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About Pearson, Pearson is the world’s learning company, with presence across 70 countries, worldwide. Our unique insights and world-class expertise comes from a long, history of working closely with renowned teachers, authors and thought leaders, as, a result of which, we have emerged as the preferred choice for millions of teachers, and learners across the world., We believe learning opens up opportunities, creates fulfilling careers and hence, better lives. We hence collaborate with the best of minds to deliver you classleading products, spread across the Higher Education and K12 spectrum., Superior learning experience and improved outcomes are at the heart of everything, we do. This product is the result of one such effort., Your feedback plays a critical role in the evolution of our products and you can, contact us – [email protected]. We look forward to it., , A01_IIT-FOUNDATION_XXXX_FM.indd 1, , 4/11/2018 11:43:12 AM

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Copyright © 2018 Pearson India Education Services Pvt. Ltd, Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128, formerly known as, TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia., No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent., This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to, remove any material in this eBook at any time., ISBN 978-93-528-6769-1, eISBN 978-93-530-6192-0, Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh,, India., Registered Office: 4th Floor, Software Block, Elnet Software City, TS 140, Block 2 & 9, Rajiv Gandhi Salai, Taramani,, Chennai - 600 113, Tamil Nadu, India., Fax: 080-30461003, Phone: 080-30461060, www.pearson.co.in, Email: [email protected]

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Contents, Prefacevii, How to Use the Practice Book, ix, Series Chapter Flow, x, CHAPTER 1 Geometry and Symmetry1.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 1.1, 1.4, 1.6, 1.10, 1.14, , CHAPTER 2 Linear Equations and, Inequations2.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 2.1, 2.3, 2.5, 2.7, 2.9, , CHAPTER 3 Mensuration3.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 3.1, 3.4, 3.6, 3.9, 3.12, , CHAPTER 4 Polynomials and Their, LCM and HCF4.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , A01_IIT-FOUNDATION_XXXX_FM.indd 5, , 4.1, 4.3, 4.5, 4.7, 4.9, , CHAPTER 5 Ratio, Proportion, and, Variations: Commercial, Mathematics5.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 5.1, 5.4, 5.6, 5.8, 5.10, , CHAPTER 6 Real Numbers, LCM, and, HCF6.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 6.1, 6.3, 6.5, 6.7, 6.9, , CHAPTER 7 Squares and Square, Roots, Cubes and Cube, Roots, and Indices7.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 7.1, 7.3, 7.5, 7.7, 7.9, , CHAPTER 8 Time and Work, Time and, Distance, and Statistics8.1, , Assessment Test I, Assessment Test II, Assessment Test III, Assessment Test IV, Answer Keys, , 8.1, 8.4, 8.6, 8.9, 8.11, , Hints and Explanations A.1, , 4/11/2018 11:42:43 AM

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Preface, Pearson IIT Foundation Practice Book Series is designed to accompany the course-books available, in this series. Developed by a team of experienced faculties, this workbook series connects, the subjective knowledge to its real world applications through various text and chapter level, problems. Each chapter has a set of assessment tests which are mapped to chapters covered in, the course-book. These worksheets will guide students step-by-step towards understanding the, central concept of that particular chapter. These tests are recommended as after class material for, further practice., Any suggestions for added or updated additional readings would also be welcome. Students can, share their feedback at [email protected]., , A01_IIT-FOUNDATION_XXXX_FM.indd 7, , 4/11/2018 11:43:12 AM

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How to Use the Practice Book, Many times, students face significant challenges in answering application level questions in Physics, Chemistry, and Mathematics. These Practice Books will enhance their problem-solving skill which will definitely lead to a, strong subject foundation. The entire practice book series are recommended to be used alongside IIT Foundation, course-books., Students can refer the following steps while using the practice books:, , Study, chapters from, IIT Foundation, Course-books, Complete, remaining, Assessment, Tests, , Check self, progress, , Solve the, chapter-end, exercises, , Solve one, Assessment, Test related to, the chapter, , While preparing for Foundation courses, students need to learn the fundamental concepts with utmost clarity. In, order to successfully complete the IIT Foundation course, one must prepare profoundly. Consistent hard work,, practice and perseverance are needed throughout the year., During any competitive examination, one must exercise clinical precision with speed since the average time, available to respond to a question is hardly a minute. The aspirants should be conceptually excellent in the subject, owing to the negative marking in the examination. A better practice to solve the paper would be to go for the, easiest questions first and then gradually progress to the more complicated ones., Regular practice of MCQs will assist the aspirants in preparing for the examination. In a nutshell, hard work,, conceptual clarity and self-assessment are the essential ingredients to achieve success in competitive examinations., IIT Foundation course-books play an important role in understanding the concepts. Student need to read-up on, all concepts/theories in a regular and systematic manner., , A01_IIT-FOUNDATION_XXXX_FM.indd 9, , 4/11/2018 11:43:13 AM

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Course-book Chapter Flow, Class 7, Expressions and, Special Products, , Indices, , 1, , 5, , 3, 2, , 4, Geometry, , Ratio and, Its Applications, , Number Systems, , Equations and their Applications, , Statistics, 10, , 8, , 6, 7, , 9, Set Theory, , Formulae, , Mensuration, , Class 9, , Linear Equations, and Inequations, , Logarithms, 3, , 1, , 5, , 2, , 4, Polynomials and Square Roots, of Algebraic Expressions, , Number, Systems, , Probability, , Geometry, , Quadratic Expressions, and Equations, , 11, , 9, , 7, , 10, , 12, Banking and, Computing, , 6, , 8, Statistics, , Matrices, Percentages, Profit and Loss,, Discount, and Partnership, , Locus, , Mensuration, , 17, , 15, , 13, , Trigonometry, , Coordinate, Geometry, Time and Work, 23, , Sales Tax and Cost, of Living Index, , Ratio, Proportion and Variation, 21, , 22, , A01_IIT-FOUNDATION_XXXX_FM.indd 10, , 18, , 16, , 14, , Time and Distance, , Sets and Relations, , Significant Figures, , 19, 20, , Shares and, Dividends, , Simple Interest and, Compound Interest, , 4/11/2018 11:43:13 AM

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xi, , Course-book Chapter Flow, , Class 8, Squares and Square, Roots and Cubes and, Cube Roots, , Polynomials,, LCM and HCF of, Polynomials, , 1, , 5, , 3, 2, , Real Numbers and, LCM and HCF, , 4, Formulae, , Indices, Simple Interest and, Compound Interest, , Percentages, 8, , 10, , 6, , 9, , 7, Profit and Loss,, Discount and Partnership, , Time and Work,, Pipes and Cisterns, Linear Equations, and Inequations, , Ratio, Proportion, and Variation, Geometry, , Statistics, 13, , 11, , 17, , 15, , 12, , 16, , 14, Sets, , Time and Distance, , Mensuration, , Matrices, , Class 10, Quadratic Equations, and Inequalities, , Polynomials and, Rational Expressions, , Sets, Relations, and Functions, , 3, , 1, , 5, 4, , 2, , 6, , Linear Equations, in Two Variables, , Number, Systems, Mensuration, , Statements, , 13, , 11, , 14, , 9, , 7, 8, , 10, , 12, Geometry, , Trigonometry, , Matrices, , Statistics, , Remainder and, Factor Theorems, , Limits, , Mathematical Induction, and Binomial Theorem, , Permutations and, Combinations, , Linear, Programming, 17, , 15, , 19, , 16, , A01_IIT-FOUNDATION_XXXX_FM.indd 11, , Banking, 23, , 25, 24, , 26, Logarithms, , Computing, , Instalments, , Partial Fractions, 27, , 20, , 18, Modular Arithmetic, , Coordinate, Geometry, , Progressions, , Shares and Dividends, , 21, 22, , Taxation, , Probability, , 4/11/2018 11:43:14 AM

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1, , Geometry and, Symmetry, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapter - Geometry; pp. 16.1–16.40, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. The following are the steps involved in constructing circumcircle of a triangle. Arrange them in sequential order., (A), (B), (C), (D), (E), (a), (c), , Draw the perpendicular bisectors of the sides of the triangle., Draw the triangle with the given measurements., Take the distance from S to any vertex of the triangle as R., Consider the point of concurrence of the perpendicular bisectors as S., Draw a circle with centre ‘S’ and radius R., ABCDE, (b) BCDAE, BADCE, (d) BACDE, , 2. In a triangle ABC, ∠A = 60° and AB = AC. If AD is a median drawn from A, to BC and AD = 6 cm, then find the altitude drawn from B to AC (in cm)., (a) 3 3, , (b) 6, , (c) 6 3, , 3. In ∆ABC, BD is a median drawn to CA. If BD =, find ∠CBD., , (d) 3, , AC, and ∠BAD = 25°, then, 2, , B, , 25, C, , (a) 45°, , D, , (b) 80°, , A, , (c) 70°, , (d) 65°, , 4. In a ∆PQR, PO and QO are the bisectors of ∠P and ∠Q, respectively. If, ∠POQ = 140°, then the measure of ∠PRQ is _________., R, O, , P, , (a) 70°, , M01_IIT-FOUNDATION_XXXX_CH01.indd 1, , (b) 80°, , Q, , (c) 90°, , (d) 100°, , 4/11/2018 4:33:43 PM

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1.2, , Chapter 1 Geometry and Symmetry, , 5. In the given figure, ∠P = 95°, ∠PQR = 30°, and ∠PSR = 35°. What will be the, value of ∠QRS?, , Space for rough work, , P, R, Q, , S, , (a) 160°, (c) 120°, , (b) 150°, (d) 110°, , 6. ABCD is an isosceles trapezium in which AB CD and ∠A = 57°. Find the, value of ∠C + ∠D., (a) 112°, , (b) 123°, , (c) 246°, , (d) 303°, , 7. PQRS is a parallelogram, if ∠PQS = 40° and PS = QS, then ∠SQR – ∠QRS =, _________., (a) 60°, , (b) 20°, , (c) 30°, , (d) 45°, , 8. KLMN is an isosceles trapezium in which LM = KL and ∠MNK = 100°. Find, the supplement of ∠KLM., (a) 80°, , (b) 60°, , (c) 90°, , (d) 100°, , 9. A triangle is inscribed inside a circle. If one of the sides of the triangle is a, diameter of the circle, then the triangle is a/an _________., (a) acute-angled triangle, (c) right-angled triangle, , (b) obtuse-angled triangle, (d) None of these, , 10. In the given figure, ∆ABC is inscribed in a circle and O is the centre of the, circle. If ∠OBC = 20°, then find the measure of ∠BAC., A, , O, C, , B, , (a) 140°, , (b) 40°, , (c) 50°, , (d) 70°, , 11. In the given figure, A, B, C, D, E and F are the points on the circle and ∠BDF =, 70°. If ‘O’ is the centre of the circle and ∠ECF = 30°, then find the measure of ∠EAB., A, F, , O, , E, , (a) 70°, , M01_IIT-FOUNDATION_XXXX_CH01.indd 2, , (b) 80°, , B, C, , D, , (c) 90°, , (d) 60°, , 4/11/2018 4:33:45 PM

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Assessment Test I, Directions for question 12 to 15: Match the statements of Column A with those of, Column B., Column A, 12. The geometric point of a triangle, which always lies inside the triangle, , 1.3, , Space for rough work, , Column B, (a)    In centre, , 13. The geometric point of a triangle, (b)   Orthocentre, which always lies outside the triangle, 14. The geometric point of a triangle, (c)     Circumcentre, which always lies on two sides of the, triangle, 15. The geometric point of a triangle, (d) Excentre, which lies only on the longest side of, the triangle, , M01_IIT-FOUNDATION_XXXX_CH01.indd 3, , 4/11/2018 4:33:45 PM

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1.4, , Chapter 1 Geometry and Symmetry, , Assessment Test II, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. The following are the steps involved in constructing an incircle of a triangle., Arrange them in sequential order., (A) Draw the bisectors of internal angles of the given triangle., (B) Let the point of concurrence of the angle bisectors be I., (C) Draw the triangle with the given measurements., (D) With ID as radius and I as the centre, draw a circle., (E) Draw ID perpendicular to any of the sides of the triangle., (a) ABCDE, (b) CBADE, (c) ACEBD, (d) CABED, 2. In a triangle PQR, the point of concurrence of medians and the point of concurrence of altitudes coincide and it is G. If altitude of the triangle is 9 cm,, then find the length of PG., (a) 3 cm, , (b) 6 cm, , (c) 9 cm, , (d) 12 cm, , 3. In a ∆PQR, PS = SR = SQ and ∠PRS = 55°. Find ∠SQP., P, , Q, , (a) 25°, , S, , R, , (b) 35°, , (c) 45°, , (d) 15°, , 4. In a ∆XYZ, I is the incentre. If ∠XIY = 100°, then what is the value of ∠XZY?, (a) 10°, (b) 20°, (c) 30°, (d) 40°, 5. In the given figure, ∠DAC = 30°, ∠CBD = 40°, ∠ADB = y°, and ∠ACB = x°., Find the difference between angles y° and x°., A, , B, , y°, D, x°, C, , (a) 30°, , (b) 80°, , (c) 60°, , (d) 70°, , 6. KLMN is an isosceles trapezium in which KL MN . If ∠NKL = 70°, then, find the measure of ∠LMN., (a) 70°, (b) 80°, (c) 100°, (d) 110°, , M01_IIT-FOUNDATION_XXXX_CH01.indd 4, , 4/11/2018 4:33:47 PM

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Assessment Test II, 7. SRMT is a parallelogram. If ∠SRM = 108°, SR = 2MR, and P is the midpoint, of TM , then find ∠SPR., (a) 90°, (b) 80°, (c) 110°, (d) 72°, , 1.5, , Space for rough work, , 8. PQRS is a trapezium in which PQ SR . T is a point on PQ such that PS = ST,, TR = SR, and ∠PST = 50°. Find the supplement of ∠SRQ., (a) 115°, (b) 88°, (c) 72°, (d) Cannot be determined, 9. A triangle PQR is inscribed in a circle. If the centre of the circle ‘O’ is outside, the triangle, then the triangle is_________., (a) acute-angled, (b) obtuse-angled, (c) right-angled, (d) equilateral, 10. ∆DEF is inscribed in a circle and ‘O’ is the centre of the circle. If ∠DEF = 70°,, then find ∠DFO., (a) 20°, (b) 25°, (c) 35°, (d) 30°, 11. In the given figure, ∠BAD = 40°, BC = CD, and AD passes through the centre, ‘O’ of the circle. If ∠ADE = 20°, then find the sum of the measures of ∠COD, and ∠ABE., E, A, O, , B, , D, C, , (a) 100°, , (b) 80°, , (c) 60°, , (d) 50°, , Direction for questions 12 to 15: Match the statements of Column A with those of, Column B., Column A, , Column B, , 12. The point of concurrence of perpendicular bisectors of the sides of a triangle, , (a) In centre, , 13. The point of concurrence of altitudes of a triangle, , (b) Orthocentre, , 14. The point of concurrence of medians of the triangle (c) Circumcentre, 15. The point of concurrence of internal angle bisectors (d) Centroid, of a triangle, (e) Excentre, , M01_IIT-FOUNDATION_XXXX_CH01.indd 5, , 4/11/2018 4:33:49 PM

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1.6, , Chapter 1 Geometry and Symmetry, , Assessment Test III, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. The following are the steps involved in proving the above theorem which, states that the opposite angles of a cyclic quadrilateral are supplementary., Arrange them in a sequential order., D, , O, , A, , C, , B, , (A) ∠AOC + reflex ∠AOC = 360°, (B) ∠B + ∠D = 180°, (C) Angle subtended by arc ADC or AB at the centre of the circle is double, the angle subtended by the arc ABC at any point on the remaining part, of the circle., (D) ABCD is a cyclic quadrilateral., (a) DBAC, (b) ABCD, (c) DCAB, (d) ACDB, 2. In the given figure, LM and PQ are parallel to each other and AB is the transversal. If ∠MCD = (3x)° and ∠QDC = (2x + 50)°, then value of x is _________., A, , L, , C, , M, (3x)°, , (2x + 50)°, P, , Q, , D, B, , (a) 26, , M01_IIT-FOUNDATION_XXXX_CH01.indd 6, , (b) 36, , (c) 40, , (d) 50, , 4/11/2018 4:33:51 PM

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Assessment Test III, , 1.7, , Space for rough work, , 3. In the given figure, ABCD . What will be the value of y?, C, x, A, 40°, , D, , y, E, , 2x, , B, , (a) 30°, , (b) 40°, , (c) 60°, , 4. Find the number of diagonals of a 12-sided polygon., (a) 48, (b) 42, (c) 54, , (d) 90°, , (d) 60, , 5. Find the sum of the interior angles of a polygon of 1947 sides., (a) 350, 100°, (b) 350, 460°, (c) 700, 200, (d) 350, 280°, 6. An equilateral triangle has _________ line(s) symmetric., (a) 1, (b) 2, (c) 3, (d) infinite, 7. Which of the following is/are point symmetry?, (a), , (b), , (c), , (d) All of these, , 8. AB and CD are two equal chords of a circle drawn on either side of the, centre ‘O’ of the circle. Both the chords are produced to meet at a point K. If, the radius of the circle is 17 cm, AB = 16 cm, and OK = 25 cm, then find the, length of KD., (a) 8 cm, (b) 12 cm, (c) 16 cm, (d) 20 cm, , M01_IIT-FOUNDATION_XXXX_CH01.indd 7, , 4/11/2018 4:33:57 PM

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1.8, , Chapter 1 Geometry and Symmetry, , 9. In the given figure, A and B are the centres of two intersecting circles and, PQ is their common chord. ∠APB = 70° and ∠PAQ = 80°. Find the measure, of ∠PBQ., , Space for rough work, , P, A, , B, , Q, , (a) 120°, , (b) 130°, , (c) 140°, , (d) 150°, , 10. In the figure, ∠BAE = ∠ECD and ∠EAD = ∠ECB. If ∠ABC = 60° and ∠ADC, = 100°, then find the measure of ∠AEC., B, 60°, E, D, 100°, A, , (a) 65°, , C, , (b) 70°, , (c) 80°, , (d) 90°, , 11. In the given figure, ‘O’ is the centre of the circle and ∠AOC = 160°. What is, the measure of ∠ABC?, B, C, , 160°, A, , O, , (a) 60°, (c) 90°, , (b) 80°, (d) All of these, , 12. In the given figure, ABCD is a parallelogram. What are the values of x° and y°?, C, , D, x°, , y° + 40°, , x° − 20°, A, , M01_IIT-FOUNDATION_XXXX_CH01.indd 8, , B, , 4/11/2018 4:34:02 PM

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Assessment Test III, (a) 50°, 90°, (c) 100°, 40°, , (b) 110°, 30°, (d) 80°, 60°, , 1.9, , Space for rough work, , 13. An equilateral triangle ABC is constructed on the diagonal of a square ADCE, as shown in the figure. What is the measure of ∠ADB?, B, , A, , E, , (a) 120°, , (b) 135°, , D, , C, , (c) 150°, , (d) 165°, , 14. Which of the following is not an exterior angle of a regular polygon?, (a) 24°, (b) 20°, (c) 12°, (d) 16°, 15. Match the statements of Column A with those of Column B., Column A, , Column B, , (i) Infinite lines of symmetry, , (A) Square, , (ii) Four lines of symmetry, , (B) Isosceles triangle, , (iii) Two lines of symmetry, , (C) Rhombus, , (iv) One line of symmetry, , (D) Circle, , (a), (b), (c), (d), , (i) → (D),, (i) → (D),, (i) → (D),, (i) → (D),, , M01_IIT-FOUNDATION_XXXX_CH01.indd 9, , (ii) → (C),, (ii) → (A),, (ii) → (C),, (ii) → (A),, , (iii) → (A), (iv) → (B), (iii) → (B), (iv) → (C), (iii) → (B),  (iv) → (A), (iii) → (C), (iv) → (B), , 4/11/2018 4:34:03 PM

