, , 9, 81,  , , P (a, 0), , 0+b, a +0, = -5, = 2,, 2, 2, \ a = 4, b = –10 \ P is (4, 0), Q is (0, –10), 21, 21, 21, 40. (b) = =, 2, 45 9 × 5 3 × 5, Clearly, 45 is not of the form 2m × 5n. So the decimal, 21, expansion of, is non-terminating and repeating., 45, 41. (d) Sample space = {HH, HT, TH, TT}, Total number of elementary events = 4, Favourable event E = HH, n (E) = 1, 1, P(E) =, 4, 42. (a) Favourable event E = {TH, HT}, n(E) = 2, 2, 1, P(E) = =, 4, 2, 43. (d) Favourable event E = {TT}, n(E) = 1, 1, P(E) =, 4, 44. (c) At most one head = {HT, TH, TT}, 3, P=, 4, 45. (b) At least one head, {HH, HT, TH}, 3, P =, 4, 46., (a), 47., (b), 48., (a), 49., (c), 50., (a), \, ,  θ , (d) Given, ,  2πr =22,  360° , , 36., , 64, 16 4k, 0, − 11. +, − 20 =, 27, 9, 3, ⇒ 128 – 176 + 12k – 180 = 0, ⇒ 12k + 128 – 356 = 0 12 k = 228 ⇒ k = 19, 3 –1 8, 38. (c) For coincident lines, = =, 6 – k 16, 1 1, = ⇒ k = 2, 2 k, x y, 39. (d) Let the line + = 1 meet x-axis at P(a, 0) and, a b, y-axis at Q(0, b). Since R is mid point at PQ., ⇒ 6.

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Sample Paper, , 9, , ANSWERKEY, 1, 11, 21, 31, 41, , (b), (a), (c), (d), (b), , 2, 12, 22, 32, 42, , (b), (a), (c), (c), (a), , 3, 13, 23, 33, 43, , (b), (b), (d), (b), (c), , 4, 14, 24, 34, 44, , (b), (b), (a), (b), (a), , 5, 15, 25, 35, 45, , (d), (b), (b), (b), (b), , 1., (b) Principal of similarity of figures., 2., (b) Let the quadratic polynomial be, ax2 + bx + c, and its zeroes be a and b., −b, c, We have α + β = −3 =, and αβ= 2=, a, a, 3., (b) Area of rectangle = 28 cm × 23 cm = 644 cm2, Radius of semicircle = 28 cm ÷ 2 = 14 cm, Radius of quadrant = 23 cm – 16 cm = 7 cm, Area of unshaded region,  1 22, , =,  × × 14cm × 14cm , 2 7, ,  1 22, , +  2 × × × 7cm × 7cm , 4, 7, , , , , cm2, , 4., , 5., , = 385, Shaded area = 644 cm2 – 385 cm2, = 259 cm2, If a = 1, then b = 3 and c = 2., So, one quadratic polynomial which fits the given, conditions is x2 + 3x + 2., (b) Let the lady has x coins of 25 p and y coins of 50 p., Then, according to problem , x + y = 40, ......... (i), 25 x + 50 y = 1250, ......... (ii), Solving for x & y we get, x = 30 (25 p coins) & y = 10 ( 50 p coins), (d) Let A(a, a), B(–a, –a) and − 3a, 3a be the given, , (, , points. Then, we have, AB =, =, , ( − a − a )2 + ( − a − a )2, , 4a 2 + 4a 2 = 2 2a, , ), , 6, 16, 26, 36, 46, , (a), (b), (b), (a), (a), , 7, 17, 27, 37, 47, , (b), (c), (d), (b), (c), , 8, 18, 28, 38, 48, , (, , BC = − 3a + a, , (c), (c), (b), (b), (b), , ) +(, , ), , 2, , (, , a2 1 − 3, , 9, 19, 29, 39, 49, , 3a + 1, , ), , 2, , + a2, , (c), (c), (b), (a), (b), , 10, 20, 30, 40, 50, , 2, , (, , ), , , ⇒, , BC =, , , ⇒, , BC= a 1 − 3, , , ⇒, , BC= a 1 + 3 − 2 3 + 1 + 3 + 2 3, , , ⇒, , =, BC a=, 8 2 2a, , and, , (, , (, , (, , 3 +1, , ) + (1 + 3 ), 2, , 3a − a, , ), , (, , 2, , 2, , ), , 2, , ), , , ⇒, , =, AC, , , ⇒, , AC, = a, , , ⇒, , AC, = a 3 +1+ 2 3 + 3 +1− 2 3, , a2, , (, , 3 + 1 + a2, , ) (, 2, , 3 +1 +, , 2, , 2, , ) +(, , AC = − 3a − a, , (b), (a), (d), (a), (a), , 3 −1, , ), , 3 −1, , 2, , 2, , a 8 = 2 2a, =, Clearly, we have AB = BC = AC, Hence, the triangle ABC formed by the given points is an, equilateral triangle., 6., (a) one, 1, 7., (b), = cos q and maximum value of cos q is 1, secθ, 1, , ⇒ Maximum value of, is 1, secθ, 8., (c) Let ABC be an isosceles triangle, where base AB = a, and equal sides AC = BC = b. Let CD be the perpendicular, on AB.