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1.10, , Chapter 1 Geometry and Symmetry, , Assessment Test IV, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. The following are the steps which are involved in proving that an exterior, angle of a cyclic quadrilateral is equal to its interior opposite angle. Arrange, them in a sequential order., D, , A, , B, , X, , C, , (A) ∠BAD + ∠BCD = ∠BCD + ∠DCX, (B) , ∠BAD + ∠BCD = 180° (∴ the opposite angles of a cyclic quadrilateral are, supplementary), (C) ∠BAD = ∠DCX, (D) ∠BCD + ∠DCX = 180° (linear pair), (a) BDAC, (b) ABDC, (c) CABD, (d) DBCA, 2. In the given figure AB and CD are parallel to each other and MN is the transversal. If ∠MEA = (4x – 60)° and ∠DFN = (2x + 20)°, then what is the value, of x?, M, (4x − 60)°, , A, , B, , E, F, , C, , (2x + 20)°, , D, , N, , (a) 30, , (b) 60, , (c) 20, , (d) 40, , 3. In the given figure, AB EF . Find the value of y., C, E, , 3x, , F, y, , 120°, 3x, B, , A, , (a) 10°, , M01_IIT-FOUNDATION_XXXX_CH01.indd 10, , (b) 30°, , (c) 60°, , (d) 20°, , 4/11/2018 4:34:06 PM

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Assessment Test IV, 4. Find the number of diagonals of a 21-sided polygon., (a) 168, (b) 189, (c) 210, , (d) 231, , 1.11, , Space for rough work, , 5. If the sum of the interior angles of an n-sided polygon is 362, 340°, then n =, ______., (a) 2013, (b) 2014, (c) 2015, (d) 2016, 6. A parallelogram has _______ line(s) of symmetry., (a) 0, (b) 1, (c) 2, , (d) 3, , 7. Which of the following is/are not line symmetric?, (a), , (b), , (c), , (d) All of these, , 8. PQ and RS are two equal chords of a circle drawn on either side of the centre, ‘O’ of the circle. Both the chords are produced to meet at a point K. Find the, radius of the circle, if RS = 48 cm, 2QK = PQ, and OK = 50 cm., (a) 2 163 cm, (b) 2 173 cm, (c) 2 193 cm, (d) 2 203 cm, 9. In the given figure, X and Y are the centres of two intersecting circles and, LM is their common chord. The measures of ∠XLM = 30° and ∠YLM = 45°., Find the measure of ∠XMY., L, , Y, , X, , M, , (a) 75°, , M01_IIT-FOUNDATION_XXXX_CH01.indd 11, , (b) 105°, , (c) 115°, , (d) 125°, , 4/11/2018 4:34:11 PM

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1.12, , Chapter 1 Geometry and Symmetry, , 10. In the figure, ∠MLO = ∠ONP and ∠MNO = ∠OLP. If ∠LON = 120° and, ∠LPN = 150°, then find the measure of ∠LMN., , Space for rough work, , M, O, P, , L, , N, , (a) 90°, , (b) 100°, , (c) 70°, , (d) 80°, , 11. In the given figure, ‘O’ is the centre of the circle, ∠OPQ = 30° and ∠ORQ = 40°., What is the measure of reflex ∠POR?, , P, , R, O, , Q, , (a) 200°, , (b) 220°, , (c) 240°, , (d) 210°, , 12. In the given figure, PQRS is a trapezium. PQ||RS and PS = QR. What are the, values of x° and y°?, S, x° − 30°, , R, y° + 40°, , x° + 20°, P, , (a) 85°, 35°, (c) 115°, 40°, , Q, , (b) 95°, 25°, (d) 145°, 70°, , 13. An equilateral triangle ABE is constructed on a side of a square ABCD. What, is the measure of ∠AED?, , M01_IIT-FOUNDATION_XXXX_CH01.indd 12, , 4/11/2018 4:34:13 PM

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Assessment Test IV, B, , A, , E, , (a) 15°, , (b) 20°, , 1.13, , Space for rough work, , D, , C, , (c) 30°, , (d) 35°, , 14. Which of the following is an exterior angle of a regular polygon?, (a) 100°, (b) 14°, (c) 45°, (d) 25°, 15. Match the values of Column A with those of Column B., Column A, (Number of Independent, Measures Required), , Column B, (To Construct a, Quadrilateral), ,     (i) Two, , (A) Square, ,   (ii) One, , (B) Trapezium, , (iii) Three, , (C) Rectangle, , (iv) Four, , (D) Parallelogram, , (a), (b), (c), (d), , (i) → (C); (ii) → (A);, (i) → (C); (ii) → (A);, (i) → (D);    (ii) → (A);, (i) → (B); (ii) → (A);, , M01_IIT-FOUNDATION_XXXX_CH01.indd 13, , (iii) → (B); (iv) → (D), (iii) → (D);   (iv) → (B), (iii) → (B);  (iv) → (C), (iii) → (D);    (iv) → (C), , 4/11/2018 4:34:13 PM

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1.14, , Chapter 1 Geometry and Symmetry, , Answer Keys, Assessment Test I, 1. (c), 11. (b), , 2. (b), 12. (a), , 3. (d), 13. (d), , 4. (d), 14. (b), , 5. (a), 15. (c), , 6. (c), , 7. (a), , 8. (d), , 9. (c), , 10. (d), , 4. (b), 14. (d), , 5. (d), 15. (a), , 6. (d), , 7. (a), , 8. (d), , 9. (b), , 10. (a), , 4. (c), 14. (d), , 5. (a), 15. (d), , 6. (c), , 7. (d), , 8. (b), , 9. (c), , 10. (c), , 4. (b), 14. (c), , 5. (c), 15. (b), , 6. (a), , 7. (c), , 8. (c), , 9. (a), , 10. (a), , Assessment Test II, 1. (d), 11. (c), , 2. (b), 12. (c), , 3. (b), 13. (b), , Assessment Test III, 1. (c), 11. (d), , 2. (a), 12. (c), , 3. (c), 13. (b), , Assessment Test IV, 1. (a), 11. (b), , 2. (d), 12. (b), , M01_IIT-FOUNDATION_XXXX_CH01.indd 14, , 3. (b), 13. (a), , 4/11/2018 4:34:13 PM

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Linear Equations and, Inequations, , 2, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapter - Linear Equations and Inequations;, pp. 12.1−12.29, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. The following are the steps involved in solving the simultaneous linear, equations 3x + 5y = 11 and 2x + 3y = 7 by elimination method. Arrange them, is sequential order., (A) We get y = 22 − 21 = 1., (B) Subtract Eq. (2) × 3 from Eq. (1) × 2., (C) 3x + 5y = 11 be the Eq. (1) and 2x + 3y = 7 be the Eq. (2)., (D) Multiply Eq. (1) by 2 and Eq. (2) by 3., (E) Substitute y = 1 in Eq. (1) and we get x = 2., (a) CBADE, (b) CBADE, (c) CDBAE, (d) DCABE, 2. Find the value of k, if the given equations 2x + 3y = 8 and 6x + ky = 24 have, infinite solutions., (a) 3, (b) 6, (c) 9, (d) 12, 3. If 201x + 102y = 504 and 102x + 201y = 405, then what will be the value of, x − y?, (a) 99, (b) 11, (c) 1, (d) 0, 3x − 2 7 8x + 1 1, + =, − , then calculate the value of x., 2, 8, 6, 8, (a) 1, (b) 3, (c) 4, , 4. If, , 5. If, , (d) 5, , 3 x 4  9x 7 , 2, 3x , + , −  − 1 =  1 +  , then x = ____., 2 7  4 8, 7, 8, , (a), , 1, 3, , (b), , 3, 2, , (c), , 2, 3, , (d) 3, , 6. Find out the number of positive integral solutions of the equation, 3 x − 8 3 x + 10, <, ., 2, 5, (a) 2, , M02_IIT-FOUNDATION_XXXX_CH02.indd 1, , (b) 3, , (c) 5, , (d) 6, , 4/11/2018 4:34:19 PM

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2.2, , Chapter 2 Linear Equations and Inequations, , 7. If, , 3 5 2, 4 3 5, − = and + = , then find the value of x + y., x y 3, x y 2, , (a) 6, , (b) 8, , (c) 4, , Space for rough work, (d) −2, , 8. The sum of the digits of a two-digit number is 2 more than the unit digit., The number formed by reversing the digits exceeds the original number by, 7 times the unit digit. Calculate the product of the digits., (a) 28, (b) 18, (c) 20, (d) 12, 9. There are 60 questions in a test. Each correct answer fetches 2 marks and, 1, for each wrong answer, mark is deducted. A student got 80 marks and, 2, he attempted all the questions. Find the number of questions answered, wrongly., (a) 32, (b) 16, (c) 18, (d) 8, 10. The sum of 3/4th of A’s salary and 5/3rd of B’s salary is `16,000. The difference of their salaries is `2000. If B’s salary is less than that of A, then what is, the salary of B?, (a) `8000, (b) `2000, (c) `6000, (d) `12,000, 11. Ashok had a total of `60 in the form of `5 coins, `2 coins and `1 coins. The, number of `2 coins is twice that of `5 coins and the number of `1 coins is, thrice that of `5 coins. Find the number of `1 coins with him., (a) 15, (b) 5, (c) 10, (d) 20, Direction for questions from 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, , 12. 3x − 4y + 1 = 0, , (a)  (5, 2), , 8 7 15, + =, x y 2, , (b)  (4, 6), , 13., , 2, x − 5y = 1, 3, 15. 4 x − 3 y = 9, 2, 14., , (c)   (3, 2), (d) (2, 2),   (e)  (5, 4),   (f)     (9, 1), , M02_IIT-FOUNDATION_XXXX_CH02.indd 2, , 4/11/2018 4:34:22 PM

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Assessment Test II, , Assessment Test II, , Time: 30 min., , 2.3, , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. The following are the steps involved in solving the linear equations 3x + y = 17, and 5x − 3y = 19 by substitution method. Find the sequential order of the, steps., (A) Find the value of y in terms of x from Eq. (1) i.e., y = 17 − 3x, (B) Substitute the value of y in Eq. (2), i.e., 5x − 3(17 − 3x) = 19, (C) 3x + y = 17 be the Eq. (1) and 5x − 3y = 19 be the Eq. (2), (D) We get the value of x, i.e., 14x = 19 + 51 ⇒ x = 5, (E) Substituting the value of x, we get the value of y, i.e., y = 17 − 15 = 2, (a) ABCDE, (b) CBADE, (c) CABDE, (d) CABED, 2. If 5x − 10y = 8 and 7x − ky = 10 have no common solution, then find the value, of k., (a) 8, (b) 10, (c) 12, (d) 14, 3. What will be the value of x − y, if 111x + 212y = 434 and 212x + 111y = 535?, (a) 0, (b) 1, (c) −1, (d) 101, 2x, 4, 5,  x 7, + 3  −  + 1 = ( x − 1) + , then x = _____., , , 5, 2 3, 5, 2, (a) 7, (b) 2, (c) 3, , 4. If, , 4+x 7 3−x 5, + =, + , then solve for x., 3, 4, 2, 4, −2, 3, (a), (b), 5, 5, , (d) 4, , 5. If, , (c), , 5, 3, , 6. The number of non-negative integers that satisfy 4 x −, (a) 4, , (b) 5, , (c) 3, , (d) −, , 5, 3, , 5, ≤ x + 7 is _______., 2, (d) 6, , 5 7, 6 9 15, + = 6 and + = , then find the value of x + y., x y, x y 2, (a) 4, (b) 3, (c) 8, , 7. If, , (d) 7, , 8. The sum of the digits of a two-digit number is eight. If eighteen is added, to the number, then the digits of the number get interchanged. Find the, number., (a) 26, (b) 17, (c) 35, (d) 62, 9. There are 48 questions in a test. Each correct answer fetches 1 mark and for, each wrong answer, 1/2 mark is deducted. A student got 21 marks and he, , M02_IIT-FOUNDATION_XXXX_CH02.indd 3, , 4/11/2018 4:34:25 PM

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2.4, , Chapter 2 Linear Equations and Inequations, attempted all the questions. What will be the number of questions answered, correctly?, (a) 28, (b) 30, (c) 12, (d) 18, , Space for rough work, , 10. The sum of 2/3rd of A’s salary and 7/3rd of B’s salary is `9000. The sum, of their salaries is `6000. Find the difference in their salaries (in `)., (a) 6000, (b) 0, (c) 12,000, (d) 8000, 11. Anand had a total of `100 in the form of `5 coins, `2 coins and `1 coins. The, number of `5 coins is 3/2 times that of `2 coins and the number of `1 coins, is half that of `2 coins. Find the number of `5 coins with him., (a) 15, , (b) 10, , (c) 5, , (d) 20, , Direction for questions from 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, , 12., , −1, > 0 ,( x ≠ 3) (a) x > 2, x−3, , 13., , 15, < 5,( x ≠ 1) (b) x > 1, x −1, , 14. 8x − 8 > 5x − 2, ,   (c) x > 4, , 15. 6x + 6 < 7x + 5, , (d) x < 3, (e) x < 1, , M02_IIT-FOUNDATION_XXXX_CH02.indd 4, , 4/11/2018 4:34:26 PM

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Assessment Test III, , Assessment Test III, , Time: 30 min., , 2.5, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. The following are the steps involved in solving the following linear equations by elimination method:, 4x − 3y = 5 and 5x + 2y = 12., Arrange them in a sequential order., (A) (4x − 3y) × 2 = 5 × 2 ⇒ 8x − 6y = 10, (5x + 2y) × 3 = 12 × 3 ⇒ 15x + 6y = 36., (B) Multiply 4x − 3y = 5 by 2 and 5x + 2y = 12 by 3., (C) 8x − 6y = 10 should be added to 15x + 6y = 36., (D) Substitute the value of x in 4x − 3y = 5, then find the value of y., (a) ABCD, (b) BACD, (c) BADC, (d) DABC, 2. If 5x + 2y = 13 and 2x + 5y = 1, then the value of x + y is ______., (a) 1, (b) 2, (c) 3, (d) 5, 4, 9, =, , then solve for the value of y is ______., 5 y − 8 5 y − 11, 2, (a) 1, (b) 5, (c) 6, , 3. If, , (d) 2, , 4. If 2(3x − 5) + 6(x − 7) = 7(x + 5) − 15, then what will be the value of x?, (a) 15.2, (b) 12.7, (c) 14.4, (d) 7, 5. If 3x + y = 10, x − 2y = 1, and 3x − 2y + 5z = 17, then the value of z is _____., (a) 2, (b) 3, (c) 4, (d) 5, 6. If 3x − 5 ≤ 5x − 9, then x is ______., (a) x ≥ 2, (b) x ≥ 5, , (c) x ≤ 2, , −7, 5x, < 1, where x ≠, ., 5x + 7, 5, −7, 7, (b) x <, (c) x <, 5, 5, , (d) x ≤ 3, , 7. Find the solution set of, (a) x >, , 7, 5, , 8. Find the solution of the inequation x + 7 > − 3 + 7x., 5, −5, 5, (a) x <, (b) x >, (c) x >, 3, 3, 3, , (d) x >, , −7, 5, , (d) x <, , −5, 3, , 9. The difference between the present ages of a person and his son is 28 years., Ten years hence their ages will be in the ratio 3:1. The sum of their present, ages is ______ years., , M02_IIT-FOUNDATION_XXXX_CH02.indd 5, , 4/11/2018 4:34:30 PM

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2.6, , Chapter 2 Linear Equations and Inequations, (a) 32, , (b) 36, , (c) 40, , (d) 48, , Space for rough work, , 10. If the system of equations 3x − 2 (k − 1) y = 20 and 5x − 3y = 25 is consistent,, then k ≠ _______., 15, 19, 9, 12, (a), (b), (c), (d), 19, 10, 10, 17, 11. A test has 100 questions, each correct answer fetches 1 mark and each wrong, 1, answer loses mark. A candidate wrote the test, attempted all the ques2, tions, and scored 61 marks. How many questions did he answer correctly?, (a) 75, (b) 72, (c) 74, (d) 80, 12. Which of the following points is not in the region y ≤ x + 2?, (a) (0, 0), (b) (1, 5), (c) (4, 2), , (d) (6, 3), , 13. Which of the following pairs of equations have no solution?, (a) 2x + 3y = 5; 3x − 2y = 7, (b) 5x − 2y = 9; 10x − 4y = 18, (c) 8x + 4y = 7; 4x + 2y = 3, (d) 3x + 5y = 9; 8x + 6y = 11, 14. Which of the following equations passes through the origin?, (a) 3x + 5 = 9, (b) 4x − 7y = 10, (d) x + y − 9 = 0, (c) 2x + 3y = 7y, Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, , Column B, ,   (i) (5, 0), , (A) A point in Q2, , (ii) (0, 5), , (B)    A point in Q4, , (iii) (10, −9), , (C)  A point on the x-axis, , (iv) (−4, 5), , (D) A point on the y-axis, , (a) (i) → (A); (ii) → (D); (iii) → (C); (iv) → (B), (b) (i) → (C); (ii) → (D); (iii) → (A); (iv) → (B), (c) (i) → (C); (ii) → (D); (iii) → (B);   (iv) → (A), (d) (i) → (D); (ii) → (C); (iii) → (A); (iv) → (B), , M02_IIT-FOUNDATION_XXXX_CH02.indd 6, , 4/11/2018 4:34:32 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 2.7, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. Solve the following linear equation by substitution method: 3x − 2y = 9 and, x + 3y = 14., The following are the steps involved in solving the above problem. Arrange, them in sequential order., (A) Substitute the value of x in terms of y in the equation 3x − 2y = 9., (B) x + 3y = 14 ⇒ x = 14 − 3y, (C) −11y = −33 ⇒ y = 3, (D) 3(14 − 3y) − 2y = 9, (E) x = 14 − 3(3) = 5, (a) ABCDE, (b) ABDEC, (c) BAEDC, (d) BADCE, 2. If 4x + 3y = 17 and 3x + 4y = 18, then the value of x − y is _______., (a) −1, (b) −2, (c) 4, (d) −5, 3. If, , 3, 7, =, , then the value of x is ______., 4 x − 5 2x + 3, , (a) 2, , (b) 3, , (c) 4, , (d) 5, , 4. If 5 (x − 2) + 7 (x − 4) = 4 (x + 5) − 14, then the value of x is ______., (a) 7.5, (b) 3.5, (c) 4.5, (d) 5.5, 5. If 4x + 5y = 2, 3x − 2y = 13, and 7x + 6y + 3z = 12, then the value of z is ______., (a) 1, (b) 2, (c) 3, (d) 5, 6. If 2x + 7 ≤ 4x − 9, then x = _______., (b) x ≤ 7, (a)x ≤ 5, 7. The solution set of, (a) x < − 4, , (c) x ≥ 8, , 5x, 4, < 5, where x ≠ − is ______., 4+x, 5, 4, (b) x >, (c) x > −4, 5, , 8. Find the solution of the inequation 3x − 5 ≤ 5x − 8., 5, 3, 5, (a) x ≤, (b) x ≥, (c) x ≤ −, 2, 2, 2, , (d) x ≥ 9, , (d) 4 <, , −4, 5, , (d) x ≥, , −3, 2, , 9. The sum of the present ages of a person and his daughter is 42 years. Six, years hence the ratio of their ages will be 7:2. The difference of their present, ages is______ years., (a) 38, (b) 20, (c) 24, (d) 30, , M02_IIT-FOUNDATION_XXXX_CH02.indd 7, , 4/11/2018 4:34:35 PM

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2.8, , Chapter 2 Linear Equations and Inequations, , 10. For what values of k is the system of equations 5x − 2ky = 40 and 3x − 4y = 24, consistent?, 10, 15, 17, 7, (a), (b), (c), (d), 3, 7, 2, 3, , Space for rough work, , 11. A test has 120 questions. Each correct answer fetches 2 marks and each wrong, answer loses 1 mark. A candidate wrote the test, attempted all the questions, and scored 81 marks. How many questions did he answer wrongly?, (a) 56, (b) 55, (c) 54, (d) 53, 12. Which of the following points lies in the region x − 2y > 5?, (a) (1, 1), (b) (3, 2), (c) (4, −1), , (d) (3, −1), , 13. Which of the following pairs of equations have unique solution?, (a) 2x + 3y = 7; 4x + 6y = 14, (b) 12x + 6y = 5; 6x + 3y = 2.5, (c) x + y = 7; x + y = 15, (d) 3x + 2y = 15; 5x − 4y = 12, 14. Which of the following does not pass through the origin?, (a) x + 4 = 2y, (b) 7y = 2x, (c) x + 2y = 8y, (d) 3x + y = 4x + 6y, Direction for question 15: Match the values of Column A with those of Column B., 15., Column B, , Column A, (a1 x + b1 y + c1 = 0;, a2 x + b2 y + c2 = 0),   (i),    (ii), (iii), , a1, a2, a1, a2, a1, a2, , =, ≠, =, , b1, b2, , ≠, , c1, c2, , b1, ,   (B) Dependent equations, , b2, b1, b2, , (A) Independent equations, , =, , c1, c2, , (iv) a21 + b12 = 0 and a22 + b22 = 0, (a) (i) → (A);, , (C) Inconsistent equations, (D) Not the linear equations, , (ii) → (B);   (iii) → (C);, , (iv) → (D), (b) (i) → (C);  , (ii) → (A); (iii) → (B);  (iv) → (D), (c) (i) → (D); (ii) → (A); (iii) → (C); (iv) → (B), (d) (i) → (B);   (ii) → (C); (iii) → (D); (iv) → (A), , M02_IIT-FOUNDATION_XXXX_CH02.indd 8, , 4/11/2018 4:34:37 PM

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Answer Keys, , 2.9, , Answer Keys, Assessment Test I, 1. (c), 11. (a), , 2. (c), 12. (e), , 3. (c), 13. (d), , 4. (a), 14. (f), , 5. (c), 15. (c), , 6. (d), , 7. (b), , 8. (b), , 9. (b), , 10. (c), , 4. (a), 14. (a), , 5. (a), 15. (b), , 6. (a), , 7. (a), , 8. (c), , 9. (b), , 10. (b), , 4. (c), 14. (c), , 5. (a), 15. (c), , 6. (a), , 7. (d), , 8. (a), , 9. (b), , 10. (b), , 4. (d), 14. (a), , 5. (a), 15. (b), , 6. (c), , 7. (c), , 8. (b), , 9. (d), , 10. (a), , Assessment Test II, 1. (c), 11. (a), , 2. (d), 12. (d), , 3. (b), 13. (c), , Assessment Test III, 1. (b), 11. (c), , 2. (b), 12. (b), , 3. (d), 13. (c), , Assessment Test IV, 1. (d), 11. (d), , 2. (a), 12. (c), , M02_IIT-FOUNDATION_XXXX_CH02.indd 9, , 3. (a), 13. (d), , 4/11/2018 4:34:37 PM

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Mensuration, , 3, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapter - Mensuration; pp. 17.1−17.20, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. The following are the steps involved in finding the area of a triangle ABC of, sides a, b, and c. Arrange them in sequential order., (A), (B), (C), (D), (a), (c), , Calculate the difference of each side with the semi-perimeter., Find the semi perimeter., Find the product of the differences and semi perimeter., Find the square root of the product., ABCD, (b) BCDA, BACD, (d) CADB, , 2. Find the area of an isosceles triangle whose base is 10 cm and the length of, each equal side is 8 cm., (a) 10 39 sq.cm., , (b) 5 11 sq.cm., , (c) 5 39 sq.cm., , (d) 6 13 sq.cm., , 3. Calculate the area of a square whose diagonal is equal to 32 × 7 × 24.5 cm, (in sq. cm)., (a) 34 × 72 × 28, (c) 34 × 72 × 29, , (b) 34 × 72 × 23.5, (d) 34 × 72 × 225, , 4. At the most, how many cakes of soap with dimensions 8 cm × 6 cm × 4 cm, can be placed in a wooden box of inner measures 28 cm × 16 cm × 12 cm?, (a) 35, , (b) 24, , (c) 28, , (d) 36, , 5. Find the area of a circle (in sq. cm) whose circumference is equal to the, perimeter of a parallelogram of sides 65 cm and 67 cm., (a) 5544, (c) 5454, , (b) 4545, (d) 4455, , 6. What is the radius of a sector of central angle 120° and the length of the arc, 924 cm (in cm)?, (a) 441, (c) 49, , M03_IIT-FOUNDATION_XXXX_CH03.indd 1, , (b) 196, (d) 121, , 4/11/2018 4:34:42 PM

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3.2, , Chapter 3 Mensuration, , 7. Find the volume of a cylinder whose base area is 154 sq. cm and the curved, surface area is 440 sq. cm (in cu cm)., (a) 1548, (c) 1540, , Space for rough work, , (b) 1234, (d) 1430, , 8. The volume of a sphere is 38,808 cm3. What is the surface area of the sphere, (in sq. cm)?, (a) 4928, (c) 5544, , (b) 5064, (d) 5644, , 9. Find the total surface area of a cuboid which can be made from 4 cubes of, edge 4 cm each joined end to end (in sq. cm)., (a) 284, (c) 196, , (b) 144, (d) 288, , 10. What is the total surface area of a solid hemisphere whose curved surface, area is 416 sq.cm?, (a) 618 cm2, (c) 624 cm2, , (b) 636 cm2, (d) 636 cm2, , 11. The product of the surface areas (in sq. cm) of three mutually adjacent faces, of a cuboid is 576. Calculate its volume (in cu cm)., (a) 24, (c) 72, , (b) 48, (d) 96, , Direction for questions 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, (Area of the shaded, region/CSA of solids), , 12., , (a) p Rr, Radius of outer circle = R, Radius of inner circle = r, 13., (b) 2p Rr, Length of the arc = R, Radius of the circle = r, , M03_IIT-FOUNDATION_XXXX_CH03.indd 2, , 4/11/2018 4:34:43 PM

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Assessment Test I, Column A, , Column B, (Area of the shaded, region/CSA of solids), , 3.3, , Space for rough work, , 14., , (c) p (R + r) (R - r), Slant height of the cone = R, Radius of the base = r, 15., , (d), , 1, p rR, 3, , (e), , Rr, 2, , Height of the cylinder = R, Radius of the base = r, , M03_IIT-FOUNDATION_XXXX_CH03.indd 3, , 4/11/2018 4:34:45 PM

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3.4, , Chapter 3 Mensuration, , Assessment Test II, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. The length of an arc of a sector of a circle is 44 cm. The following are the, steps involved in finding the area of the circle, if the central angle is 120°., Arrange them in a sequential order., x°, × 2p r, 360, (B) Recall the formula for length of an arc of a sector., (C) Recall the formula for the area of a circle., (D) Substitute the value of r in p r2., , (A) Substitute l = 44, x° = 120° in l =, , (E) Find the value of the radius of the sector., (a) CBAED, (b) BAECD, (c) BACED, (d) BACDE, 2. One of the equal sides of an isosceles triangle is 13 cm and length of the altitude drawn to unequal side is 5 cm. What is the area of the triangle?, (a) 50 cm2, (c) 60 cm2, , (b) 38 cm2, (d) 28 cm2, , 3. The area of a square is 24 × 36 × 510 sq. cm. Find the length of its diagonal, (in cm)., (a) 22 × 33 × 56· 2, (c) 22 × 33 × 55· 2, , (b) 22 × 34 × 55· 2, (d) 23 × 33 × 55· 2, , 4. How many solid lead balls of diameter 4 cm each can be made from a solid, lead ball of radius 8 cm?, (a) 64, (c) 8, , (b) 32, (d) 26, , 5. Find the circumference of a circle whose area in sq. cm is equal to its, circumference., (a) p cm, (c) 4p cm, , (b) 2p cm, (d) 8p cm, , 6. Three sectors are cut from a circle of radius 12 cm. If the length of the arc of, each sector is 22 cm, then find the angle of the sector formed with the, remainder of the circle., (a) 60°, (c) 15°, , (b) 45°, (d) 35°, , 7. The base radius of a cylinder is 20 cm and its height is 2016 cm. Find its, curved surface area in sq. cm., (a) 253,440, (c) 332,550, , M03_IIT-FOUNDATION_XXXX_CH03.indd 4, , (b) 234,550, (d) 372,440, , 4/11/2018 4:34:47 PM

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Assessment Test II, 8. If the surface area of a sphere is 5544 sq. cm, then what will be the volume of, the sphere (in cm3)?, (a) 38,808, (c) 310,464, , 3.5, , Space for rough work, , (b) 47,704, (d) 55,088, , 9. Find the lateral surface area of a cube formed when 27 cubes of edge 5 cm, each are stacked (in sq. cm)., (a) 400, (c) 900, , (b) 600, (d) 1350, , 10. What will be the total surface area of a solid hemisphere whose area of the, circular face is 670 sq. cm (in sq. cm)?, (a) 2010, (c) 2013, , (b) 2016, (d) 2014, , 11. The product of the surface areas (in sq. cm) of three mutually adjacent faces, of a cuboid is 3600. Find its volume (in cu cm)., (a) 30, (c) 75, , (b) 45, (d) 60, , Direction for questions 12 to 15: Match the statements of Column A with values of, Column B., Column A, 12. Volume of a cylinder whose height, is equal to its base diameter (2R)., 13. Volume of a cone whose height is, , Column B, (a) p R3, (b) 2p R3, , equal to 3 times its base radius (R)., 14. Volume of a sphere whose radius, 3, is R., 2, 15. Volume of a hemisphere whose, radius is 3R., ,   (c), , 9, p R3, 2, , (d) 10p R3, (e) 18p R3, , M03_IIT-FOUNDATION_XXXX_CH03.indd 5, , 4/11/2018 4:34:48 PM

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3.6, , Chapter 3 Mensuration, , Assessment Test III, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. The following are the steps involved in finding the area of an equilateral, triangle ABC whose side is ‘a’ units. Arrange them in a sequential order., 1, (A) Area of a triangle = × base × height., 2, (B) Let AD be the altitude ⇒ D is the midpoint of BC., 1, 3, ×a×, a., 2, 2, (D) In DADB, AD2 = AB2 - BD2, (C) Area of the triangle =, , 2, ,  a, ⇒ h2 = a2 −   .,  2, (a) BDAC, (c) DBAC, , (b) BCDA, (d) BADC, , 2. Find the area of a regular hexagon of perimeter 36 cm., (a) 36 3 cm2, (c) 72 3 cm2, , (b) 54 3 cm2, (d) 12 3 cm2, 12 6, 3. If the area of a right isosceles triangle is 54 sq. cm, then find its hypotenuse, 3 36, 12 6, (in cm)., 12, 3 36, (a) 12, (b) 63 33, (c) 63 33, (d) 66 63, 4. The angle, of the corresponding circle, 6 6, 66 63 of a sector is 84° and the circumference, is 330 cm. Calculate the area of the sector., 6 6, (a) 2021.25 sq. cm, (b) 2121.5 sq. cm, (c) 2221.25 sq. cm, (d) 2112.5 sq. cm, 5. The area of a trapezium is 49 sq. cm. If one of the parallel sides is 4 cm and, the distance between the parallel sides is 7 cm, then what will be the length, of the other parallel side?, (a) 5 cm, (c) 10 cm, , (b) 7 cm, (d) 12 cm, , 6. A welder made a frame of an equilateral triangular prism with iron bars as, edges. If the side of the base is 60 cm and the height is 1.95 m, then find the, total length of the iron bar (omit the overlapping)., (a) 8.5 m, (c) 8.95 m, , M03_IIT-FOUNDATION_XXXX_CH03.indd 6, , (b) 9.25 m, (d) 9.45 m, , 4/11/2018 4:34:50 PM

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Assessment Test III, 7. The surface area of a hemispherical shell of inner diameter is 14 cm and, thickness 7 cm is ______ sq. cm (approximately)., (a) 624p, (c) 648p, , 3.7, , Space for rough work, , (b) 637p, (d) 652p, , 8. A sector of central angle 120° and radius 42 cm was made into a cone. Find, the height of the cone., (a) 24 2 cm, (c) 12 3 cm, , (b) 28 2 cm, (d) 16 2 cm, , 9. What is the volume of a cylinder having base diameter 6 cm and height is, 14 cm?, (a) 308 cm3, (c) 402 cm3, , (b) 396 cm3, (d) 428 cm3, , 10. ABCD is a rectangle with AB = 14 cm and BC = 7 cm. DP and BQ are the arcs, of the sectors drawn with ‘A’ as centre. Find the area of the shaded region, (in sq. cm)., Q, , D, , C, , 30°, A, , P, , B, , (, , ), , (, , ), , (, , ), , (, , ), , 7, 7, 11 + 21 3, 11 + 21 3, (b), 3, 6, 7, 7, 11 + 12 3, 11 + 12 3, (c), (d), 6, 3, 11. Total surface area of a solid cube is 96 sq. cm. A smaller cube from the corner, of the initial cube is carved out. Find the total surface area of the remainder., (a), , (a) 92 sq. cm, (c) 90 sq. cm, , (b) 93 sq. cm, (d) 96 sq. cm, , 12. ABCD is a parallelogram AB = 6 cm, AD = 8 cm, and ∠DAB = 60°. What will, be the area of the parallelogram (in sq. cm)?, (a) 12 3, , (b) 16 3, , (c) 24 3, , (d) 32 3, , 13. Find the area of an isosceles triangle of sides 5 cm, 5 cm, and 6 cm (in sq. cm)., (a) 9, , (b) 10, , (c) 12, , (d) 16, , 14. A rectangular grassy lawn is 20 m by 16 m. It has a gravel path 1.8 m wide, all around it inside. Calculate the area of the path (in sq. m)., (a) 120.28, (c) 112.42, , M03_IIT-FOUNDATION_XXXX_CH03.indd 7, , (b) 116.64, (d) 108.26, , 4/11/2018 4:34:54 PM

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3.8, , Chapter 3 Mensuration, , 15. Match the statements of Column A with those of Column B., Column A, , Column B, , (i) Area of a rhombus whose, diagonals are equal and each, diagonal is equal to d cm., , (A) Area of a square of side, d cm., , (ii) Area of a rectangle whose, length and breadth are equal, to d cm., ,   (B) Area of a square of diagonal, is d cm., , (iii) Area of a trapezium whose, parallel sides are d, 2d and, distance between them is d., , (C) Area of a parallelogram, whose each of longer sides, , (iv) Area of a quadrilateral whose, diagonal is d and the altitudes, d, drawn on to it are d and, 2, (a), (b), (c), (d), , Space for rough work, , is 3d and distance between, d, them is ., 4, (D) Area of a triangle whose, base 3d and corresponding, height is d., , (i) → (A); (ii) → (B);   (iii) → (C);  , (iv) → (D), (i) → (B); (ii) → (A); (iii) → (D); (iv) → (C), (i) → (D); (ii) → (C);  , (iii) → (D); (iv) → (A), (i) → (B); (ii) → (D); (iii) → (A); (iv) → (C), , M03_IIT-FOUNDATION_XXXX_CH03.indd 8, , 4/11/2018 4:34:55 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 3.9, , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. The following are the steps involved in finding the area of the triangle, if, ABC is an isosceles triangle in which AB = AC = a units and BC = b units., Arrange them in a sequential order., (A) In DADB, AD2 = AB2 - BD2, b, (B) Draw AD ^ BC ⇒ BD = CD =, 2, 2, , 1, 4 a − b2 , (C) Area of DABC = (b) , , 2 , 2, , (D) AD =, ,  b, a2 −  ,  2, , 2, , (E) Area of triangle ABC =, (a) ABCDE, (c) ABDEC, , 1, × base × height, 2, (b) BADEC, (d) BAEDC, , 2. Find the area of a regular hexagon whose side is 12 cm (in sq. cm)., (a) 36 3, (c) 196 2, , (b) 96 3, (d) 216 3, , 3. If the hypotenuse of a right isosceles triangle is 13 2cm, then what will be, the area of the triangle?, (a) 80 sq. cm, (c) 169 sq. cm, , (b) 84.5 sq. cm, (d) 169 2 sq. cm, , 4. The area of a sector is 770 sq. cm. The arc of a sector of a circle makes an, angle of 72° at the centre of the circle. Find the circumference of the circle., (a) 220 cm, (c) 330 cm, , (b) 110 cm, (d) 55 cm, , 5. In a trapezium, the sum of the lengths of the parallel sides is 32 cm and, the area of the trapezium is 128 sq. cm. Calculate the distance between the, parallel sides., (a) 7 cm, (c) 10 cm, , (b) 8 cm, (d) 12 cm, , 6. The total length of the edges of an equilateral triangular based prism is, 8.64 m. The side of the base is half of the height in length. What will be the, perimeter of the base?, (a) 2.20 m, (c) 2.32 m, , M03_IIT-FOUNDATION_XXXX_CH03.indd 9, , (b) 2.16 m, (d) 2.18 m, , 4/11/2018 4:34:58 PM

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3.10, , Chapter 3 Mensuration, , 7. Find the inner surface area of a hemispherical shell whose outer diameter is, 16.4 cm and the thickness is 1.2 cm., (a) 208 sq. cm, (c) 308 sq. cm, , Space for rough work, , (b) 312 sq. cm, (d) 324 sq. cm, , 1, of the circle is removed, 3, and the remaining is made as a cone. Find the base radius of the cone formed., , 8. From a circle of radius 21 cm, a sector of area of, (a) 21 cm, (c) 14 cm, , (b) 17 cm, (d) 12 cm, , 9. The volume of a cylinder is 2156 cm3. If the height of the cylinder is twice its, base radius, then what will be its height?, (a) 14 cm, (c) 21 cm, , (b) 7 cm, (d) 24 cm, , 10. In the figure, ABCD is a square. A sector is drawn with A as centre. If AB =, 7 3 cm, then find the area of the shaded region (in sq. cm)., D, , C, , A, , B, , (a) 31.5, (c) 42.5, , (b) 32.5, (d) 36.5, , 11. Three identical cubes are joined as shown in the figure. If the total surface, area of each cube is 6 sq. cm, then what will be the total surface area of the, cuboid formed (in sq. cm)?, , (a) 18, , (b) 14, , (c) 16, , (d) 12, , 12. PQRS is a parallelogram. If PS = 10 cm and ∠SPQ = 30° and the area of the, parallelogram is 55 3 sq. cm, then find the length of PQ (in cm)., (a) 11 3, , (b) 12 3, , (c) 9 3, , (d) 15 3, , 13. Area of an isosceles triangle is 48 sq. cm. Each of its equal side is 2 cm less than, its base. If its perimeter is an integer in cm, then calculate the perimeter (in cm)., (a) 32, , M03_IIT-FOUNDATION_XXXX_CH03.indd 10, , (b) 36, , (c) 42, , (d) 24, , 4/11/2018 4:35:02 PM

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Assessment Test IV, 14. A rectangular play ground is 75 m by 60 m. It has a cement path of width 2 m, all around and outside of the ground. Find the perimeter of the order edge, of the path (in m)., (a) 286, , (b) 244, , (c) 320, , 3.11, , Space for rough work, , (d) 275, , 15. Match the statements of Column A with those of Column B., Column A, , Column B, , (i) P,  erimeter of an isosceles triangle whose, base is b cm, each, equal sides is a cm., , (A) Perimeter of a quadrilateral, whose sides are a, 2a, b,, and b., , (ii) Perimeter of a rectangle whose length, is a cm and breadth, is 2b cm., , (B) P,  erimeter of a pentagon, whose sides are a, b, 2a, a,, and b., , (iii) Perimeter of a parallelogram whose, adjacent sides are 2a, cm and b cm., , (C) Perimeter of a hexagon whose sides are, b b a a a a, a, , , , , , , and, 2 2 2 2 2 2, 2, ,  erimeter of an, (iv) P, isosceles trapezium whose parallel, sides are a, 2a, and, non-parallel sides, each is b cm., , (D) Perimeter of a scalene, triangle whose sides are, 2a, 3b, and b., , (a), (b), (c), (d), , (i) → (C);, (i) → (B);, (i) → (C);, (i) → (B);, , M03_IIT-FOUNDATION_XXXX_CH03.indd 11, , (ii) → (B);, (ii) → (C);, (ii) → (D);, (ii) → (C);, , (iii) → (A); (iv) → (D), (iii) → (D); (iv) → (A), (iii) → (B); (iv) → (A), (iii) → (A); (iv) → (D), , 4/11/2018 4:35:03 PM

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3.12, , Chapter 3 Mensuration, , Answer Keys, Assessment Test I, 1. (c), 11. (a), , 2. (c), 12. (c), , 3. (a), 13. (e), , 4. (c), 14. (a), , 5. (a), 15. (b), , 6. (a), , 7. (c), , 8. (c), , 9. (d), , 10. (c), , 4. (a), 14. (c), , 5. (c), 15. (e), , 6. (b), , 7. (a), , 8. (a), , 9. (c), , 10. (a), , 4. (a), 14. (b), , 5. (c), 15. (b), , 6. (d), , 7. (b), , 8. (b), , 9. (b), , 10. (b), , 4. (a), 14. (a), , 5. (b), 15. (c), , 6. (b), , 7. (c), , 8. (c), , 9. (a), , 10. (a), , Assessment Test II, 1. (b), 11. (d), , 2. (c), 12. (b), , 3. (c), 13. (a), , Assessment Test III, 1. (a), 11. (d), , 2. (b), 12. (c), , 3. (d), 13. (c), , Assessment Test IV, 1. (b), 11. (b), , 2. (d), 12. (a), , M03_IIT-FOUNDATION_XXXX_CH03.indd 12, , 3. (b), 13. (a), , 4/11/2018 4:35:03 PM