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Mathematics, , S-34, , C, , 14., , (b) AB =, BC =, , b, , b, , D, , A, , CD =, , 2, (−9 − 9)2 + (6 − 0), =, , 2, , 2, , (−9 − 9) + (0 − 6), =, , 324 + 36, =, 360 6 10, , 324 + 36, =, , =, 360 6 10, , B, , N, , BC =, , (0 − 4)2 + (3 − 0)2, , (8=, )2, , 8, , 25= 5, 2, , (−4 − 0) + (0 − 3) =, , 16 + 9=, , 25 = 5, , BC = CA ⇒ DABC is isosceles., (a) The circle is divided into 18 equal sectors, 360°, Central angle in each sector =, = 20°, 18, θ, × πr 2, Area of each sector =, 360°, 20°, = × 3.14 × 4 × 4 = 2.79, 360°, Area of shaded portion = 9 × 2.79 = 25.12, 12. (a) Product, 13. (b) n(S) = 6 × 6 = 36, E = {(1, 1), (2, 2), (3, 3), (4, 4),, (5, 5), (6, 6)}, n(E) = 6, n(E) 6 1, =, = =, P(E), n(S) 36 6, 11., , AC =, , 24 m, , 2, (4 + 4)2 + (0 − 0)=, , CA =, , (9 + 9)2 + (0 − 0)2 =, 18, , N, , AB =, , 2, , DA =, , BC =, , 1, 4b 2 − a 2, 2, , 16 + 9=, , (−9 + 9)2 + (0 − 6)2 =6, , =, , 1, Area of the ∆ABC =, base × altitude, 2, 1, 1, a, 4b 2 − a 2 =, 4b 2 − a 2 ., =× a ×, 2, 2, 4, 9., (c) We have, p(x) = x2 –10x –75, = x2 – 15x + 5x – 75, = x(x – 15) + 5(x –15) = (x –15) (x + 5), , \ p(x) = (x –15) (x + 5), So, p(x) = 0 when x = 15 or x = –5. Therefore required, zeroes are 15 and –5., 10. (b) A(–4, 0), B(4, 0), C(0, 3), , =, , (−9 − 9)2 + (6 − 6)2 =, 18, , B, , 1, a, So, AD = DB = AB =, 2, 2, Altitude, CD = height of the, , DABC is given by, , h = AC 2 − AD 2, , =, ⇒h, , (9 − 9)2 + (6 − 0)2 =, 6, , 15., , (b), W, , O, , 10 m, , A, , E, , S, , Let the initial position of the man be at O and his final, position be B, since the man goes to 10 m due east and, then 24 m due north., Therefore DAOB is a right angled triangle at angle A., \ DAOB, OB2 = OA2 + AB2 = (10)2 + (24)2, = 100 + 576, OB, =, 676, , OB = 26m, , Hence, the man is at a distance of 26 m from the starting, point., 16. (b) Let length and breadth be x cm and y cm respectively., According to problem,, 2 (x + y) = 40, .... (i), y 2, =, and, .... (ii), x 3, on solving, x = 12, y = 8, , \ Length = 12 cm and breadth = 8cm., P, 3, 17. (c) tan A =, =, b, 4, h=, , 18., , P 2 + b 2 = 9 + 16, P 3, sin A = =, h 5, 3, (c), = 0.6 where as other numbers have non-terminating, 5, decimals.

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S-35, , Solutions, 19., , (c) Let the medians through C meets AB at D., A(2, 2), , D, , C(5, –8), , B(–4, –4), Coordinates of D are,  −4 + 2 −4 + 2 , ,, ,  = ( −1, −1), 2 ,  2, , , , 36 + 49 =, 85., , Length of CD =, , ( 5 + 1)2 + ( −8 + 1)2, , =, , 20. (a) Let the number of blue balls = x, , \ Total number of balls = 5 + x, x, P (blue ball) =, 5+ x, , , P (red ball) =, , 5, 5+ x, , Given that P (blue) = 2 • P (red), x, 5, , =2×, 5+ x, 5+ x, x, 10, , =, 5+ x, 5+ x, (c) Total number of outcomes are {HH, HT, TH, TT}., The outcomes favourable to the event ‘atmost one head’, are HT, TH and TT., n(E) 3, ∴ P(E) =, =, , n(s) 4, (c), , 27., , (d) The L.C.M. of 16, 20 and 24 is 240., The least multiple of 240 but it is not a perfect square., Similarly 2400 is also ruled out because it is also not a, perfect square. 1600 is divided by 16 and 20 but not by 24., Therefore 3600 is least number which is a perfect square, and divisible by 16, 20, 24., 28. (b) The centre of the circle is the midpoint of the diameter., So coordinates of centre = midpoint of AB,  −2 + 4 3 − 5   2 −2 , ,, = ,  =  ,  = (1, – 1), 2  2 2 ,  2, , 21., , 22., , , \ ∠BCA = ∠DEA = q, Clearly, DABC and DADE are similar, AC AB, , \, =, AE AD, 24 15, 15 × 16 240, , ⇒, =, ⇒h=, =, = 10, 16 h, 24, 24, Hence, height of the telephone pole = 10 cm., 24. (a) Polynomial p(x) has four real zeros., 25. (b) sin q + 2 cos q = 1 ⇒ (sin q + 2 cos q)2 = 1, , ⇒ sin2 q + 4 cos2 q + 4 sin q cos q = 1, , ⇒ 1 – cos2 q + 4 (1 – sin2 q + 4 sin q cos q = 1, , ⇒ 4 sin2 q + cos2 q – 4 sin q cos q = 4, , ⇒ (2 sin q – cos q)2 = 4, , ⇒ 2 sin q – cos q = 2, [Q 2 sin q – cos q ≠ –2], 26. (b) The system of simultaneous equations a1x + b1y + c1, = 0 and a2x + b2y + c2 = 0, have exactly one (unique), a1 b1, ≠ ., solution if, a2 b2, , a1 b1, 3, −2, ≠, ⇒, ≠, 2m − 5 7, a2 b2, , , , , or –4m + 10 ≠ 21, or –4m ≠ 11, 11, m≠−, or, 4, 23. (d) Let h metres be the height of the telephone pole. Since, time is the same in both the cases., B, , 29., , (b) Since, the graph of y = f(x) is a parabola, therefore, f(x) is quadratic., 30. (d) 1+ sin2 q = 3 sin q cos q, [Q 1 = cos2 q + sin2 q], , ⇒ cos2 q + 2 sin2 q = 3 sin q cos q, , ⇒ cos2 q – 3 sin q cos q + 2 sin2 q = 0, , ⇒ cos q – sin q) (cos q – 2 sin q) = 0, , ⇒ cos q – sin q = 0 or cos q – 2 sin q = 0, , ⇒ sin q = cos q or 2 sin q = cos q, sin θ, sin θ, or 2, 1, = 1=, , ⇒, cos θ, cos θ, , ⇒ tan q = 1 or 2 tan q = 1, 1, Thus, tan q = 1 or tan q = ., 2, 31. (d) By Pythagoras theorem in DBAC, we have, C, , D, , 15 m, , h, , C, , , , , E, , 16 m, 24 m, , O, , A, , r, , , , A, , B