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4, , Polynomials and, Their LCM and HCF, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapter - Polynomials and their LCM and HCF;, pp. 4.1−4.14, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., 2  4, 2 2, 4,  2, 1. The following are the steps involved in simplifying  x 3 − y 3   x 3 + x 3 y 3 + y 3  ., , , , 2  4, 2 2, 4,  2, 3, 3, 3, 3, 3, 3, Arrange, them, in, sequential, order., x − y  x + x y + y  ., , , , , Space for rough work, , 2, 3, 3, 2   2  2, 2 2,  2,  2   2  2, , , 3, 3, 3, 3, 3, 3, 3, 3, (A) \  x − y   x  + x y +  y  =  x  −  y , ,   ,       , , , (B) Given expression can be written as, 2, 2   2  2, 2 2,  2,  2 , , 3, 3, 3, 3, 3, 3,  x − y   x  + x y +  y  , ,   ,  , , , (C) We have (a − b)(a2 + ab + b2) = a3 − b3, 2, , ×3, , 2, , (D) ⇒ x 3 − y 3, (a) ABCD, (c) BADC, , ×3, , = x2 − y2, (b) DBAC, (d) BCAD, , 2. Factorize: (16x2 + 40xy + 25y2) − (4x2 − 4xy + y2), (a) 4(3x + 2y) (x + 3y), (c) 4(3x − 2y) (x + 3y), , (b) 8(3x + y) (9x + 3y), (d) 4(3x + y) (x − 3y), , 3. Find the LCM of 42x (x − 1)2 (x − 3)2 and 14x2 (x2 − 1) (x − 3)., (a) 42x2 (x2 − 1) (x − 3)2, (c) 14x2 (x2 − 1) (x − 1) (x − 3), , (b) 14x (x − 1)2 (x − 3)2, (d) 42x2 (x2 − 1) (x −1) (x − 3)2, , 4. If x2 + y2 + xy = 50 and x + y = 7, then find the value of, (a) 3, (c) 1, , (b) −3, (d) −1, , x2 + y 2, ., 17 xy, , 5. A = 6x3 − 3x2 − 2x + 1 and B = 2x − 1. If A is divided by B, then the quotient is, Q. Find the value of B + Q., (a) 3x2 − 2x + 1, (c) 3x2 + 2x − 2, , M04_IIT-FOUNDATION_XXXX_CH04.indd 1, , (b) 2x − 1, (d) 3x2 − 2x + 2, , 4/11/2018 4:35:27 PM

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4.2, , 6., , Chapter 4 Polynomials and Their LCM and HCF, , ( a2 − b2 ) ( a − b)2, , Space for rough work, , = _______, , ( a + b )3, , (a), , a2 + b2, a+b, ,  a − b, (b) ,  a + b , , (c), , ( a − b )2, ( a + b )3, , (d), , ( a − b )3, ( a + b )2, , (b), , x−y, x+y, , 7. Solve for:, (a), , ( x + y ) ( x2 + xy + y2 ), , ( x3 − y 3 ), , 2, , ., , (x + y), (x − y), , (c) x − y, , (d) x + y, , 8. If (x + 4) is a factor of 3x2 + 13x + k, then the value of k is ______., (a) 1, (c) 3, , (b) 2, (d) 4, , 9. If the HCF of f(x) = (x − 3) (x2 + x + m) and g(x) = (x − 2) (x2 + x + n) is (x − 3), (x − 2), then m + n = ______., (b) −12, (d) 12, , (a) 6, (c) −18, , 10. The degree of a polynomial f(x) is 3 and the degree of another polynomial, g(x) is 4. The degree of the polynomial [f(x) + g(x)] is ______., (a) 3, (c) 7, , (b) 4, (d) 12, , 11. (3x2 + 2x) (9x4 − 6x3 + 4x2) = _______, (a) 27x6 + 8x2, (c) 27x6 − 8x2, , (b) x3 (27x3 + 8), (d) x3(27x3 − 8), , Direction for questions from 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, , 12., , 4x4−, , (a) Not a polynomial, , 13., , x2, , 6x3, , + 3x + 5, , − 9x − 6, , (b) Quadratic polynomial, , 14. 2 x + 5, 3, 2, , (c) Zero degree polynomial, , 15. 92 − 72, , (d) Linear polynomial, (e) Biquadratic polynomial, , M04_IIT-FOUNDATION_XXXX_CH04.indd 2, , 4/11/2018 4:35:32 PM

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4.4, , Chapter 4 Polynomials and Their LCM and HCF, , 8. If (x + 3) is a factor of x2 − 2x − k, then the value of k is _______., (a) 9, , (b) −7, , (c) 15, , (x2 −, , 9. If the HCF of f(x) = (x + 3), then find the value of p − q., (a) 9, , 4x + p) and g(x) = (x − 3), , (b) 10, , (c) 8, , Space for rough work, (d) 5, , (x2, , + x + q) is x2 − 9,, (d) 7, , 10. The degree of a polynomial p(x) is 2 and that of another polynomial q(x) is 3,, then the degree of p(x) × q(x) is_____., (a) 3, , (b) 2, , (c) 6, , (d) 5, , 11. (4x2 − 1) (16x4 + 4x2 + 1) = _______, (a) 64x6 − 1, (c) (8x3)3 − 1, , (b) (8x3)2 + 1, (d) 64x3 + 1, , Direction for questions from 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, , 12. Polynomial in one, variable, , (a) 100x°, , 13. Not a polynomial, , (b), , 14. Constant polynomial, 15. Zero polynomial, , M04_IIT-FOUNDATION_XXXX_CH04.indd 4, , 3 x2 + 7 x − 1, , 1, x, (d) ax2 + bx + c, where, a=b=c=0, (c) x +, , 4/11/2018 4:35:37 PM

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Assessment Test III, , Assessment Test III, , Time: 30 min., , 4.5, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1, 1, = 5, then find x3 + 3 . The following are the steps involved in solving, x, x, the above problem. Arrange them in sequential order., 1, , 1, (A) x3 + 3 + 3  x +  = 27, , x, x, 1, , (B)  x +  = 3, , x, , 1. If x +, , (C) x3 +, , 1, + 3 × 3 = 27, x3, 3, , 1, , (D)  x +  = 27, , x, 1, (E) x3 + 3 = 18, x, (a) BCADE, (c) BDEAC, 1, 1, 2. If x − = 4, then x2 + 2 = ______., x, x, (a) 16, , (b) 12, , (b) BDACE, (d) BACDE, , (c) 18, , (d) 8, , 3. Simplify: (2x − 3y) (4x2 + 6xy + 9y2)., (a) 8x3 − 27y3, (c) 4x2 − 9y2, , (b) 2x3 − 3y3, (d) 8x3 + 27y3, , 4. If x2 − 3x − 28 = (x − A) (x − B), then the value of A + B = ______., (a) 3, 5. If, , 3x2, , (b) 11, , (c) −3, , (d) 7, , + 13x + 4 = (Ax + B) (Cx + D), then find the value of AC + BD., , (a) 15, , (b) 12, , (c) 7, , (d) 20, , 6. If a, b and c are consecutive even numbers, and abc + ab + bc + ac + a + b + c, = 692, then what will be the value of a + b + c?, (a) 20, , (b) 18, , (c) 30, , (d) 24, , 7. (5x2 + 12x + 7) ÷ (5x + 7) = ______, (a) x + 5, (c) x + 5, , (b) x + 3, (d) x + 1, , 8. If x6 + 64y6 = (Ax2 + By2) (Cx4 + Dx2 y2 + Ey4), then A + B + C + D + E = ______., (a) 16, 9. Solve for:, , (b) 18, a2, , +, , b2, , −, , c2, , (d) 20, , − 2ab., , (a) (a + b + c) (a + b − c), (c) (a − b − c) (a − b + c), , M04_IIT-FOUNDATION_XXXX_CH04.indd 5, , (c) 26, (b) (a − b + c) (a + b + c), (d) (a − b + c) (a − b − c), , 4/11/2018 4:35:40 PM

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4.6, , Chapter 4 Polynomials and Their LCM and HCF, , 10. Calculate the LCM of (x + m)2 and (x2 − m2)., (a) x + m, (c) (x + m)2, , Space for rough work, , (b) x2 − m2, (d) (x2 − m2) (x + m), , 11. The HCF of (x2 − 1)2, (x − 1), and (x2 − 1) is ______., (a) x + 1, (c) x – 1, , (b) x2 − 1, (d) x2 + 1, , 12. What will be HCF of 6x2 − 7x + 2 and 4x2 − 1?, (b) 2x − 1, (d) 3x − 2, , (a) 1, (c) 2x + 1, , 13. The LCM of (x + 1) (x − 2)2 (x + 3)2 and (x + 1) (x − 2) (x + 3)3 is _______., (a) (x − 2)2 (x + 3)3 (x + 1), (c) (x + 3)3, , (b) (x − 2) (x + 3) (x + 1), (d) (x − 2), , 14. If the HCF of f(x) = (x − 4) (3x2 − 8x + m) and g(x) = (x − 3) (4x2 − 15x + n) is, (x − 4) (x − 3), where m and n are constants, then find the value of m + n is, _______., (a) 0, , (b) 7, , (c) −7, , (d) 12, , Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (Polynomial), , Column B, (Product of Factors), , (i) 2x2 + x − 6, , (A) (2x − 3) (x + 2), , (ii) 2x2 − x − 6, , (B) (3x − 2) (x − 3), , (iii) 3x2 + 11x + 6, , (C) (3x + 2) (x + 3), , − 11x + 6, , (D) (2x + 3) (x − 2), , (iv), (a), (b), (c), (d), , 3x2, , (i) → (A); (ii) → (B);  (iii) → (C); (iv) → (D), (i) → (A); (ii) → (D); (iii) → (C); (iv) → (B), (i) → (D); (ii) → (B);  (iii) → (A); (iv) → (C), (i) → (B);  (ii) → (C);  (iii) → (D); (iv) → (A), , M04_IIT-FOUNDATION_XXXX_CH04.indd 6, , 4/11/2018 4:35:40 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 4.7, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1, 1, = 5, then find a3 + 3 . The following are the steps are involved in, a, a, solving the above problem. Arrange them in sequential order., , 1. If a −, , (A) a3 −, , 1, , 1, − 3  a −  = 125, 3, , a, a, , (B) a −, , 1, , 1, = 5 ⇒  a −  = 125, , a, a, , 3, , 2, , 2,  3 1, 1, , (C)  a3 + 3  −  a − 3  = 4, , a, a , 1, (D) a3 − 3 = 140, a, 1, (E) a3 + 3 = 19604, a, (a) BADCE, (c) BDACE, 1, 1, 2. If x + = 6, then x2 + 2 = ______., x, x, , (a) 34, , (b) ABCDE, (d) AEBCD, , (b) 36, , 3. (4x + 3y), , (16x2, , − 12xy +, , 9y2), , (c) 32, , (d) 16, , = __________., , (a) 4x3 + 3y3, (c) 12x3 + 9y3, , (b) 16x2 + 9y2, (d) 64x3 + 27y3, , 4. If x2 + 3x + 2 = (x − A) (x − B), then A + B = ______., (a) −1, , (b) 1, , (c) 3, , (d) −3, , 5. If 6x2 + 11x − 10 = (Px + Q) (Ax + B), then find the value of AP + BQ., (a) −4, , (b) 6, , (c) 4, , (d) 20, , 6. If x, y, and z are consecutive odd integers and xyz + xy + yz + zx + x + y + z =, 959, then xyz = ______., (a) 495, 7., , (6x2, , (b) 315, , (c) 693, , (d) 429, , (c) 3x − 3, , (d) 2x + 3, , + x − 15) ÷ (3x + 5) = ______, , (a) 2x − 3, , (b) 2x − 5, , 8. If −, =, +, P + Q + R − S + T?, a6, , 729b6, , (a) 60, , (Pa2, , Qb2), , (Ra4, , (b) 74, , +, , Sa2b2, , +, , Tb4),, , then what will be the value of, , (c) 55, , (d) 65, , 9. x2 + y2 − z2 + 2xy = ______, (a) (x + y − z) (x − y + z), (c) (x + y − z) (x + y + z), , M04_IIT-FOUNDATION_XXXX_CH04.indd 7, , (b) (x − y + z) (x − y + z), (d) (x + y − z) (x − y − z), , 4/11/2018 4:35:46 PM

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4.8, , Chapter 4 Polynomials and Their LCM and HCF, , 10. The LCM of (x − n)2 and (x2 + n2) is ______., (a) x2 − n2, (c) (x − n)2 (x2 + n2), , Space for rough work, , (b) (x2 − n2) (x + n), (d) (x2 − n2)2, , 11. Find the HCF of (x + 1)2, (x2 − 1), and (x + 1)., (a) x + 1, (c) x − 1, , (b) x2 − 1, (d) x2 + 1, , 12. What will be the HCF of (x2 − 9) and (x2 − 6x + 9)?, (a) x + 3, (c) x − 3, , (b) (x + 3)2, (d) 1, , 13. The LCM (x + 2)2 (x − 2) (x2 − 1) and (x − 2)2 (x + 2) (x + 1) is ______., (a) (x − 2)2 (x + 2)2 (x + 1), (c) (x − 2) (x + 2) (x + 1) (x − 1), , (b) (x − 2)2 (x + 2) (x2 − 1), (d) (x − 2)2 (x + 2)2 (x2 − 1), , 14. If the HCF of (x − 2) (2x2 + 7x + P) and (x + 2) (2x2 − 7x + Q) is x2 − 4, where P, and Q are integers, then P + Q = _______., (a) 0, , (b) 6, , (c) 12, , (d) 8, , Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (Polynomials), , Column B, (HCF), , (i) x2 − y2, (x + y)2, (ii) (x −, , y)2,, , (iii), , x3, , +, , y3,, , (iv), , x2, , + xy +, , (a), (b), (c), (d), , (x3, , x2, , (i) → (A);, (i) → (C);, (i) → (B);, (i) → (C);, , M04_IIT-FOUNDATION_XXXX_CH04.indd 8, , −, , (A) 1, , − xy +, , y2, , ,, , (B) x − y, , y3), x2, , (C) x + y, , y2, , − xy +, , y2, , (D) x2 − xy + y2, , (ii) → (B); (iii) → (C); (iv) → (D), (ii) → (B); (iii) → (D); (iv) → (A), (ii) → (C); (iii) → (D); (iv) → (A), (ii) → (B); (iii) → (A); (iv) → (D), , 4/11/2018 4:35:46 PM

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Answer Keys, , 4.9, , Answer Keys, Assessment Test I, 1. (d), 11. (b), , 2. (a), 12. (e), , 3. (d), 13. (b), , 4. (b), 14. (d), , 5. (c), 15. (c), , 6. (d), , 7. (a), , 8. (d), , 9. (c), , 10. (b), , 4. (b), 14. (a), , 5. (c), 15. (d), , 6. (a), , 7. (a), , 8. (c), , 9. (a), , 10. (d), , 4. (a), 14. (c), , 5. (c), 15. (b), , 6. (d), , 7. (d), , 8. (b), , 9. (d), , 10. (d), , 4. (d), 14. (c), , 5. (a), 15. (b), , 6. (c), , 7. (a), , 8. (d), , 9. (c), , 10. (c), , Assessment Test II, 1. (b), 11. (a), , 2. (c), 12. (b), , 3. (a), 13. (c), , Assessment Test III, 1. (b), 11. (c), , 2. (c), 12. (b), , 3. (a), 13. (a), , Assessment Test IV, 1. (a), 11. (a), , 2. (a), 12. (c), , M04_IIT-FOUNDATION_XXXX_CH04.indd 9, , 3. (d), 13. (d), , 4/11/2018 4:35:46 PM

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Ratio, Proportion, and, Variations: Commercial, Mathematics, , 5, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapters - Ratio Proportion Variations; Percentages;, Profit and Loss, Discount and Partnership; Simple Interest and Compound Interest; pp. 6.1–6.14; 7.1–7.9; 8.1–8.11; 9.1–9.13, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., 1 1, 1. If two numbers are in the ratio : , then the following steps are involved, 7 5, in calculating by what percentage is the second number more than the first., Arrange them in sequential order., (A) i.e., the second number is 2x more than the first number., (B) Difference between the two numbers is 7x - 5x = 2x., 1 1, (C) Given that the ratio of two numbers = : = 5:7., 7 5, ∴ Let the numbers be 5x and 7x., (D) The percentage by which the second number is more than the first number, 2x, =, × 100% = 40%, 5x, (a) ABCD, 2. If 2.7 p = 0.09q, then, (a), 3. If, , 4, 3, , (b) CBAD, , (c) BCAD, , Space for rough work, , (d) CABD, , q + 6p, = ______., q − 6p, (b), , 3, 2, , (c), , 5, 4, , (d), , 6, 5, , a c, e, a+c, = = , then what will be the value of, ?, b d, f, b+d, , (a), , a+e, b+ f, , (c) Both (a) and (b), , (b), , c+e, d+ f, , (d) Neither (a) nor (b), , 4. In which of the following situations can it be concluded that A varies directly, with B?, 3, A,  A, (a) 3, is a constant., (b)   is constant.,  B, B, (c) Both (a) and (b), , M05_IIT-FOUNDATION_XXXX_CH05.indd 1, , (d) Neither (a) nor (b), , 4/11/2018 4:35:58 PM

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5.2, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , 5. The ratio of the incomes of P and Q is 2:3. The ratio of their expenditures is, 5:6. Who saves a greater part of his income?, (a) P, (b) Q, (c) Both save equal parts of their incomes., (d) Cannot say, , Space for rough work, , 6. P, Q and R had some stamps among them. The ratio of the stamps with, 1 1 1, them was : : . If the ratio of the stamps with them had been 2:3:6 instead,, 2 3 6, which of the following can be concluded?, (a), (b), (c), (d), , Q would have less stamps than he actually does., P would have less stamps than he actually does., Both (a) and (b)., Neither (a) nor (b)., , 7. In an exam, a student gets 30% of the maximum marks and failing by 30, marks. If 40% of the maximum marks is required for passing, then find the, maximum marks., (a) 200, , (b) 250, , (c) 100, , (d) 300, , 8. A trader professes to sell his goods at cost price but sells 900 g weight for, every 1 kg. What is the profit percentage?, (a) 20%, (b) 15%, 1, (c) 10%, (d) 11 %, 9, 9. A trader marks his product 30% above the cost price and then gives a discount of 10%. Find his profit percentage., (a) 21%, (b) 15%, (c) 13%, (d) 17%, 10. Trader A gives a single discount of 30% on an article whose marked price, is `x and trader B gives two discounts of 25% and 5% on an article whose, marked price is `x. Which is a better discount offered to the customers?, (a) Discount offered by trader A., (b) Discount offered by trader B., (c) Both are same., (d) Cannot be determined., 11. A merchant gets `1160 by selling article A at 30% profit and B at 20% loss. He, gets the same amount if he sells A at 20% loss and B at 5% profit. What is the, cost price of article B?, (a) `400, (b) `800, (c) `500, (d) `600, , M05_IIT-FOUNDATION_XXXX_CH05.indd 2, , 4/11/2018 4:35:59 PM

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Assessment Test I, Directions for questions 12 to 15: Match the statements of Column A with the values, of Column B., Column A, 12., , A number when increased by 60%, becomes 16. The number is ______., 13. A man sells 150 mangoes at the cost price, of 225 mangoes. His profit is _______%., 14. A certain sum doubles in 6 years under, simple interest. The time taken to it to, become 4 times itself is_______ years., 15. If `125 amounts to `216 under compound, interest in 3 years, then the rate of interest, is_______%., , 5.3, , Space for rough work, , Column B, (a) 8, (b) 12,   (c) 20, , (d) 10, , (e) 50, (f) 15, , M05_IIT-FOUNDATION_XXXX_CH05.indd 3, , 4/11/2018 4:35:59 PM