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Mathematics, , S-36, , BC2 = AB2 + AC2 = 62 + 82 = 100 ⇒ BC = 10 CM, Now,, Area of DABC = Area of DOAB + Area of, DOBC + Area of DOCA, 1, 1, 1, 1, , ⇒ AB × AC = AB × r + BC × r + CA × r, 2, 2, 2, 2, 1, 1, (6 × r) + (10 × r) + (8 × r), 2, 2, , ⇒ 48 = 24 r ⇒ r = 2 cm, 32. (c) (– 1)n + (– 1)4n = 0 will be possible, when n is any odd, natural number i.e.,, , ⇒, , 33., , A, , (b) –2, 1, 3, , A, , 38., , (b), , ×6×8=, , O, 28, 28, 45, , (b), , 37., , B, , C, Area of sector OACB, 45 22, = ×, × 28 × 28 = 308 cm27, 360 7, 1, Area (DAOB) = (28) (28) sin 45°, 2, = 14 × 28 × = 277.19, Area (minor segment) = 308 – 277.19, = 30.81, 34. (b) Let the required ratio be K : 1, \ The co–ordinates of the required point on the y–axis is, K(−4) + 3(1) y = K(2) + 5(1), x =, ;, K +1, K +1, Since, it lies on y – axis, , \ Its x–cordinates = 0, −4K + 3, 0 ⇒ –4K + 3 = 0, ∴, =, , K +1, 3, , ⇒ K=, 4, 3, , ⇒ Required ratio = : 1, 4, , \ ratio = 3 : 4, 35. (b) Consistent system, tan θ + sec θ − 1, 36. (a), tan θ − sec θ + 1, (tan θ + sec θ) − (sec2 θ − tan 2 θ), tan θ − sec θ + 1, (tan θ + sec θ)[1 − (sec θ − tan θ)], =, tan θ − sec θ + 1, (tan θ + sec θ)(1 − sec θ + tan θ), =, (1 + tan θ − sec θ), =, , = tan q + sec q, 1 + sin θ, =, cos θ, , E, D, B, In DAEB, ∠AEB = 90°, , \ AB2 = AE2 + BE2, In DAED, ∠AED = 90°, , \ AD2 = AE2 + DE2, , ⇒ AE2 = (AD2 – DE2), , \ AB2 = (AD2 – DE2) + BE2, = (AD2 – DE2) + (BD – DE2), = (AD2 – DE2) +  1 BC − DE , , , 2, , , C, ....(i), , 2, , 1, = AD2 + BC2 – BC × DE, 4, 39. (a) The numbers common to given numbers are 22, 5 and, 72., \ H.C.F. = 22 × 5 × 72 = 980., 40. (a) Perimeter of sector = 25 cm, θ, 2r +, × 2r = 25, 360, 25, 90, , ⇒ 2r +, ×2×, × r = 25, 7, 360, 22, 11, , ⇒, 2r + r = 25 ⇒, r = 25 ⇒ r = 7, 7, 7,  πθ Sinθ  2, −, Area of minor segment = , r, 2 ,  360°,  22 90 sin 90  2, −, =,  ×,  (7), 2 ,  7 360°, 4,  11 1 , × 49 = 14 cm2., =,  −  × 49 =, 14, 14, 2, , , 41., 44., 46., , 47., , 48., , 49., , (b), 42., (a), 43., (c), (a), 45., (b), (a) As three faces are marked with number ‘2’, so number, of favourable cases = 3., 3 1, \ Required probability, P(2)= =, 6 2, (c) No. of favourable cases = No. of events of getting the, number 1 + no. of events of getting the number 3 = 2 + 1 = 3, 3 1, \ Required probability, P(1 or 3)= =, 6 2, (b) Only 1 face is marked with 3, so there are 5 faces, which are not marked with 3., 5, \ Required probability, P (not 3) =, 6, (b) , 50. (a)

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Sample Paper, , 10, , ANSWERKEY, 1, 11, 21, 31, 41, , 1., , (c), (b), (b), (a), (c), , 2, 12, 22, 32, 42, , (b), (d), (c), (a), (c), , 3, 13, 23, 33, 43, , (a), (a), (d), (d), (c), , 4, 14, 24, 34, 44, , (, , AB =, , ( −6 )2 + ( 6 )2, , = 36 + 36, =, , ) + ( −3, , 3 − 3=, , (, , ) + ( −3, , 3 +3=, , AC = −3 3 − 3, , 2, , 2, , ), , 2, , ), , 2, , ), , =, 72 6 2, , (, , BC = −3 3 + 3, , 4., , 5, 15, 25, 35, 45, , (c), (b), (b), (a), (c), , (c) Let a and b be the zeroes of the quadratic polynomial., we have a = 8 and b = 10, Sum of zeroes = a + b = 8 + 10 = 18, Product of zeroes = ab = 8 × 10 = 80., \ The required quadratic polynomial, = x2 – (Sum of the zeroes)x + Product of the zeroes, = x2 – 18x + 80, Any other quadratic polynomial that fits these condition, will be of the form, k (x2 – 18x + 80), where k is a real., , 2., (b) A(3, –3), B(–3, 3), −3 3, −3 3, , , 3., , (b), (a), (d), (c), (d), , =, 72 6 2, =, 72 6 2, , \ DABC is equilateral triangle., (a) Let the two numbers be x and y (x > y). Then,, x – y = 26 , ...