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5.4, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , Assessment Test II, , Time: 30 min., , Space for rough work, , Directions for questions 1 to 11: Select the correct answer from the given options., 1 1, : , then the following steps are involved, 15 20, in finding by what percentage is the second number less than the first., Arrange them in sequential order., (A) Difference between the two numbers is 4x - 3x = x., 1 1, (B) Given that the ratio of two numbers = :, = 20:15 = 4:3., 15 20, , Let the numbers be 4x and 3x., (C) The percentage by which the second number is less than the first numx, ber =, × 100% = 25%., 4x, (D) i.e., the second number is x less than the first number., (a) BADC, (b)BDAC, (c) DBAC, (d) ABDC, 1. If two numbers are in the ratio, , 2. If x:y = 4:9, then find the value of, , 4 x +5 y, 6 x +7 y, , ., , 21, 19, 27, 23, (b), (c), (d), 31, 29, 37, 33, b, a, c, 3. If, =, =, , then each of these ratios can be equal to ____., a+b−c a+c−b b+c−a, 1, 2, 1, (a) 1, (b), (c), (d), 2, 3, 3, (a), , 4. In which of the following situations can we conclude that P varies inversely, with Q?, (a), (c), , PQ is a constant., (PQ)3, , is a constant., , (b) 2PQ is a constant., (d) All of the above, , 5. The ratio of the incomes of X and Y is 3:4. The ratio of their expenditures is, 4:5. Who spends a greater part of his income?, (a) X, (b) Y, (c) Both spend equal parts of their incomes., (d) Cannot say, 6. A, B and C intended to divide a certain sum among themselves in the ratio, 1 1 1, 2:3:4. By mistake, they divided it in the ratio : : . How many of A, B and, 2 3 4, C lost due to this mistake?, (a) 3, (b) 0, (c) 1, (d) 2, , M05_IIT-FOUNDATION_XXXX_CH05.indd 4, , 4/11/2018 4:36:13 PM

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Assessment Test II, 7. Amar gets 20% of the total marks in an examination and fails by 20 marks., Ajay gets 40% of the total marks and gets 10 marks more than the passing, marks. Find the total marks., (a) 120, (b) 300, (c) 150, (d) 125, , 5.5, , Space for rough work, , 8. A trader professes to sell his goods at cost price and still earns a profit of, 25%. What weight has he used for 1 Kg?, (a) 1250 g, (b) 750 g, (c) 800 g, (d) 600 g, 9. A trader marks his product 60% above the cost price and then gives a discount of 30%. Find his profit percentage., (a) 30%, (b) 24%, (c) 48%, (d) 12%, 10. Three traders A, B and C sell an article of the same marked price. Trader A, gives a single discount of 30%. Trader B gives two successive discounts of, 20% and 10%. Trader C gives successive discounts of 15%, 10% and 5%. Who, gives the maximum discount?, (a) A, (b) B, (c) C, (d) B or C, 11. A trader presently earns 15% profit by selling his product. If he increases the, price of the product by `24, then his gain per cent increases to 27%. Find the, cost price of the product., (a) `20, (b) `200, (c) `150, (d) `160, Direction for questions 12 to 15: Match the statements of Column A with the values, of Column B., Column A, , Column B, , 12., , A number when decreased by 40%, becomes 12. The number is _______., , (a) 20, , 13., , A man sells 150 mangoes at the cost, price of 100 mangoes. His loss is, _______ percentage., , (b) 20, , 14., , A certain sum triples in 3 years, under simple interest. The time, taken to it to become 5 times itself is, _______ years., ,   (c) 6, , 15., , If `1000 amounts to `1331 under, compound interest in 3 years, then, rate of interest is _______%., , (d) 10, , (e) 5,    (f) 33, (g) 16, , M05_IIT-FOUNDATION_XXXX_CH05.indd 5, , 1, 3, , 4/11/2018 4:36:14 PM

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5.6, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , Assessment Test III, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. If x:y = 3:4, then the following steps are involved in finding the value of (3x, + y):(x + 3y). Arrange them in sequential order., (A) x = 3a, y = 4a, (B) (9a + 4a):(3a + 12a), (C) (3x + y):(x + 3y), (D) 13a:15a = 13:15, (a) ABCD, (b) BACD, (c) CBAD, (d) ACBD, 2. If A:B = B:C = 4:5, then find the ratio of A:B:C., (a) 8:20:25, (b) 20:15:25, (c) 4:5:6, (d) 16:20:25, 1 1, 3. The ratio of incomes of A and B is : . By what percentage A’s income less, 3 2, than that of B?, 2, 1, 3, (a) 20%, (b) 16 %, (c) 33 %, (d) 23 %, 3, 3, 7, 4. The compounded ratio of (x + 5):25 and 9:x is 12:25. Find the value of x., (a) 5, (b) 15, (c) 20, (d) 25, 5. Two successive discounts of 25% and 28% are equal to a single discount of, _________., (a) 33%, (b) 40%, (c) 46%, (d) 53%, 6. A certain sum amounts to `4300 in 3 years and to `5100 in 4 years at simple, interest. Find the initial sum., (a) `1900, (b) `2700, (c) `3100, (d) `3300, 7. What will be the value of x, if, 52, (a), 53, , 26, (b), 27, , 9 + 8x + 9 − 8x, 9 + 8x − 9 − 8x, (c), , =, 53, 56, , 3, ?, 2, (d), , 27, 26, , 8. What is the least number that should be subtracted from each of the numbers 14, 11, 17 and 13 so that the resulting numbers are in proportion?, (a) 4, (b) 6, (c) 5, (d) 3, 9. 250 men consume 2250 kg of rice in 60 days. In how many days will 150 men, consume 720 kg of rice?, (a) 60, (b) 32, (c) 45, (d) 30, , M05_IIT-FOUNDATION_XXXX_CH05.indd 6, , 4/11/2018 4:36:20 PM

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Assessment Test III, 10. If y is mean proportion of x and z, then (x − y)4:(y − z)4 = ________., (a) x3:y3, (b) x4:y4, (c) x2:y2, (d) x:y, , 5.7, , Space for rough work, , 11. X, Y and Z together completed a work and earned `86,000. But X worked, for 3 days, Y for 4 days and Z for 5 days. If their daily wages are in the ratio, 6:7:8, then how much amount will be received by X?, (a) `10,000, (b) `28,000, (c) `20,000, (d) `18,000, 12. A shopkeeper sells two articles at the same selling price, one at 15% profit, and other 15% loss. What is the overall profit or loss percentage?, (a) 2.25% profit, (b) 22.5% loss, (c) 2.25% loss, (d) 0%, 13. The selling price of 50 articles is equal to cost price of 75 articles. Find profit, or loss percentage., (a) 50% profit, (b) 50% loss, 1, (c) 33 % profit, 3, , 1, (d) 33 % loss, 3, , 14. In a school, there are 1320 students. The ratio of the number of boys to the, number of girls is 5:6. If the number of boys is increased by 18%, then what, will be the ratio of the number of boys to the number of girls?, (a) 59:69, (b) 69:59, (c) 59:60, (d) 60:59, 15. Match the values of Column A with those of Column B., Column A, , Column B, , (i) C,  .P. = `200;, S.P. = `250, , (A) Profit% = 20%, , (ii) C.P. = `200;, S.P. = `150, ,   (B) Profit% = 25%, , (iii) C.P. = `150;, S.P. = `180, , (C) Loss% = 25%, , (iv) C.P. = `150;, S.P. = `120, , (D) Loss% = 20%, , (a), (b), (c), (d), , (i) → (C);, (i) → (B);, (i) → (D);, (i) → (B);, , M05_IIT-FOUNDATION_XXXX_CH05.indd 7, , (ii) → (D); (iii) → (A); (iv) → (B), (ii) → (C); (iii) → (A); (iv) → (D), (ii) → (B);  (iii) → (C); (iv) → (A), (ii) → (C); (iii) → (D); (iv) → (A), , 4/11/2018 4:36:21 PM

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5.8, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , Assessment Test IV, , Time: 30 min., , Space for rough work, , Directions for questions 1 to 15: Select the correct answer from the given options., 1. By selling an article for `49, a shopkeeper earns a profit of 40%. The following steps are involved in finding at what price should he sell it to earn a, profit of 50%. Arrange them in a sequential order.,  100 × S.P. , = `35, (a) C.P. = ,  100 + Profit% , C.P. (100 + Profit %), = `52.5, 100, (c) Profit% = 40% and S.P. = `49, (d) Profit% = 50% and C.P. = `35, (a) DACB, (b) CDAB, (c) CADB, (d) ACBD, (b) S.P. =, , 2. If A:B = 2:5 and B:C = 4:3, then find the ratio of A:C., (a) 2:3, (b) 5:3, (c) 8:15, , (d) 16:9, , 1 1, 3. The ratio of the incomes of P and Q is : . By what percentage is Q’s income, 3 4, less than that of P?, 2, 1, (a) 20%, (b) 25%, (c) 16 %, (d) 33 %, 3, 3, 4. If the compound ratio of (x + 6):x and 3:7 is 12:7, then what will be the value, of x?, (a) 1, (b) 2, (c) 3, (d) 4, 5. Two successive discounts of 20% and 25% are equal to a single discount of, _______., (a) 30%, (b) 45%, (c) 35%, (d) 40%, 6. A certain sum amounts to `4300 in 4 years and to `5200 in 6 years at simple, interest. Find the amount of interest for 3 years., (a) `1350, (b) `1500, (c) `1225, (d) `1200, 7. Find the value of x, if, (a), , 21, 41, , 7 + 3x + 7 − 3x, , 7 + 3x − 7 − 3x, 41, (b), 21, , = 9., (c), , 41, 9, , (d), , 1, 3, , 8. What is the number that must be added to each of the numbers 5, 2, 12, and, 6 so that the resulting numbers will be in proportion?, (a) 4, (b) 3, (c) 1, (d) 2, , M05_IIT-FOUNDATION_XXXX_CH05.indd 8, , 4/11/2018 4:36:25 PM

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Assessment Test IV, 9. 30 men working 6 hours per day can complete a work in 24 days. In how, many days, 18 men can complete the same work, working 8 hours per day?, (a) 33, (b) 28, (c) 35, (d) 30, , 5.9, , Space for rough work, , 10. If b is the mean proportional of a and c, then (a + b)3:(b + c)3 = _______., (a) a3:b3, (b) a3:c3, (c) b3:c3, (d) Both (a) and (c), 11. A, B and C completed a work for `90,000. A worked for 6 days, B for 7, days and C for 9 days. If their daily wages are in the ratio 8:6:5, how much, amount will be received by B?, (a) `28,000, (b) `32,000, (c) `30,000, (d) `26,000, 12. If a trader sold two articles at `2250 each and gains 9% on the first and loses, 9% on the second, then his profit or loss per cent on the whole is _____., (a) 0.81% loss, (b) 0.81% profit, (c) 0%, (d) 8.1% loss, 13. If the cost price of 125 articles is equal to the selling price of 150 articles, then, the profit or loss percentage is _________., 2, (a) 20% profit, (b) 16 % profit, 3, 2, (c) 16 % loss, (d) 20% loss, 3, 14. In a school, there are 800 students. The ratio of the number of boys to girls, is 2:3. By what percentage of the number of the boys, should the strength of, the boys increase so that the ratio of boys to girls would be 3:2?, (a) 80%, (b) 125%, (c) 120%, (d) 100%, 15. Match the values of Column A with those of Column B., Column A, , Column B, , (i) Sub triplicate ratio of a:b, , (A) a2:b2, , (ii) Duplicate ratio of a:b, ,   (B)  a: b, , (iii) Triplicate ratio of a:b, , (C) a3:b3, , (iv) Sub duplicate ratio of a:b, , (D), , (a), (b), (c), (d), , 3, , a:3 b, , (i) → (C);  (ii) → (A); (iii) → (D); (iv) → (B), (i) → (D); (ii) → (B);  (iii) → (C);  (iv) → (A), (i) → (C);  (ii) → (B);  (iii) → (A); (iv) → (D), (i) → (D); (ii) → (A); (iii) → (C); (iv) → (B), , M05_IIT-FOUNDATION_XXXX_CH05.indd 9, , 4/11/2018 4:36:28 PM

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5.10, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , Answer Keys, Assessment Test I, 1. (b), 11. (b), , 2. (b), 12. (d), , 3. (d), 13. (e), , 4. (d), 14. (a), , 5. (b), 15. (c), , 6. (d), , 7. (d), , 8. (d), , 9. (d), , 10. (a), , 4. (d), 14. (c), , 5. (a), 15. (d), , 6. (d), , 7. (c), , 8. (c), , 9. (d), , 10. (a), , 4. (b), 14. (c), , 5. (c), 15. (b), , 6. (a), , 7. (d), , 8. (c), , 9. (b), , 10. (b), , 4. (b), 14. (b), , 5. (d), 15. (d), , 6. (a), , 7. (a), , 8. (d), , 9. (d), , 10. (d), , Assessment Test II, 1. (a), 11. (b), , 2. (d), 12. (a), , 3. (a), 13. (f), , Assessment Test III, 1. (d), 11. (d), , 2. (d), 12. (c), , 3. (c), 13. (a), , Assessment Test IV, 1. (c), 11. (a), , 2. (c), 12. (a), , M05_IIT-FOUNDATION_XXXX_CH05.indd 10, , 3. (b), 13. (c), , 4/11/2018 4:50:36 PM

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Real Numbers, LCM,, and HCF, , 6, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapter - Real Numbers, LCM and HCF; pp. 1.1–1.22, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. The following steps are involved in finding the number of factors of 360., Arrange them in sequential order., (A) The number of factors = (3 + 1) (2 + 1) (1 + 1), (B) If x = pa × qb × rc (where p, q, and r are distinct primes), then the number, of factors of x = (a + 1) (b + 1) (c + 1)., (C) 360 = 23 × 32 × 51, (D) 4 × 3 × 2 = 24, (a) ABDC, (b) BCAD, (c) CBDA, (d) BACD, 2. The units digit of the product 1315 × 1517 × 1719 is _________., (a) 3, (b) 5, (c) 7, 3. 2.648 = _________, 88, 98, (a), (b), 35, 37, , (c), , 211, 99, , (d) 9, , (d), , 89, 39, , 4. The GCD of 145 and 256 is A and the LCM of 144 and 1728 is B. Find the, product of AB., (a) 5296, (b) 1296, (c) 256, (d) 1728, 5. The product of the HCF and the LCM of, (a), , 2, 5, , (b), , 2, 3, , 2 2, 3, , , and ., 3 5, 5, 3, (c), 5, , (d), , 1, 15, , 6. Find a natural number which leaves a remainder of 7 when divided by 8, 18,, or 28., (a) 647, (b) 159, (c) 727, (d) 511, 7. Find a natural number which leaves remainder 7, 17, and 27 when divided, by 15, 25, and 35, respectively., (a) 532, (b) 547, (c) 572, (d) 517, , M06_IIT-FOUNDATION_XXXX_CH06.indd 1, , 4/11/2018 4:36:35 PM

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6.2, , Chapter 6 Real Numbers, LCM, and HCF, , 8. The HCF of two numbers A and B is 18 and their LCM is 144. If B > A, then, find the greatest possible value of B., (a) 72, (b) 144, (c) 288, (d) 432, , Space for rough work, , 9. Which of the following numbers has two pairs of twin primes as their, factors?, (a) 7700, (b) 3080, (c) 5005, (d) 3003, 10. 984x6y is a six-digit number, which is a multiple of 9, and x and y are nonzero digits. Find the minimum value of the product xy., (a) 2, (b) 6, (c) 8, (d) 14, 11. Which of the following is a perfect number?, (a) 27(28 − 1), (b) 25(26 − 1), (c) 24 (25 − 1), (d) 23(24 − 1), Direction for questions 12 to 15: Match the statements of Column A with the values, of Column B., Column A, , Column B, , 12. The number of factors of 48 (a) 12, 13. Perfect number, , (b) 10, 887, , 14. The units digit of, 15. The least composite, number, , (c) 8, (d) 6, (e) 4, (f) 2, , M06_IIT-FOUNDATION_XXXX_CH06.indd 2, , 4/11/2018 4:36:35 PM

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Assessment Test II, , Assessment Test II , , Time: 30 min., , 6.3, , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. The GCD of two numbers is 4 and their LCM is 72. How many pairs of numbers exist? Find the sequential order of steps in solving the above problem., (A) 4x × 4y = 72 × 4, (B) 18 = 1 × 18 or 2 × 9, (C) Let the numbers be 4x, 4y, where x and y are coprimes., (D) xy = 18, (E) There are two pairs of numbers., (a) CDEAB, (b) CDEAB, (c) CADBE, (d) CADEB, 2. The units digit of the sum 415 × 516 + 617 × 718 is_________., (a) 4, (b) 5, (c) 8, , (d) 2, , 3. 3.345 = _________, (a), , 1711, 333, , (b), , 335, 111, , (c), , 1192, 333, , (d), , 1114, 333, , 4. The HCF of two numbers P and Q is A, and their LCM is B. Which of the, following is true?, (a) A > B, (b) P + Q = A + B, A Q, (c) AP = BQ, (d), =, P B, 5. The product of LCM and HCF of, (a), , 4, 5, , (b) 3, 7, , 4 6, 7, , and is _________., 5 7, 9, (c) 4, 15, , (d) 5, 7, , 6. Find a number which leaves a remainder of 6 when divided by 8, 12, or 15., (a) 54, (b) 78, (c) 102, (d) 126, 7. Find a natural number which leaves remainders 6, 8, and 10 when divided, by 12, 14, and 16, respectively., (a) 246, (b) 78, (c) 330, (d 336, 8. The HCF of two numbers P and Q (where P < Q) is one of the numbers and, their LCM is 94. If P ≠ 1, then what can be the value of P × Q?, (a) 94, (b) 188, (c) 28, (d) 376, , M06_IIT-FOUNDATION_XXXX_CH06.indd 3, , 4/11/2018 4:36:43 PM

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6.4, , Chapter 6 Real Numbers, LCM, and HCF, , 9. Which of the following is/are false statement(s)?, (A) 17 and 19 are twin primes., (B) 15 and 13 are coprimes., (C) 24 is a perfect number., (D) The least prime number is an even number., (a) (A), (b) (B), (c) (C), (d) Both (B) and (C), , Space for rough work, , 10. 2k − 1 (2k − 1) is a perfect number for which of the following values of k?, (a) 8, (b) 9, (c) 6, (d) 3, 11. If 684m7n is divisible by 11, where m and n are non-zero digits, then find the, maximum value of the product of m and n., (a) 28, , (b) 18, , (c) 14, , (d) 20, , Direction for questions 12 to 15: Match the statements of Column A with the values, of Column B., Column A, , Column B, , 12. The number of primes, between 50 and 100, , (a) 10, , 13. The units digit of 777, , (b) 9, , 14. An odd number but not, prime, , (c) 8, , 15. The number of factors of 62 (d)  7, (e) 6, (f) 5, (g) 4, , M06_IIT-FOUNDATION_XXXX_CH06.indd 4, , 4/11/2018 4:36:43 PM

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Assessment Test III, , Assessment Test III, , Time: 30 min., , 6.5, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. The following steps are involved in finding the HCF of 72 and 96. Arrange, them in sequential order., (A) 24, (B) 72 = 22 × 32 ; 96 = 25 × 31, (C) The common prime factors are 2 and 3., (D) HCF is 23 × 31., (a) BCAD, (b) ABDC, (c) BCDA, (d) ACDB, 2. 0.6 + 0.5 = _________, (a) 1.1, (b) 1.2, 3. If 5 × 125 =, (a) 324, , (c) 1.3, , (d) 1.4, , x + 6 , then x = _________., (b) 361, (c) 625, , (d) 684, , 4. If the sum of two integers is negative, then which of the following must be, true?, (a) At least one of the integers is positive., (b) Both are positive., (c) At least one of the integers is negative., (d) Both are negative., 5. Find the sum of the factors of 36., (a) 90, (b) 91, , (c) 96, , (d) 97, , 6. Find the number of factors of 64., (a) 5, (b) 6, , (c) 7, , (d) 8, , 7. What will be the LCM of 32 and 48?, (a) 48, (b) 96, , (c) 192, , (d) 384, , 8. The LCM of x and y is a. Then LCM of kx and ky is _________., (a) k, (b) a, (c) ka, , (d), , 9. Which of the following is a perfect number?, (a) 12, (b) 24, (c) 28, , (d) 32, , M06_IIT-FOUNDATION_XXXX_CH06.indd 5, , k, a, , 4/11/2018 4:36:46 PM

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6.6, , Chapter 6 Real Numbers, LCM, and HCF, , 10. The HCF of, , 7 7, 2, ,, and is _________., 3 9, 5, , (a) 45, , (b), , 1, 45, , Space for rough work, (c) 90, , (d), , 1, 90, , 11. The HCF of two numbers is 9 and their sum is 72. What will be the number, of pairs of numbers that satisfy the above condition?, (a) 1, (b) 2, (c) 3, (d) 4, 12. The product of the LCM and the HCF of two numbers is 896 and one of the, numbers is 28, then the other number is _________., (a) 22, (b) 32, (c) 42, (d) 52, 13. Find the greatest four-digit number which leaves remainders of 10, 8, and, 12, respectively when divided by 13, 11, and 15., (a) 8567, (b) 8561, (c) 8568, (d) 8577, 14. How many natural numbers less than 1000 have exactly three factors?, (a) 11, (b) 12, (c) 31, (d) 16, Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (Numbers), , Column B, (Units Digit of, the Number), , (i) 725, , (A) 6, , (ii) 239, , (B) 7, , (iii) 6100, , (C) 8, , (iv) 99, , (D) 9, , (a), (b), (c), (d), , (i) → (A); (ii) → (B);, (i) → (B);  (ii) → (C);, (i) → (C); (ii) → (B);, (i) → (B); (ii) → (C);, , M06_IIT-FOUNDATION_XXXX_CH06.indd 6, , (iii) → (C);   (iv) → (D), (iii) → (A);  (iv) → (D), (iii) → (A);  (iv) → (D), (iii) → (D);  (iv) → (A), , 4/11/2018 4:36:47 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 6.7, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. The following steps are involved in finding the HCF of 24 and 36. Arrange, them in sequential order., (A) 24 = 23 × 31 and 36 = 22 × 32, (B) HCF = 22 × 31, (C) The common prime factors are 2 and 3., (D) 12, (a) ACBD, , (b) ABCD, , (c) CADB, , (d) CDAB, , (c) 1.7, , (d) 1.6, , (c) 900, , (d) 625, , 2. 2.3 − 0.4 = _________, (a) 1.9, , (b) 1.8, , 3. If 6 × 216 = y − 4 , then y = _________., (a) 1600, , (b) 1225, , 4. If the product of eleven integers is negative, then which of the following can, be true?, (a) Two integers are positive., (b) Four are positive., (c) Six are positive., (d) All of these, 5. Find the sum of the factors of 56., (a) 118, (b) 119, , (c) 120, , (d) 121, , 6. Find the number of factors of 512., (a) 8, (b) 9, , (c) 10, , (d) 11, , 7. What will be the LCM of 128 and 142?, (a) 9088, (b) 18176, , (c) 4544, , (d) 2272, , 8. If the HCF of a and b is k, then HCF of, (a) 1, , (b) k, , a, b, and, is _________., k, k, (c) m, , 9. Which of the following is a perfect number?, (a) 412, (b) 456, (c) 476, , M06_IIT-FOUNDATION_XXXX_CH06.indd 7, , (d), , m, k, , (d) 496, , 4/11/2018 4:36:52 PM

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6.8, , Chapter 6 Real Numbers, LCM, and HCF, , 10. The HCF of, (a), , 40, 77, , 5 4, 2, ,, and is _________., 7 11, 7, 1, (b), 77, , Space for rough work, (c), , 20, 77, , (d), , 10, 77, , 11. The HCF of two numbers is 11 and their sum is 209. Find the number of, pairs of numbers that satisfy the above condition., (a) 1, (b) 2, (c) 3, (d) 9, 12. The HCF and the LCM of two numbers are 12 and 72, respectively. If one of, the numbers is 24, then what is the other number?, (a) 26, (b) 36, (c) 16, (d) 14, 13. What is the greatest four-digit number, which leaves remainders of 4, 5, and, 8 when divided by 7, 8, and 11, respectively?, (a) 9856, (b) 9982, (c) 9859, (d) 9853, 14. How many natural numbers between 50 and 500 have odd number of, factors?, (a) 13, (b) 14, (c) 15, (d) 16, Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (Numbers), , Column B, (Units Digit of the Number), , (i) 489, , (A) 3, , (ii) 536, , (B) 4, , (iii) (14)18, , (C) 5, , (iv) (33)89, , (D) 6, , (a), (b), (c), (d), , (i) → (D);    (ii) → (C);, (i) → (B); (ii) → (C);, (i) → (D);    (ii) → (B);, (i) → (B); (ii) → (C);, , M06_IIT-FOUNDATION_XXXX_CH06.indd 8, , (iii) → (A); (iv) → (B), (iii) → (A); (iv) → (D), (iii) → (A); (iv) → (C), (iii) → (D); (iv) → (A), , 4/11/2018 4:36:54 PM

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Answer Keys, , 6.9, , Answer Keys, Assessment Test I, 1. (b), 11. (c), , 2. (b), 12. (b), , 3. (b), 13. (d), , 4. (d), 14. (f), , 5. (a), 15. (e), , 6. (d), , 7. (d), , 8. (b), , 9. (c), , 10. (c), , 4. (d), 14. (b), , 5. (c), 15. (g), , 6. (d), , 7. (c), , 8. (b), , 9. (c), , 10. (d), , 4. (c), 14. (a), , 5. (b), 15. (b), , 6. (c), , 7. (b), , 8. (c), , 9. (c), , 10. (b), , 4. (d), 14. (c), , 5. (c), 15. (d), , 6. (c), , 7. (a), , 8. (a), , 9. (d), , 10. (b), , Assessment Test II, 1. (c), 11. (d), , 2. (a), 12. (a), , 3. (d), 13. (d), , Assessment Test III, 1. (c), 11. (b), , 2. (b), 12. (b), , 3. (b), 13. (d), , Assessment Test IV, 1. (a), 11. (d), , 2. (b), 12. (b), , M06_IIT-FOUNDATION_XXXX_CH06.indd 9, , 3. (a), 13. (d), , 4/11/2018 4:36:54 PM

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Squares and Square, Roots, Cubes and Cube, Roots, and Indices, , 7, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapters - Squares and Square Roots, Cubes and Cube, Roots; Indices; pp. 2.1–2.19; 3.1–3.8, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. Find the sequential order of steps to find the square root of n by prime factorization method., (A) The product of these primes is the square root of n., (B) Express n as the product of prime factors., (C) Choose one prime from each pair and multiply all such primes., (D) Pair the factors such that primes in each pair, thus, formed are equal., (a) ABCD, (b) BCAD, (c) CDBA, (d) BDCA, 2. What will be the smallest number by which 10,584 should be divided so that, the quotient is a perfect square?, (a) 2, (b) 3, (c) 14, (d) 6, 3. If A is a natural number and it can be expressed as the sum of the first B, consecutive odd natural numbers, then the value of A − B2 is _______., (a) A, (b) 1, (c) B, (d) 0, 4., , 50.41 + 34.81 = _______, (a) 17, (b) 18, , (c) 13, , 5 2, 4, x − 8 = x2 + 34 , then what can be the value of x?, 8, 7, (a) 28, (b) 14, (c) 27, , (d) 12, , 5. If, , 6., , 3, , (d) 35, , −4913 + 3 1728 = ________, , (a), , 3, , −125, , (b), , 3, , −64, , (c), , 3, , 3185, , (d), , 3, , 27, , 7. What is the least positive integer that should be subtracted form 3388 so that, difference is a perfect cube?, (a) 8, (b) 25, (c) 13, (d) 23, , M07_IIT-FOUNDATION_XXXX_CH07.indd 1, , 4/11/2018 4:37:03 PM

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7.2, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, , 8. If 3 x +, , 36, 3, , x, , = 13, then how many integer values of x satisfy the equation?, , (a) 1, 9., , (b) 2, , (c) 3, , (d) 4, , (c) (x3)2, , (d), , Space for rough work, , x6 × 3 x6 = _______, x10, , (a), , (b) (x3), , 3, , x12, , 10. If (729)x + 3 = (27)5 + x, then find the value of xx., (a) 1, (b) 0, (c) −1, (d) Undefined, 11. Which is the greatest among a = 21458, b = 3972, c = 5729, and d = 6486?, (a) d, (b) c, (c) b, (d) a, Direction for questions 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, 1, 2, , 12. (7x)2 > 435, , (a), , 13. (3x)3 = 729, , (b)   1, , 14. 5x = 3 125, , 2, , (c) 2, , 15. (7x)2 < 23, , (d)   3, , M07_IIT-FOUNDATION_XXXX_CH07.indd 2, , 4/11/2018 4:37:10 PM

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Assessment Test II, , Assessment Test II, , Time: 30 min., , 7.3, , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. Find the sequential order of steps to find the length of a diagonal of the, square plot with area of 6561 sq. m., (A) Area of the square plot = x2 = 6561 (given), (B) Length of the diagonal = 2 × x = 81 2 m, (C) Let the side of the square plot be x m., (D) ∴ Side of the square plot, x = 6561 = 81 m, (a) DCBA, (b) BCAD, (c) CADB, (d) ABDC, 2. What will be the smallest positive integer by which 2916 should be divided, so that the quotient is a perfect cube?, (a) 3, (b) 2, (c) 4, (d) 9, 3. Find the maximum number of consecutive positive odd numbers to give a, sum of 3844., (a) 44, (b) 62, (c) 53, (d) 31, 4., , 3, , 4096 − 3 1331 = ________, , (a), , 16, , (b), , 25, , (c), , 3, , 2765, , 7 2, 7, x − 10 = x2 + 200 , then what can be the value of x?, 3, 5, (a) 5, (b) 10, (c) 15, , (d), , 3, , 15, , 5. If, , 6. If, , 3, , −27 + 729 − 3 216 = x, then, , (a) 24, , (b) 33, , (d) 20, , 32x = _____., (c) 28, , (d) 18, , 7. What is the least positive integer that should be added to 4904 so that the, sum is a perfect cube?, (a) 8, (b) 9, (c) 27, (d) 36, 42, = 289, then which of the following can, 8. For a positive integer x, if x +, x, be the value of x?, (a) 9, (b) 4, (c) 49, (d) 36, 9., , 3, , 343 × 784 = _______, (a) 7, (b) 14, , M07_IIT-FOUNDATION_XXXX_CH07.indd 3, , (c) 21, , (d) 28, , 4/11/2018 4:37:20 PM

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7.4, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, x, , −x, , 2 , 5, , 3, 10. If  3 −  ⋅  1 +  = (8 ) , then x = ______., , 7   14 , (a) 3, (b) 4, (c) 5, , Space for rough work, (d) 9, , 11. Which is the greatest among a = (2401)7, b = (49)15, c = (343)9, and d = 729?, (a) a, (b) b, (c) c, (d) d, Direction for questions 12 to 15: Match the values of Column A with those of, Column B., Column A, , Column B, , 12. (8x)3 = (3x)2, ,   (a) 1, , 13. (125)x = (625)3, 14. (216)x = 83 × 39, , (b) 2,   (c) 3, , 15. [(14)x]x = 42 × (49)2, ,   (d) 4,    (e) 0, , M07_IIT-FOUNDATION_XXXX_CH07.indd 4, , 4/11/2018 4:37:21 PM

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Assessment Test III, , Assessment Test III, , , , Time: 30 min., , 7.5, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. Arrange the following steps in sequential order to find the square of 57., (A) 502 + 72 + 2(50) (7), (B) (a + b)2 = a2 + b2 + 2ab, (C) (57)2 = (50 + 7)2, (D) 2500 + 49 + 700 = 3249, (a) CDBA, (b) CBAD, (c) BCAD, (d) BDCA, 2. What will be the positive value of x, when 4(x + 9)2 = 676?, (a) 5, (b) 8, (c) 4, , (d) 6, , 3. The sum of the cube roots of the perfect cubes 46,656 and 42,875 is ______., (a) 75, (b) 69, (c) 76, (d) 71, 4. What is the maximum number of digits in the cube of a 3-digit number?, (a) 6, (b) 7, (c) 8, (d) 9, 5. What is the smallest number by which 720 should be multiplied so that the, product is perfect square?, (a) 5, (b) 4, (c) 3, (d) 2, x°, × p r2, then find the value of r., 360, (b) 15, (c) 16, , 6. If A = 32p, xo = 80o, and A =, (a) 12, , (d) 7, , 7. Find the least whole number that should be added to 3965 so that the sum, obtained is a perfect square., (a) 4, (b) 5, (c) 6, (d) 7, 8. If 87,880 = 23 × 51 × 13x, then what is the value of x?, (a) 1, (b) 2, (c) 3, 9. If x = 6y and y = 12z, then find the value of, (a) 12, 10. Evaluate:, , (b) 7, , (, , 4, , 5+43, , (a) 2, , M07_IIT-FOUNDATION_XXXX_CH07.indd 5, , )(, , 4, , 5−43, , (b) 3, , )(, , 5+ 3, , 3, , (d) 4, , x3, ., 144 z2 y, , (c) 6, , (d) 8, , (c) 4, , (d) 5, , ), , 4/11/2018 4:37:23 PM

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7.6, , 11., , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, , ( pa+ b )6 . ( pb + c )6 . ( pc+ a )6 = __________, ( pa .pb .pc )12, , (a) 0, , (b) 1, , 8, , Space for rough work, (d) pa + b + c, , (c) p, 8, , 8, , 8, ,  x9  y9,  x  11  y  11, 12. If   +   = 2, then find the value of   +   .,  x,  x,  y,  y, (a) 2, , (b) 1, , (c) 8, , (d) 11, , 13. If (33x − 11) (27y − 9) = 2592 where x and y are non-negative integers, then find, the value of 3x + 2y., (a) 20, (b) 3, (c) 19, (d) 6, 14. If (2401)3x − 5 = (49)3y − 1 = 75x − 6, then find the value of x + y., (a) 1, (b) 2, (c) 3, , (d) 7, , Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (i), , 4, , Column B, (A) 1, , 256, , (ii) Unit’s digits of, , 3, , 1, 40 , 608, , (B) 2, , (iii) Unit’s digits of 52, 27 , 229, , (C) 3, , (iv) 5°, , (D) 4, , (a), (b), (c), (d), , (i) → (A);, (i) → (B);, (i) → (D);, (i) → (D);, , M07_IIT-FOUNDATION_XXXX_CH07.indd 6, , (ii) → (B);, (ii) → (C);, (ii) → (B);, (ii) → (C);, , (iii) → (C); (iv) → (D), (iii) → (D); (iv) → (A), (iii) → (C); (iv) → (A), (iii) → (B); (iv) → (A), , 4/11/2018 4:37:26 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 7.7, , Space for rough work, , Direction for questions 1 to 14: Select the correct answer from the given options., 1. Find the square root of 729 using division method. Find the sequence of, steps in solving the above problem., (A) Now, the new divided is 329., (B) Place a bar over 29 and select the divisor as 2, since 22 < 7., (C) Subtract 4 from 7., (D) Double the quotient (2), it is ten’s digit of the divisor., (E) Write 47 × 7 = 329 under the new dividend and subtract., (F) The unit’s digit of the new divisor 7., (a) BCADFE, (b) BACDFE, (c) BCADEF, (d) BCFDAE, 2. Find the positive value of x, when 9(x − 5)2 = 1296., (a) 12, (b) 18, (c) 16, , (d) 17, , 3. What is the difference of the cube roots of the perfect cubes 17,576 and 4913?, (a) 5, (b) 7, (c) 9, (d) 11, 4. The minimum number of digits in the cube of a 4-digit number is _________., (a) 9, (b) 8, (c) 10, (d) 11, 5. The smallest number by which 432 should be multiplied so as to get a perfect square is _________., (a) 4, (b) 3, (c) 5, (d) 6, x, × 2p r, then find the value of x., 360, (b) 144, (c) 150, (d) 132, , 6. If l = 12p, r = 15 cm, and l =, (a) 120, , 7. What is the least whole number that should be added to 4091 so that the, sum obtained is a perfect square?, (a) 3, (b) 12, (c) 5, (d) 9, 8. If 40,824 = 23 × 3x × 71, then the value of x is _________., (a) 2, (b) 3, (c) 7, 9. If x = 7y and y = 7z, then find the value of, (a) 14, , M07_IIT-FOUNDATION_XXXX_CH07.indd 7, , (b) 8, , 8 x3, ., 49 z2 y, (c) 7, , (d) 6, , 3, , (d) 12, , 4/11/2018 4:37:27 PM

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7.8, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, , 10. Evaluate:, (a) 2, 11., , (, , 4, , 8−45, , )(, , 8+ 5, , (b) 3, , ) .( ) .(, ( xa .xb .xc )15, , 8, xa + b, , (a) 0, , (, , 8, xb + c, , 5, , ), , 8, xc + a, , )(, , 4, , 8+45, , ), , Space for rough work, (c) 7, , (d) 5, , (c) x, , (d) xa + b + c, , = _________, , (b) 1, , 5, , 7, , 7, ,  a  12  b  12,  b  10,  a 8, 12. If   +   = 2, then find the value of   −   .,  b,  a,  a,  b, (a) 0, (b) 2, (c) 1, , (d) 10, , 13. If (210x - 6) × (55y - 5) = 50,000, then find the value of 2x + 5y., (a) 12, , (b) 5, , (c) 10, , 14. If (81)3x − 5 = (9)3y + 2 = 35x − 6, then what is the value of x − y?, (a) 1, (b) 2, (c) 3, , (d) 6, , (d) 5, , Direction for question 15: Match the values of Column A with those of Column B., 15., Column A, (i), , 3−5, , (ii), , 2−3, , (iii), , 4−3, , (iv), , 52, , (a), (b), (c), (d), , Column B, , ×, , 93, , ×, , 3−3, , ×, , 64, , ×, , 9−2, , ×, , 363, , ×, , 62, , ×, , (i) → (A);, (i) → (D);, (i) → (C);, (i) → (D);, , M07_IIT-FOUNDATION_XXXX_CH07.indd 8, , (A) 10, , (30)−1, , (B) 9, (C) 6, ×, , 3−1, , (ii) → (B);, (ii) → (C);, (ii) → (D);, (ii) → (B);, , (D) 3, (iii) → (C); (iv) → (D), (iii) → (B); (iv) → (A), (iii) → (A); (iv) → (B), (iii) → (C); (iv) → (A), , 4/11/2018 4:37:30 PM

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Answer Keys, , 7.9, , Answer Keys, Assessment Test I, 1. (d), 11. (b), , 2. (d), 12. (d), , 3. (d), 13. (c), , 4. (c), 14. (b), , 5. (a), 15. (a), , 6. (a), , 7. (c), , 8. (b), , 9. (a), , 10. (c), , 4. (b), 14. (c), , 5. (c), 15. (b), , 6. (a), , 7. (b), , 8. (a), , 9. (b), , 10. (d), , 4. (d), 14. (c), , 5. (a), 15. (c), , 6. (a), , 7. (a), , 8. (c), , 9. (c), , 10 (a), , 4. (c), 14. (b), , 5. (b), 15. (b), , 6. (b), , 7. (c), , 8. (d), , 9. (a), , 10. (b), , Assessment Test II, 1. (c), 11. (b), , 2. (c), 12. (e), , 3. (b), 13. (d), , Assessment Test III, 1. (b), 11. (b), , 2. (c), 12. (a), , 3. (d), 13. (c), , Assessment Test IV, 1. (a), 11. (d), , 2. (d), 12. (a), , M07_IIT-FOUNDATION_XXXX_CH07.indd 9, , 3. (c), 13. (a), , 4/11/2018 4:37:30 PM

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Time and Work,, Time and Distance,, and Statistics, , 8, , Reference: Coursebook - IIT Foundation Mathematics Class 8; Chapters - Time and Work; Time and Distance; Statistics;, pp. 10.1–10.9; 11.1–11.13; 14.1–14.14, , Assessment Test I, , Time: 30 min., , Direction for questions 1 to 11: Select the correct answer from the given options., , Space for rough work, , 1. In a pie diagram, a certain component is represented by 60° and the total, value of the components is 2160. Find the sequential order to calculate the, value of the component., x, × 360° = 60°, (A), 2160, Component value, (B) Degree of the component =, × 360°, Total value, (C) Let the component value be x., (D) x = 360, (a) ABCD, (b) BCAD, (c) CABD, (d) CBAD, 2. If 13 men can dig a trench of 26 m long in 18 days, working 8 h a day, then how, many more men should be engaged to dig a similar trench of 30 m long in, 6 days, working 6 h a day?, (a) 51, , (b) 47, , (c) 61, , (d) 72, , 3. A is twice as good a workman as B and C is thrice as good a workman as B. If, A, B and C together can complete a work in 2 days, then in how many days, can A alone complete the work?, (a) 9, , (b) 3, , (c) 12, , (d) 6, , 4. A and C started doing a work. After working for 6 days, both left and B, alone completed the remaining work in 4 days. In how many days can C, alone complete the work if A and B independently take 12 and 24 days,, respectively, to complete the work?, (a) 36, , (b) 18, , (c) 20, , (d) 30, , 5. Two pipes can fill a tank in 18 min and 24 min, respectively. Both are opened, simultaneously. After how many minutes should the first pipe be closed so, that the tank becomes full in 16 min?, (a) 6, , M08_IIT-FOUNDATION_XXXX_CH08.indd 1, , (b) 7, , (c) 8, , (d) 9, , 4/11/2018 4:37:34 PM