(i), x = 3y ...(ii), Substituting value of x from (ii) in (i), 3y – y = 26, 2y = 26, y = 13, Substituting value of y in (ii) x = 3 × 13 = 39, Thus, two numbers are 13 and 39., (b) Area of equilateral triangle , , 3 2, a, 4, , 6, 16, 26, 36, 46, , (d), (d), (b), (b), (c), , 7, 17, 27, 37, 47, , (d), (b), (d), (b), (a), , 8, 18, 28, 38, 48, , (b), (c), (c), (c), (d), , 9, 19, 29, 39, 49, , (a), (b), (a), (a), (a), , 10, 20, 30, 40, 50, , (d), (d), (b), (d), (b), , 3 2, 121 3, ⇒, a =, 4, , ⇒ a2 = 484, ⇒ a = 22 cm, , Perimeter of equilateral D = 3a, , = 3 (22), , = 66 cm, , Since the wire is bent into the form of Q circle, So, perimeter of circle = 66 cm, ⇒ 2pr = 66, 22, 66, ⇒ 2× × r =, 7, , 1 7, ⇒ r = 66 × ×, 2 22, , ⇒ r = 10.5 cm, , So Area enclosed by circle = pr2, , , 22, 10.510.5, 7, , , = 22 × 1.5 × 10.5, , = 346.5 cm2, 5. (c) 1st wheel makes 1 revolutions per sec, 6, 2nd wheel makes, revolutions per sec, 10, 3rd wheel makes, , 4, revolutions per sec, 10, , In other words 1st, 2nd and 3rd wheel take 1,, seconds respectively to complete one revolution., , 5, and, 3

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Mathematics, , S-38, , L.C.M of 1,, , 5 L.C.M of 1, 5, 5, 5, =5, and =, 2 H.C.F of 1, 3, 2, 3, , Hence, after every 5 seconds the red spots on all the three, wheels touch the ground., a 2 − b2, 6., (d) sin θ = 2, a + b2, , , A, , , , , , , , 9., , = (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3) = 6, No. of ways of getting 11 = (5, 6), (6, 5) = 2, No. of favourable ways = 1 + 2 + 4 + 6 + 2 = 15, No. of exhaustive ways = 6 × 6 = 36, Probability of the sum as a prime , 15 5, = =, 36 12 , , (a) Given, AB = 2DE and DABC ~ DDEF, Hence,, , b², , a² – b², , +, a², , B, , perpendicular, base, , 2, 2, \ AC = a − b, 2, 2, AB a + b, , Now in D ABC ,, ∠ B = q and ∠ C = 90°, (a2 + b2)2 = BC2 + (a2 – b2)2, \ BC = 2ab, , cosec q =, , a 2 + b2, ,, a 2 − b2, , BC, 2ab, θ, = 2, cot=, AC, a, − b2, , , 1, P (7, –6), , , , 2, , a +b, 2ab, a+b, =, + 2, 2, 2, 2, a−b, a −b, a −b, 2, Q (3,4), R, , cosec q +=, cot q, (d), , 2, , Coordinate of R,  7 (2) + 3 (1) −6(2) + 4(1) , ,, = , , 1+ 2, 1+ 2, , ,  17 −8 , =  , ,  3 3 , , 56, 4 DE 2, 4, = =, area(∆DEF ), DE 2, , =, area (DDEF), , C, , Since, sin θ =, , 7., , or, , area(∆ABC ) AB 2, =, area(∆DEF ) DE 2, , 10., , [Q AB = 2DE], , 56, = 14 sq.cm., 4, , (d) Given : length of the sheet = 11 cm, Breadth of the sheet = 2cm, Diameter of the circular piece = 0.5 cm, Radius of the circular piece, 0.5, =, = 0. 