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8.2, , Chapter 8 Time and Work, Time and Distance, and Statistics, , 6. The distance between Mumbai and Pune is 192 km. A car started from, Mumbai towards Pune at 32 kmph. Another car started 2 1 h later from, 2, Mumbai and reached Pune 1 an hour earlier than the first car. Find the, 2, speed of the second car (in kmph)., (a) 64, , (b) 68, , (c) 72, , Space for rough work, , (d) 76, , 7. A person takes 20 h to travel a certain distance. If his speed is increased by, 25%, then what time will he take to travel the same distance?, (a) 24 h, , (b) 16 h, , (c) 22 h, , (d) 12 h, , 8. In a 500 m race, A beats B by 20 m and 5 s. What is the speed of A?, (a) 4 m/s, (c) 25/6 m/s, , (b) 25/3 m/s, (d) 5 m/s, , 9. A boat can cover 48 km downstream in 3 h. It can cover a 40 km upstream in, 4 h. What is the speed of the boat in still water (in kmph)?, (a) 13, , (b) 15, , (c) 17, , (d) 19, , 10. If the mode of a distribution is 12 and the mean is 3, then the median is, ______., (a) 8, , (b) 36, , (c) 4, , (d) 6, , 11. If the arithmetic mean of the following table is 12.95, then what is the value of, p?, x, , 4, , 7, , 9, , 13, , 18, , 20, , F, , 3, , 10, , 15, , p, , 12, , 2p – 4, , (a) 6, , (b) 8, , (c) 10, , (d) 14, , Direction for questions 12 to 15: Match the statements of Column A with those of, Column B., Column A, 12. A can do a piece of work in, 12 days and B can do the same, work in 6 days., 13. 8 men or 12 women can do a piece, of work in 10 days., 14. P travels at a speed of 32 km/h and, Q travels at a speed of 10 m/s., 15. The pipes D and E together can fill, the tank in 3 h, whereas the pipes D, and F fill it in 4 h., , Column B, (a) P is faster than Q., , (b) The working rate of 1 man is, equal to the working rate of, 2 women., (c) F takes less time to fill the tank, when compared with E., (d) Q is faster than P, , (e) The working rate of 2 men is equal, to the working rate of 3 women., , M08_IIT-FOUNDATION_XXXX_CH08.indd 2, , 4/11/2018 4:37:35 PM

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Assessment Test I, Column A, , Column B, , 8.3, , Space for rough work, , (f) E takes less time to fill the tank, when compared with F., (g) A is twice as efficient as B., (h) B is twice as efficient as A., , M08_IIT-FOUNDATION_XXXX_CH08.indd 3, , 4/11/2018 4:37:35 PM

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8.4, , Chapter 8 Time and Work, Time and Distance, and Statistics, , Assessment Test II, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 11: Select the correct answer from the given options., 1. X can do a piece of work in 6 days, whereas Y can do the same work in, 8 days. Arrange the following steps in sequential order to find in how many, days X and Y together can do the same work?, (A) One day’s work of X is 1/6 and one day’s work of Y is 1/8., (B) X can do a piece of work in 6 days and Y can do the same work in 8 days., 3, (C) X and Y together can do the same work in 3 days., 7, 7, (D) One day’s work of X and Y =, ., 24, (a) ABCD, (c) BDAC, , (b) BADC, (d) ABDC, , 2. If 18 women can manufacture 900 items in 10 days, then how many women, will be required to manufacture 660 items of the same kind in 12 days?, (a) 14, , (b) 15, , (c) 11, , (d) 10, , 3. A is twice as good a workman as B and working together they finish a work, in 14 days. In how many days can A alone finish the work?, (a) 19, , (b) 17, , (c) 43, , (d) 21, , 4. R and S can do a work in 10 and 15 days, respectively. They work together, for 5 days and then S leaves. In how many days will the remaining work be, completed?, (a) 3, , (b) 4, , (c) 6, , (d) 5/3, , 5. Pipe A can fill a tank in 12 min and pipe B can empty the tank in 8 min. If, both the pipes are opened when tank is half full, then in how many minutes, will the tank become empty?, (a) 18, , (b) 24, , (c) 6, , (d) 12, , 6. At 11:00 am, a thief escapes from a police station running at 30 kmph. At, 1:00 pm, police start chasing the thief with a speed of 40 kmph from the police, station. At what time will the police catch the thief?, (a) 5:30 pm, (c) 7:30 pm, , (b) 6:00 pm, (d) 7:00 pm, , 7. Travelling at 4/7 of his usual speed, a man gets late by 9 min. What time, does he take when he travels at his usual speed?, (a) 9 min, (c) 15 min, , (b) 12 min, (d) 25 min, , 8. In a 300 m race, Rohan beats Mohan by 60 m or 15 s. Find Rohan’s speed (in, m/s)., (a) 3.5, , M08_IIT-FOUNDATION_XXXX_CH08.indd 4, , (b) 4, , (c) 4.5, , (d) 5, , 4/11/2018 4:37:35 PM

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Assessment Test II, 9. The speed of a boat moving upstream is 28 kmph and the speed of the boat, moving downstream is 44 kmph. What is the speed of the boat in still water, and the speed of the stream? (in kmph), (a) 32, 12, (c) 24, 20, , 8.5, , Space for rough work, , (b) 36, 8, (d) 40, 4, , 10. If the difference between median and mean is 8, then the difference between, mode and median is ______., (a) 16, , (b) 18, , (c) 24, , (d) 20, , 11. Find the value of x, if for the given data mean = 10 and Sfi = 22., X, , 4, , 8, , 9, , 12, , 15, , F, , 5, , x, , 2, , 5, , y, , (a) 2, , (b) 4, , (c) 5, , (d) 6, , Direction for questions 12 to 15: Match the statements of Column A with those of, Column B., Column A, , Column B, , 12. A can do a piece of work in 8 days,, whereas B can do the same piece of, work in 10 days., , (a) Car takes more time when, compared with train to cover, 100 km., , 13. 5 men can do a work in 9 days., , (b) Train takes more time when, compared with car to cover, 50 km., , 14. A car covers a certain distance with, a speed of 5 m/s, whereas the train, covers the same distance with a, speed of 20 kmph., , (c) B is more efficient than A., , 15. The pipes P and Q can fill the tank, in 40 min, whereas Q and R can fill, the tank in 45 min., , (d) A is more efficient than B., , (e) 1 man can do the work in, 45 days., (f)     1 man can do the work in, 40 days., (g) R takes more time to fill the tank, when compared with P., (h) R takes less time to fill when, compared with P., , M08_IIT-FOUNDATION_XXXX_CH08.indd 5, , 4/11/2018 4:37:35 PM

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8.6, , Chapter 8 Time and Work, Time and Distance, and Statistics, , Assessment Test III, , Time: 30 min., , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. In a pie diagram, a certain component is represented by 80° and the total, value of all the components is 3240. Arrange the following steps in sequential order, involved in finding the value of the component., 80° × 3240, (A) x =, 360°, (B) x = 20° × 36, (C) Let the value of the component be x., (D) Degree of the component =, (a) CDAB, , (b) CDBA, , x, × 360° = 80°, 3240, (c) DCAB, , (d) CADB, , 2. P and Q can do a work in 24 days, Q and R in 18 days and P and R in, 12 days. In how many days can Q do the work?, (a) 144, (b) 150, (c) 100, (d) 128, 3. A certain number of men can do a work in 12 days working 6 h a day. If the, 1, number of men is decreased by , then in how many days can thrice the, 4, previous work be completed by the remaining men working 8 h per day?, (a) 32, (b) 36, (c) 24, (d) 40, 4. M and N can do a piece of work in 8 days and 12 days, respectively, and, with the help of P, they complete the work in 3 days and earn `1200. Find, the share of N., (a) `200, (b) `500, (c) `400, (d) `300, 5. Two pipes X and Y can fill a tank in 3 h and 4 h, respectively, whereas another, pipe Z can empty the tank in 6 h. If all the three pipes are opened at the same, time, then find the time in which the tank can be filled (in hours)., 2, 2, 1, 1, (a) 2, (b) 2, (c) 2, (d) 2, 3, 5, 2, 4, 6. Three men and four women can complete a job in 12 days. 12 men and, 4 women can complete it in 4 days. How much time does a woman take to, complete it (in days)?, (a) 144, (b) 120, (c) 98, (d) 84, 7. A boat travels a distance of 25 km upstream in 5 h. If the speed of the boat in, still water is 6 km/h, then find the speed of the current., , M08_IIT-FOUNDATION_XXXX_CH08.indd 6, , 4/11/2018 4:37:38 PM

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Assessment Test III, (a) 1 km/h, (c) 3 km/h, , (b) 2 km/h, (d) 4 km/h, , 8.7, , Space for rough work, , 8. In 1000 m race, P beats Q by 100 m and Q beats R by 80 m. By how much, distance can P beat R in the race?, (a) 180 m, (b) 192 m, (c) 172 m, (d) 200 m, 9. A train 300 m long crosses a pole in 15 s. Find the time a train takes to cross, a platform of length 180 m (in seconds)., (a) 20, (b) 24, (c) 30, (d) 26, 10. Leaving his house and travelling at 6 kmph a person reached his office 12, min late. Had he travelled at 8 kmph, he would have reached 10 min early., Find the distance from his house to the office., (a) 9 km, (b) 12 km, (c) 8.8 km, (d) 9.6 km, 11. How long will a train 120 m long travelling at a speed of 92 kmph take to, overtake another train 240 m long travelling at a speed of 74 kmph starting, from the same point on parallel tracks in the same direction?, (a) 24 s, (b) 36 s, (c) 48 s, (d) 72 s, 12. A bar graph is drawn to the scale of 1 cm = K units. The length of the bar, representing a quantity 720 units is 6 cm. Find the value of K., (a) 100, (b) 120, (c) 150, (d) 160, 13. If the mean of the following data is 6.5, then which of the following is the value, of P?, x, , 2, , 3, , 6, , 7, , P, , f, , 2, , 4, , 5, , 7, , 4, , (a) 12, , (b) 8, , (c) 10, , (d) 9, , 14. If the mode of a data is 14 and the mean is 17, then find the median of the, data., (a) 15, (b) 16, (c) 18, (d) 15.5, 15. Match the values of Column A with those of Column B., Column A, , Column B, , (i) 12, 14, 13, 12, 14, 12, 14, 12, , (A) Mode is ill-defined., , (ii) 14, 13, 12, 12, 13, 13, 14, 14, , (B) Median is 13., , (iii) 11, 12, 13, 14,, , (C) Mode is 12., , (iv) 11, 11, 12, 13, 14, 14, 13, 13, , (D) Mean is, , M08_IIT-FOUNDATION_XXXX_CH08.indd 7, , 101, ., 8, , 4/11/2018 4:37:39 PM

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8.8, , Chapter 8 Time and Work, Time and Distance, and Statistics, (a), (b), (c), (d), , (i) → (A); (ii) → (B); (iii) → (C); (iv) → (D), (i) → (C);   (ii) → (B);  (iii) →(A);   (iv) → (D), (i) → (C);   (ii) → (A);    (iii) → (B); (iv) → (D), (i) → (B);   (ii) → (C);  (iii) → (A); (iv) → (D), , M08_IIT-FOUNDATION_XXXX_CH08.indd 8, , Space for rough work, , 4/11/2018 4:37:39 PM

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Assessment Test IV, , Assessment Test IV, , Time: 30 min., , 8.9, , Space for rough work, , Direction for questions 1 to 15: Select the correct answer from the given options., 1. Construct a histogram for the frequency distribution given below:, CI, , 0–10, , 10–20, , 20–30, , 30–40, , 40–50, , F, , 3, , 4, , 7, , 9, , 5, , Arrange the following steps in the sequential order, involved in the construction of histogram., (A) Find a suitable scale on X-axis and Y-axis., (B) Analyse the data given in the table., (C) Construct the rectangles with bases along the X-axis and the heights, along the Y-axis., (D) Represent the class boundaries along the X-axis and frequencies along, the Y-axis., (a) BCAD, (b) BADC, (c) BCDA, , (d) BDAC, , 2. X and Y can do a piece of work in 20 days, Y and Z in 24 days, and X and Z, in 30 days. In how many days can X do the work?, (a) 38, (b) 48, (c) 42, (d) 54, 3. A certain number of men can do a piece of work in 15 days working 9 h a day., 1, If the number of men is increased by , then in how many days can twice the, 4, previous work be completed by the total men working 6 h per day?, (a) 30, (b) 32, (c) 36, (d) 40, 4. A and B can do a piece of work in 6 days and 12 days, respectively. With the, help of ‘P’, all the three of them, together can complete the work in 2 days, and earn `2,400. Find the share of ‘P’., (a) `800, (b) `600, (c) `1,200, (d) `1,000, 5. Two pipes M and N can fill a tank in 20 min and 30 min, respectively, whereas, another pipe ‘P’ can empty the tank in 24 min. If all the three pipes are turned, on at the same time, then find the time in which the empty tank can be filled., (a) 20 min, (b) 18 min, (c) 30 min, (d) 24 min, 6. Five men and three women can complete a piece of work in 20 days. Nine, men and 8 women can complete it in 8 days. Find out in how much time a, man can complete it (in days)., (a) 480, (b) 520, (c) 260, (d) 500, , M08_IIT-FOUNDATION_XXXX_CH08.indd 9, , 4/11/2018 4:37:39 PM

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8.10, , Chapter 8 Time and Work, Time and Distance, and Statistics, , 7. A boat travels a distance of 24 km downstream in 3 h. If the speed of the boat, in still water is 5 km/h, then find the speed of the current (in kmph)., (a) 1, (b) 2, (c) 3, (d) 4, , Space for rough work, , 8. In a 1000m race, A beats B by 125 m and B beats C by 100 m. By how much, distance can A beat C in the race (in m)?, (a) 212.5, (b) 225, (c) 180.5, (d) 206.5, 9. A train 450 m long crosses a platform of length 150 m in 20 s. Find the time, taken by the train to cross a pole (in seconds)., (a) 15, (b) 20, (c) 25, (d) 10, 10. A person leaves his house and travelling at 8 kmph reaches his office 6 min, early. Had he travelled at 5 kmph, he would have reached 9 min late. Find, the distance from his house to the office., 1, (a) 3 km, (b) 5 km, 2, 1, (c) 4 km, (d) 3 km, 3, 11. How long will a train 160 m long and travelling at a speed of 100 kmph, crosses another train 200 m long, travelling at a speed of 44 kmph in the, opposite direction?, (a) 19 s, (b) 12 s, (c) 15 s, (d) 9 s, 12. A bar graph is drawn to the scale of 1 cm = 40 units. Find the length of the, bar representing a quantity of 580 units., (a) 15.5 cm, (b) 12 cm, (c) 14.5 cm, (d) 18 cm, 13. The mean of the following data is 9. Find the value of K., x, , 6, , 7, , 8, , 9, , 10, , 12, , f, , 10, , 8, , 12, , 10, , K, , 15, , (a) 11, , (b) 12, , (c) 13, , (d) 14, , 14. If the mode of a data is 16 and median is 15, then find the mean of the data., (a) 14.5, (b) 16, (c) 17, (d) 13.5, 15. Match the statements of Column A with those of Column B., Column A, , Column B, , (i) Speed, , Total distance covered, (A)   , Total time taken, , (ii) Average speed, , M08_IIT-FOUNDATION_XXXX_CH08.indd 10, , (B) Sum of the speeds, , 4/11/2018 4:37:40 PM

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Answer Keys, Column A, , Column B, , (iii) R,  elative speed (Opposite, direction), , (C), , 8.11, , Space for rough work, , Distance, Time, , (iv)   Relative speed (same direction) (D) Difference of the speeds., (a), (b), (c), (d), , (i) → (A); (ii) → (B);  (iii) → (C); (iv) → (D), (i) → (C); (ii) → (B);   (iii) → (D); (iv) → (B), (i) → (C); (ii) → (A); (iii) → (B); (iv) → (D), (i) → (C); (ii) → (A); (iii) → (D); (iv) → (B), , Answer Keys, Assessment Test I, 1. (d), 11. (b), , 2. (b), 12. (h), , 3. (d), 13. (e), , 4. (b), 14. (d), , 5. (a), 15. (f), , 6. (a), , 7. (b), , 8. (c), , 9. (a), , 10. (d), , 4. (d), 14. (a), , 5. (d), 15. (g), , 6. (d), , 7. (b), , 8. (d), , 9. (b), , 10. (a), , 4. (d), 14. (b), , 5. (b), 15. (b), , 6. (a), , 7. (a), , 8. (c), , 9. (b), , 10. (c), , 4. (c), 14. (a), , 5. (d), 15. (c), , 6. (b), , 7. (c), , 8. (a), , 9. (a), , 10. (d), , Assessment Test II, 1. (b), 11. (b), , 2. (c), 12. (d), , 3. (d), 13. (e), , Assessment Test III, 1. (a), 11. (a), , 2. (a), 12. (b), , 3. (b), 13. (a), , Assessment Test IV, 1. (b), 11. (d), , 2. (b), 12. (c), , M08_IIT-FOUNDATION_XXXX_CH08.indd 11, , 3. (c), 13. (c), , 4/11/2018 4:37:41 PM

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Hints and Explanation, CHAPTER 1, Geometry and Symmetry, Assessment Test I, R, + 140 ° = 180° [from Eqs (1) and (2)], 2, , 1. BADCE is the sequential order of steps., Hence, the correct option is (c)., , ⇒ 90 ° −, , 2. AB = AC and ∠A = 60°, ⇒ ∠A = ∠B = ∠C = 60°, ⇒ ∆ABC is an equilateral triangle., ⇒ The medians and altitudes are equal in length., ∴ The length of the altitude = 6 cm, Hence, the correct option is (b)., , ⇒ 230° − 180° =, , AC, 3. BD =, 2, , R, 2, , R, 2, ∴ ∠R = 100°, Hence, the correct option is (d)., ⇒ 50° =, , 5. PQRS is a concave quadrilateral., Reflex ∠QRS = 360° − (30° + 95° + 35°), = 360° − 160°, , B, , = 200°, ∴ ∠QRS = 360° − 200° = 160°, 25°, C, , D, , Hence, the correct option is (a)., , A, , 6. In an isosceles trapezium, angles made by a parallel side with the non−parallel sides are equal., , ⇒ BD = AD, ∴ In ∆ABD, ∠DBA = ∠DAB = 25°, ⇒ ∠BDC = 25° + 25° = 50°, 180 ° − 50 °, BD = CD ⇒ ∠CBD =, 2, , ⇒ ∠C = ∠D, But ∴ ∠A + ∠D = 180°, D, , ∠CBD = 65°, Hence, the correct option is (d)., 4., , 57°, A, , P Q R, + + = 90°, 2 2 2, , B, , ∠D = 180 − 57 = 123°, , ⇒, ,  R, P Q, = 90° −  , +,  2, 2 2, , (1), , ⇒, , P Q, + + ∠POQ = 180°, 2 2, , (2), , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 1, , C, , ∠C = ∠D = 123°, ∴ ∠C + ∠D = 246°, Hence, the correct option is (c)., , 4/11/2018 3:27:51 PM