25 cm, 2, Now, area of the sheet = length × breadth, = 11 × 2 = 22 cm2., Area of a circular disc = pr2, 22, =, × (0.25)2 cm2, 7, Number of circular discs formed, Area of the sheet, =, Area of one disc, =, , 22, 22 × 7, = 112, =, 22, 2, 22, ×, 0.0625, × ( 0.25 ), 7, , Hence, 112 discs can be formed., 11. (b) a = x3y2, , =x×x×x×y×y, b = xy3, , =x×y×y×y, ⇒ HCF (a, b) = xy2, tan q, cot q, 12. (d), +, 1 − cot q 1 − tan q, , , Thus, the point R lies in IV quadrant., 8., (b) The sum of the two numbers lies between 2 and 12., So the primes are 2, 3, 5, 7, 11., , No. of ways for getting 2 = (1, 1) = 1, sin q, cos q, , No. of ways of getting 3 = (1, 2), (2, 1) = 2, cos q + sin q, =, , No. of ways of getting 5 = (1, 4), (4, 1),, cos q, sin q, 1−, 1−, , (2, 3), (3, 2) = 4, sin q, cos q, , No. of ways of getting 7

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S-39, , Solutions, , A, , =, , sin 2 q, cos 2 q, −, cos q(sin q − cos q) sin q(sin q − cos q), , 2x, , =, , sin 2 q× sin q − cos 2 q× cos q, sin q cos q(sin q − cos q), , =, , sin 3 q − cos3 q, sin q cos q(sin q − cos q), , =, , (sin q − cos q)(sin 2 q + cos 2 q + sin q cos q), sin q cos q(sin q − cos q), , =, , 1 + sin q cos q, sin q cos q , , So,, , =+, 1 sec θ cosec θ, , –k, =1, 2, , \ k=2, 13. (a) Let present age of Nuri = x years, , Let present age of Sonu = y years, , Five years ago,, , x – 5 = 3(y – 5), , x – 5 = 3y – 15, , x – 3y = –10 , ...(i), , Ten years later,, , (x + 10) = 2(y + 10), , x + 10 = 2y + 20, , x – 2y = 10 , ...(ii), , Subtracting (ii) from (i), we get, , – y = – 20, ⇒, y = 20, , Substituting y = 20 in (ii), we get, , x – 2 × 20 = 10, ⇒, x = 50, , So, present age of Nuri is 50 years and present age, of Sonu is 20 years, 14. (a) Using Pythagoras theorem in DABL we have, AL = 8cm,, Also, DBPQ ~ DBAL, \, , BQ BL, 6− x 6, =, ⇒, =, PQ AL, y, 8, , or, , x= 6 −, , 10, cm, , sin 2 q, cos 2 q, +, cos q(sin q − cos q) sin q(cos q − sin q), , 3, y, 4, , Q, , m, , =, , y, , P, , c, 10, , sin q, cos q, +,  sin q − cos q ,  cos q − sin q , cos q. ,  sin q. , , sin q, cos q, , , , , , =, , 2x, y, , B, , C, R, 12 cm, 15. (b) Let the common factor be x – k, , we have,, , f(k) = g(k) = 0, ⇒, k2 + 5k + p = k2 + 3k + q, q−p, k=, 2, , S, , , substituting “k” in x2 + 5x + p = 0, x2 + 5x + p = 0, 2, , q−p q−p, 0, ,  +, +p =,  2   2 , \ (p – q)2 = 2 (3p – 5q), 16. (d) X = (April, June, September, November), , Hence, n(X) = 4, 5, 1, 2, ±   5 = irrational, 17. (b) x = ⇒ x =, 9, 3, , ( ), , 18. (c) PQ = 13 ⇒ PQ2 = 169, ⇒, (x – 2)2 + (–7 – 5)2 = 169, ⇒, x2 – 4x + 4 + 144 = 169, ⇒, x2 – 4x – 21 = 0, ⇒, x2 – 7x + 3x – 21 = 0, ⇒, (x – 7) (x + 3) = 0, ⇒, x = 7, –3, 19. (b), B, , D, A, C, P, , We have two chord AB and CD when produced, meet outside the circle at P., , Since in a cyclic quadrilateral the exterior angle is, equal to the interior opposite angle,, \, ∠PAC = ∠PDB , ....(i), , From (1) and (2) and using AA similarity we have, DPAC ~ DPDB

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Mathematics, , S-40, , \ Their corresponding sides are proportional., PA PC, ⇒, =, , PD PB, ⇒, 20., , PA.PB = PC.PD., a sin φ, (d) we have, tan q =, 1 − a cos φ, , 1, − cot φ, a sin φ, 1, ⇒ cot q + cot φ =, , a sin φ, , b sin q, tan φ =, 1 − b cos q, cot q, ⇒ =, , Thus, , Area (DABE) 1, =, Area (DACD) 2, , , Thus reqd. ratio is 1 : 2., 23. (d) Area of circle A = 3.14 × 10 × 10 = 314, , Area of circle B = 3.14 × 8 × 8 = 200.96, 1, , Area of Q =, × Area of B, 8, =, , ...(i), , 1, × 200.96 = 25. 12, 8, , Now, =, , Area of P, Area of Q, , 1, cot φ, − cot q, ⇒ =, b sin q, , ⇒ Area of P =, , 1, ⇒ cot φ + cot q =, , b sin q, , =, , ...(ii), , , , From (i) and (ii), we have, 1, 1, =, a, sin, φ, b, sin, q, , , 24., 25., , a sin q, ⇒ =, b sin φ, 21. (b), , , , 22. (c), , Let f(x) = xn + yn., Divisible by (x + y) means f(–y) = 0., So, (–y)n + yn = 0., This is possible only when “n” is an odd number., D, 26., , E, , A, , x, 2x, x, C, B, , Let, AB = BC = x., Since DABC is right-angled with, ∠B = 90°, \, AC2 = AB2 + BC2 = x2 + x2 = 2x2, ⇒ AC = 2x, Since DABE ~ DACD, \, , Area ( DABE ) AB, x, 1, = = =, ., 2, 2, Area ( DACD ) AC, 2, 2x, 2, , 2, , 27., , 5, × Area of Q, 4, , 5, × 25.12 = 31.4, 4, Area of square = 7 × 7 = 48, Required Area, = (314 + 200.96 + 49 – 25.12 – 31.4), = 507.44 cm2, (d) All the properties are satisfied by real numbers., (b) A(0, 4), B(0, 0), C(3, 0), AB =, , (0 − 0)2 + (0 − 4)2 =, 4, , BC =, , (3 − 0)2 + (0 − 0)2 =, 3, , CA =, , (0 − 3)2 + (4 − 0)2 =, 5, , AB + BC + CA = 12, (b) Let the two parts be x and y., We have,, x + y = 62 , x, 4 =2, 2y 3, 5, 15x – 16y = 0 , By solving (i) and (ii) we get x = 32, y = 30, 1, , (d) Here, ( p + 2 )  q −  = pq − 5 , 2, , , ...(i), , ...(ii), ...(i), , ⇒, , 1, pq − p + 2q − 1= pq − 5 ...(ii), 2, , ⇒, , p, − + 2q =−4 ...(iii), 2, , ⇒, , p, 4 , − 2q =, 2, , 1, , also, ( p − 2 )  q −  = pq − 5, 2, 

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S-41, , Solutions, ⇒, , 1, pq − p − 2q + 1= pq − 5, 2, , ⇒, , 1, − p − 2q =, −6 ...(iv), 2, , By adding (iii) and (iv), we get, p = 10, p, 4, =, − 2q =, 2, or, , 32., 33., , 10, 4, − 2q =, 2, , ⇒, , ⇒ 5 – 4 = 2q ⇒ q =, , 1, 2, , ⇒, ,  1, Hence, solution set (p,q) = 10, ,  2, 28., , ⇒, , (c) Let cos q + 3 sin, =, q 2sin q, , , , p = 4 and 2q = 2p – 4, ⇒ 2q = 8 – 4 = 4, Now, q = 2, ⇒ p+q=4+2=6, ⇒ p–q=4–2=2, (a) An irrational number., cos 2 q, cos 2 q, =3, 3, ⇒, =, (d), cos 2 q, cot 2 q − cos 2 q, 2, − cos q, sin 2 q, , Multiplying both sides by 2 + 3 , we get, ⇒ cos q= 2sin q − 3 sin q= (2 − 3)sin q, , 34., , cos 2 q× sin 2 q, cos 2 q − sin 2 q cos 2 q, , sin 2 q cos 2 q, cos 2 q(1 − sin 2 q), sin 2 q, 2, , cos q, , 3, =, , 3 ⇒ tan2q = 3 ⇒ tan q =, =, , 3, , tan q = tan 60º ⇒ q = 60º (acute angle), (c), , (2 + 3) cos q= (2 + 3)(2 − 3)sin q, 2, , 3, =, , d, , 5m, , 2, , ⇒ (2 + 3) cos, =, q {(2) − ( 3) }sin q, , 6m, , ⇒ 2cos q + =, 3 cos q (4 – 3)sin q, ⇒ sin q − 3 cos, =, q 2cos q, , Using pythagoras theorem,, , (a) Substituting the given zeros in (x – a) (x – b), we get, 1, 2,  x − ,  x + , 3, 5, , , , 29., , 30., , 31., , 1 , 15x 2 + x − 2 , , 15 , , 35., , (b) S = {(1, 1), ....., (1, 6), (2, 1),.....,(2, 6), (3, 1), ...., (3, 6),, (4, 1), ....., (4, 6), (5, 1),...., (5, 6), (6, 1),....., (6, 6)}, n(S) = 36, Let E be the event that both dice show different numbers., E {(1, 2), (1, 3),...., (1, 6), (2, 1), (2, 3), (2, 4),....,, (2, 6), (3, 1), (3, 2), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4,, 3), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (6, 1),, (6, 2), (6, 3), (6, 4), (4, 5)}, n(E) = 30, n ( E ) 30 5, P ( E=, = =, ), n ( S) 36 6, (a), , A (3p, 4), , P (5, p), , d=, , B (–2, 2q), , Since, P (5, p) is the mid point of AB, 3p - 2, 4 + 2q, and p =, \ 5=, 2, 2, , 122 + 52=, , 144 + 25=, , 169, , \ d = 13 m, So, distance between the tops of poles is 13 m., (a) Radius of circle = 14 cm ÷ 2 = 7 cm, One side of the figure opposite to, 35cm = 35cm –7cm = 28cm, 35 m, , 7 cm, , =, , 11 m, , 12 m, , ⇒ 2cos q + 3 cos q= (4 − 3)sin q, , 7 cm, O, , 28 cm, 7 cm, , 14 cm, , 14 cm, Perimeter of the two sectors of circle, 1 22, = ×, × 14cm = 22cm, 2 7, \ Total perimeter = 134 cm, The perimeter of the given figure is 134 cm.

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Mathematics, , S-42, , 36., , 37., , 161, =7, 23, ⇒, 2 × product of zeroes = 14p, ⇒, 2 × 7 = 14p, 14, p=, ⇒ p= 1, \, 14, (b) Suppose the required ratio is m1 : m2 Then, using the, section formula, we get, m (4) + m 2 ( –3), –2 = 1, m1 + m 2, (b) Product of zeroes =, , ⇒ – 2m1 – 2m 2 = 4m1 – 3m 2, , 46., , ⇒ m2 = 6 m1 ⇒ m1 : m2 = 1:6, 38., , 39., , (c) If the sum of 3 prime is even, then one of the numbers, must be 2., Let the second number be x. Then as per the given, condition,, x + (x + 36) + 2 = 100 ⇒ x = 31, So, the number are 2, 31, 67., Hence largest number is 67., (a), , Also 27a × 27b = 27 × 162., , ⇒, ab = 6 , ...(ii), (a – b)2 = (a + b)2 – 4ab, , ⇒, a–b=1, Solving (i) and (iii), we get, a = 3, b = 2, So numbers are 27 × 3, 27 × 2 i.e., 81, 54, 43. (c) H.C.F. of two co-prime natural number is 1., 44. (d) LCM =HCF, , ⇒, two numbers are equal., 45. (c) Clearly, LCM = (LCM of p and p3), (LCM of q2 and q) = p3q2, , ar (DABC) BC2, =, ar (DDEF) EF2, , 47., , 48., , 2, ,  2.1 , ⇒ ar (DABC), = , = 9cm 2,  × ar (DDEF),  2.8 , 40., , k −1, ≠, ⇒ k ≠ 3., (d), 6 −2, , (c) H.C.F. = 16 and Product = 3072, Pr oduct 3072, L.C.M. =, = = 192, H.C.F., 16, 42. (c) H.C.F. of two numbers is 27, So let the numbers are 27a and 27b, Now 27a + 27b = 135, , ⇒, a + b = 5 , 41., , 49., , 50., , ...(i), , (c) Area of minor sector OAPB, q, 90, =, × p=, r2, × 3.14 × (10)2, 360, 360, = 78.5 cm2, (a) Area of minor segment APB, q, q,  pq, = , − sin cos  r 2, 2, 2,  360, 90, , , =  3.14 ×, − sin 45° cos 45°  (10)2, 360, , , = 28.5 cm2, (d) Area of the major sector OAQB, = Area of circle – Area of minor sector OAPB., = (314 – 78.5)cm2 = 235.5 cm2, (a) Area of major segment AQB, = Area of the circle – Area of the minor segment APB, = (3.14 × 10 × 10 – 28.5) cm2, = 285. 5 cm2, (b) Length of arc APB, 90, 22, × 2 × × 10, =, 360, 7, = 15.71 cm