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A.2, , Chapter 1 Geometry and Symmetry, 11. Join AF, , 7., S, , R, , A, 30° 80°, F, , O, , 40°, P, , Q, , E, , PS = SQ (given), ⇒ SPQ = ∠PQS = 40°, ⇒ PSQ = 180° − 40° − 40° = 100°, ⇒ SQR = 100° PS = SQ = QR, ∴ ∠SPQ = ∠SRQ = 40°, ∴ ∠SQR − ∠QRS, = 100° − 40° = 60°, Hence, the correct option is (a)., 8. ∠MNK + ∠NKL = 180° ⇒ ∠NKL = 80°, N, , M, , 100°, , 80°, L, , K, , In an isosceles trapezium, the angles made by, a parallel side with the non−parallel side are, equal., ⇒ ∠NKL = ∠KLM = 80°, ∴ Supplement of ∠KLM = 100°, Hence, the correct option is (d)., 9. Angle in a semicircle is 90°., ∴ The triangle is right angled triangle., Hence, the correct option is (c)., , 70°, D, , 30°, , B, C, , AFDB is a cyclic quadrilateral., ∠BDF = 70° (given), ⇒ ∠FAB = 180 − 70° = 110°, ∠FCE = ∠FAE = 30°, ⇒ ∠EAB = 110° − 30° = 80°, Hence, the correct option is (b)., 12. Option (a): Incentre always lies inside the, triangle., 13. Option (d): Excentre always lies outside the, triangle., 14. Option (b): Orthocentre can be the vertex of the, triangle., 15. Option (c): Circumcentre lies on the hypotenuse., , Assessment Test II, 1. CABED is the sequential order of steps., Hence, the correct option is (d)., 2. The centroid and the orthocentre of a triangle, coincide with an equilateral triangle., P, , 10. ∠OCB = ∠OBC = 20°, , G, , A, Q, O, B, , 20° 20°, , ∴ PG =, C, , ⇒ BOC = 180° − (20° + 20°) = 140°, 140°, ∠BAC =, = 70°, 2, Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 2, , T, , R, , 2, (PT ), 3, , 2, × 9 = 6 cm, 3, Hence, the correct option is (b)., =, , 3. PS = SR and ∠PRS = 55°, ⇒ ∠SPR = ∠PRS = 55°, , 4/11/2018 3:27:54 PM

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Hints and Explanation, ⇒ ∠N = ∠M and ∠K = ∠L, And ∠N + ∠K = 180°, ⇒ ∠M + ∠K = 180°, ⇒ ∠LMN = 180° − 70° = 110°, Hence, the correct option is (d)., , P, 55°, 110°, Q, , 55°, , S, , R, , ⇒ ∠QSP = 55° + 55° = 110°, In ∆QPS, ∠QPS = ∠SQP =, , 180° − 110°, 2, , A.3, , 7., T, , P, , M, , Z, , S, , Q, , R, , I, , SR = TM, SR TM, =, 2, 2, ⇒ MR = PM, , = 35°, ∴ ∠SQP = 35°, Hence, the correct option is (b)., 4., , 100, X, , Y, , ∠X + ∠Y + ∠Z = 180, , Draw PQ parallel to MR. PQRM is a, rhombus., , ∠X ∠Y ∠Z, ⇒, = 90, +, +, 2, 2, 2, ∠Z, = 90°, ⇒ 80° +, 2, ∠Z, ⇒, = 10°, 2, ⇒ ∠Z = 20°, , ∠SRM = 108° (given), 108°, = 54°, ⇒ ∠QRP = ∠PRM =, 2, ⇒ ∠QPR = 54°, In ∆SQP, ∠SQP = 108° = ∠SRM, , ∴ ∠XZY = 20°, Hence, the correct option is (b)., 5. In quadrilateral ADBC,, ∠A + ∠B + ∠C + ∠D = 360°, ⇒ 30° + 40° + x + 360° − y = 360°, ⇒ 70° + x − y = 0°, ∴ y − x = 70°, Hence, the correct option is (d)., , SQ = QP ⇒ ∠SPQ =, , ∴ ∠SPR = 54° + 36° = 90°, Hence, the correct option is (a)., 8., S, , K, , 50°, , 65°, 65° 65°, P, , M, , L, , In an isosceles trapezium, angles made by a, parallel side with the non−parallel sides are, equal., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 3, , R, , 65°, 50°, , 6. Given ∠NKL = 70°, N, , 180° − 108° 72°, = 36°, =, 2, 2, , 50°, , T, , Q, , ∴ ∠SRQ cannot be determined., Hence, the correct option is (d)., 9. Circumcentre lies outside only for an obtuse−, angled triangle., Hence, the correct option is (b)., , 4/11/2018 3:27:57 PM

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A.4, , Chapter 1 Geometry and Symmetry, , 10., , Given, ∠LCD = (3x)°, ∠PDC = (2x + 50)°, ⇒ 3x + (2x + 50)° = 180°, ∴, ( LM  PQ, ∠LCD and ∠PDC are, supplementary), ⇒5x + 50° = 180°, ⇒5x = 130°, ⇒x = 26°, Hence, the correct option is (a)., , D, O, F, , E, , ∠DEF = 70° ⇒ ∠DOF = 140°, 180 − 140, ⇒ ∠DFO =, = 20°, 2, Hence, the correct option is (a)., 11., , 3., , E, , A, 40, , O, , C, 20, x, , D, , B, , A, , C, , Join BO ⇒ ∠BOD = 2 × ∠BAD, ⇒ ∠BOD = 2 × 40 = 80°, 80°, ⇒ ∠BOC = ∠COD =, = 40°, 2, ∠ABE = ∠ADE = 20° (∴ angles in the same, segment), ∴ ∠COD + ∠ABE = 40° + 20° = 60°, Hence, the correct option is (c)., 12. Option (c): Circumcentre, , 40°, , D, , y, E, , 2x, , B, , Given, AB CD, Clearly, x = 40° (alternate angles), , 13. Option (b): Orthocentre, 14. Option (d): Centroid, , ∴ 40° + y + 2 (40°) = 180°(∠AEB is a straight, angle.), , 15. Option (a): Incentre, , y + 120° = 180°, y = 60°, , Assessment Test III, , Hence, the correct option is (c)., , 1. DCAB is the required sequential order., Hence, the correct option is (c)., 2., C, , L, , M, , (3x)°, , P, , ∴The number of diagonals =, =, , n(n − 3), 2, , 12(12 − 3), = 54, 2, , Hence, the correct option is (c)., , (2x + 50)°, , Q, , D, , 5. Given n = 1947., Sum of the interior angles = (2n − 4) 90°, = [2(1947) − 4] 90° = 350, 100°, , B, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 4, , 4. Given, n = 12, , Hence, the correct option is (a)., , 4/11/2018 3:27:59 PM

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A.6, , Chapter 1 Geometry and Symmetry, Given, ∠ABC = 60° and ∠ADC = 100°, , 12., C, , D, , Let ∠BAE = ∠ECD = x° and ∠EAD = ∠ECB = y°, ∴ Reflex ∠ADC = 360° − 100° = 260°, , x°, , y° + 40°, , ∴ ∠BAD + Reflex ∠ADC + ∠DCB + ∠CBA, = 360°, (x° + y°) + 260° + (x° + y°) + 60° = 360°, , x° − 20°, , 2 (x° + y°) = 40°, , A, , x° + y° = 20°, , B, , Given, ABCD is a parallelogram., , ∴ ∠EAD + Reflex ∠ADC + ∠DCE + ∠CEA =, 360°, , ∠A + ∠D = 180°, , ⇒ y°+ 260°+ x° + ∠AEC = 360°, , x° + (x° − 20°) = 180°, , ⇒ ∠AEC = 360° − 260° − 20°, , 2x° = 200°, , ⇒ ∠AEC = 80°, , ⇒x° = 100°, , Alternatively:, , And ∠A° = ∠C°, y° + 40° = x − 20°, , 180° − 60° − 2(x° + y°)(1), , y° = 40°, , 180° −∠AEC − (x° + y°)(2), , ∴ x° = 100°; y° = 40°, , = 180° − ∠DAC + ∠DCA, , Hence, the correct option is (c)., , 180° − 100° = 80°(3), From Eqs (1) and (3), x° + y° = 20°, From Eqs (2) and (3), ∠AEC = 180° − 80° − 20°, = 80°, Hence, the correct option is (c)., , 13., B, , A, , 11., , D, , B, 100°, , A, , C, , 160°, O, 80°, , Given, ∠AOC = 160° Reflex ∠AOC = 360°, − 160° = 200°, We have, reflex ∠AOC = 2 ∠ABC ⇒ 200° = 2, ∠ABC, ⇒∠ABC = 100°, Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 6, , E, , C, , We have ∠BAC = 60° and ∠EAD = 90°, ⇒∠DAC = 45° and ∠ABD = 30°, Now, ∠ADB = 180° − ∠BAD − ∠ABD, = 180° − (60° − 45°) − 30°, = 135°, Hence, the correct option is (b)., 14. The measure of each exterior angle of a regular, polygon must be a factor of 360°., However, 16 is not a factor of 360., Hence, the correct option is (d)., , 4/11/2018 3:28:02 PM

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Hints and Explanation, 15. (i) → (D), (ii) → (A), (iii) → (C), (iv) → (B), , ∴ y = 3 (10°), y = 30°, Hence, the correct option is (b)., , Hence, the correct option is (d)., , Assessment Test IV, 1. The required sequential order is BDAC., Hence, the correct option is (a)., , A.7, , 4. Given, n = 21, n(n − 3), ∴ The number of diagonals =, 2, 21(21 − 3) 21(18), =, = 189, 2, 2, Hence, the correct option is (b)., =, , 2., M, (4x − 60)°, A, , E, , F, , C, , 5. Sum of the interior angles = (2n − 4) 90°, ⇒ (2n − 4) 90° = 362, 340°, ⇒ 2n − 4 = 4026, ⇒ 2n = 4030, ⇒ n = 2015, Hence, the correct option is (c)., , B, , D, , (2x + 20)°, , 6. A parallelogram is not line symmetric., Hence, the correct option is (a)., , N, , Given, ∠MEA = (4x − 60)°, and ∠DFN = (2x + 20)°, ⇒ (4x − 60)° = (2x + 20)°, (∴Alternate exterior angles), ⇒ 4x − 2x = 20° + 60°, ⇒ 2x = 80°, ⇒x = 40°, Hence, the correct option is (d)., , 7. O,  ptions (a) and (b) are line symmetric. However, ‘R’ is not line symmetric., Hence, the correct option is (c)., 8., 24 M, , P, , O, , 24 Q, , 24, , 50, , K, , S, R, , 3., F, C, E, , y, , 3x, 120°, 3x, B, A, , Given, AB CD, y = 3x (Alternate angles), We have ∴ 3x +y + 120° = 180°, 3x + 3x = 60°, 6x = 60°, x = 10°, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 7, , Given, PQ = RS = 48 cm, 2 QK = PQ ⇒ QK = 24 and OK = 50 cm, Let ‘M’ be the midpoint on PQ., ⇒ OM ⊥ PQ, In ∆OMK, OK2 = OM2 + MK2, 502 = OM2 + 482, ⇒ OM2 = 196, ⇒ OM = 14 cm, In ∆OMQ, OQ2 = OM2 + MQ2, ⇒ OQ2 = 142 + 242, ⇒ OQ2 = 196 + 576, ⇒ OQ2 = 772, ∴ OQ = 2 193, Hence, the correct option is (c)., , 4/11/2018 3:28:05 PM

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Hints and Explanation, ∴PQRS is an isosceles trapezium., S, x° − 30°, , 13. We have AD = AE., ∠DAE = 90° + 60° = 150°, , R, , 180° − 150°, = 15°, 2, Hence, the correct option is (a)., ⇒∠ADE = ∠AED =, , y° + 40°, , x° + 20°, P, , ∴x° − 30° + x° + 20° = 180°, ⇒ 2x° = 190° ⇒ x° = 95°, and y° + 40° = x° − 30°, y° = 95° − 30° − 40°, ∴ y° = 25, , A.9, , Q, , 14. The measure of each exterior angle of a regular, polygon must be a factor of 360°., ∴ Option (c) follows., Hence, the correct option is (c)., 15. (i) → (C); (ii) → (A); (iii) → (D); (iv) → (B), Hence, the correct option is (b)., , Hence, the correct option is (b)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 9, , 4/11/2018 3:28:06 PM

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A.10, , Chapter 2 Linear Equations and Inequations, , CHAPTER 2, Linear Equations and Inequations, Assessment Test I, 1. The following are the steps involved in solving, the simultaneous linear equations 3x + 5y = 11, and 2x + 3y = 7 by elimination method:, (C) 3x + 5y = 11 be the Eq. (1) and 2x + 3y = 7, be the Eq. (2)., (D) Multiply Eq. (1) by 2 and Eq. (2) by 3., (B) Subtract Eq. (2) × 3 from Eq. (1) × 2., (A) We get y = 22 − 21 = 1., (E) Substitute y = 1 in Eq. (1) and we get x = 2., , ⇒ 6 = −x + 7 ⇒ x = 1, Hence, the correct option is (a)., 5., , 3 x 72 x − 28, 16 + 6 x, +, −1 =, 2, 56, 56, 84 x + 72 x − 28 − 56 16 + 6 x, =, 56, 56, 156x − 84 = 16 + 6x, 156x − 6x = 16 + 84, 150x = 100, , \ CDBAE is the sequential order of steps., Hence, the correct option is (c)., 2. 2x + 3y = 8, , (1), , 6x + ky = 24, (2), Equations (1) and (2) have infinite solutions., 2 3, 8, \ = =, 6 k 24, 1 3, = ⇒ k = 9., 3 k, Hence, the correct option is (c)., , 10 2, =, 15 3, Hence, the correct option is (c)., x=, , 6., , 3. 201x + 102y = 504, 102x + 201y = 405, −,   −, 99x − 99y = 99, 99 (x − y) = 99,     x − y = 1, Hence, the correct option is (c)., 4., , 3x − 2 7 8x + 1 1, + =, −, 2, 8, 6, 8, 7 1 8x + 1 3x − 2, + =, −, 8 8, 6, 2, 8 8 x + 1 − (9 x − 6 ), =, 6, 8, −x + 7, 1=, 6, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 10, , 3 x 4  9x 7 , 2, 3x , + , −  − 1 = 1+ , , , , 2 7 4 8, 7, 8, 3 x 4  18 x − 7 , 2  8 + 3x , + , −1 = , , , 2 7 8 , 7 8 , , 7., , 3 x − 8 3 x + 10, <, 2, 5, 15x − 40 < 6x + 20, 9x < 60, 60, x<, 9, 6, x<6, 9, x = 1, 2, 3, 4, 5, 6,, Hence, the correct option is (d)., 3, −, x, 4, +, x, , 5 2, = (1), y 3, 3 5, = (2), y 2, , Let, , 1, 1, = a and = b, x, y, , Eqs (1) and (2) changes as, 3a − 5b =, , 2, (3), 3, , 4/11/2018 3:28:11 PM

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Hints and Explanation, 5, 4a + 3b = (4), 2, 1, 1, Solving Eqs (3) and (4), we get a = and b = ., 2, 6, 1, 1, \ =a= ⇒x=2, x, 2, 1, 1, =b= ⇒y=6, y, 6, x+y=8, Hence, the correct option is (b)., 8. Let the unit’s digit be x and the ten’s digit be y., x+y=x+2⇒y=2, Number formed by reversing the digits is 10x + y, and the original number is 10y + x., 10x + y = 10y + x + 7x, 10x + y − 8x − 10y = 0, 2x − 9y = 0, 2x − 9 × 2 = 0, 2x = 18 ⇒ x = 9, Product of the digits xy = 9 × 2 = 18, Hence, the correct option is (b)., 9. Let the number of questions answered correctly, be x and that of those answered wrongly be y., x + y = 60, , (1), , y,  −1, 2x +   y = 80 ⇒ 2 x − = 80,  2, 2, 4x − y = 160 , (2), Eq. (1) × 4 – Eq. (2), 4x + 4y = 240, 4x − y = 160, − +,   5y = 80,   y = 16, \ Number of questions answered wrongly = 16, Hence, the correct option is (b)., 10. Let A’s salary = x and B’s salary = y, 3, 5, Given x + y = 16,000 and x − y = 2000, 4, 3, ⇒ 9x + 20y = 192,000 and x − y = 2000, Solving the above equations, we get x = 8000, and y = 6000., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 11, , A.11, , \ B’s salary is `6000., Hence, the correct option is (c)., 11. Let the number of `5 coins be x., ∴ The number of `2 coins = 2x, and the number of `1 coins = 3x, Given 5 × x + 2 × 2x + 1 × 3x = 60, 12x = 60 ⇒ x = 5, ∴ The number of `1 coins = 3x, = 3 × 5 = 15, Hence, the correct option is (a)., 12. Option (e): (5, 4); 3(5) − 4(4) + 1 = 16 − 16 = 0, 8 7 15, + =, 2 2 2, 2, 14. Option (f): (9, 1); × 9 − 5 × 1 = 6 − 5 = 1, 3, , 13. Option (d): (2, 2);, , 15. Option (c): (3, 2); 4 × 3 −, , 3, × 2 = 12 − 3 = 9, 2, , Assessment Test II, 1. The following are the steps involved in solving, the linear equations 3x + y = 17 and 5x - 3y = 19, by substitution method:, (C) 3x + y = 17 be the Eq. (1) and 5x − 3y = 19, be the Eq. (2), (A) Find the value of y in terms of x form Eq., (1) i.e., y = 17 - 3x, (B) Substitute the value of y in Eq. (2), i.e.,, 5x - 3(17 − 3x) = 19, (D) We get the value of x, i.e., 14x = 19 + 51, ⇒x=5, (E) , Substituting the value of x, we get the, value of y, i.e., y = 17 - 15 = 2, \ CABDE is the sequential order of steps., Hence, the correct option is (c)., 2. 5x − 10y = 8, 7x − ky = 10 have no solution., 5 −10 8, =, ≠, 7 −k 10, k = 14, Hence, the correct option is (d)., , 4/11/2018 3:28:13 PM

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A.12, , Chapter 2 Linear Equations and Inequations, , 3. 212x + 111y = 535, 111x + 212y = 434, −, −, −, 101x − 101y = 101, x−y=1, Hence, the correct option is (b)., 4., , 2x, 4, 5,  x 7, + 3  −  + 1 = ( x − 1) +,  2 3, 5, 5, 2, 2x, 3x, 4x 4 5, +, =, − +, 5 2−7 +1 5 5 2, 2 x 4 x 3 x −4 5, −, +, =, + +6, 5, 5, 2, 5 2, 4 x − 8 x + 15 x −8 + 25 + 60, =, 10, 10, 11x = 77, ⇒x=7, Hence, the correct option is (a)., , 4+x 7 3−x 5, + =, +, 3, 4, 2, 4, 16 + 4 x + 21 6 − 2 x + 5, =, 12, 4, 4 x + 37 11 − 2 x, =, 12, 4, 4x + 37 = 33 − 6x, 10x = −4, x = −2/5, Hence, the correct option is (a)., 5, 6. 4 x − ≤ x + 7, 2, 5, 4x − x ≤ 7 +, 2, 14 + 5, 3x ≤, 2, 19, 3x ≤, 2, 19, x≤, 6, 1, x≤3, 6, Non-negative integer values of x are 0, 1, 2, and 3., Hence, the correct option is (a)., 5 7, 7., + = 6, (1), x y, 5., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 12, , 6 9 15, + = , x y 2, 1, 1, = a and = b, x, y, 5a + 7b = 6, , (2), , Let, , (3), , 15, 6a + 9b = (4), 2, Solving Eqs (3) and (4), we get, 1, 1, a = and b =, 2, 2, 1, 1, = 2; y = = 2, a, b, \x+y=2+2=4, Hence, the correct option is (a)., \x=, , 8. Let the two-digit number be xy., Sum of digits, x + y = 8, (1), Its numerical value is 10x + y., Number formed by interchanging the digits, is yx., Its numerical value is 10y + x., 10x + y + 18 = 10y + x ⇒ 9y - 9x = 18, y - x = 2, (2), Solving Eqs (1) and (2), we get, x = 3 and y = 5, ∴ The number is 35., Hence, the correct option is (c)., 9. Let the number of questions attempted correctly be x and the number of questions, attempted wrongly be y.,  1, Given x + y = 48 and x(1) − y   = 21,  2, 2x − y, ⇒, = 21, 2, 2x − y = 42, Solving x + y = 48 and 2x − y = 42, we get x = 30, and y = 18., \ Number of questions attempted correctly = 30., Hence, the correct option is (b)., 10. Let A’s salary be x and B’s salary be y., 2, 7, Given x + y = 9000 and x + y = 6000, 3, 3, 2x + 7y = 27,000 , , (1), (2), , 4/11/2018 3:28:18 PM

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Hints and Explanation, Solving Eqs (1) and (2), we get x = 3000 and, y = 3000., \ Their difference of salaries is x − y = 3000, − 3000 = 0., Hence, the correct option is (b)., , ⇒x+y=2, Hence, the correct option is (b)., 3. Given,, , x, 2, , 3x, x, + 2 × x + 1 × = 100, 2, 2, 15x + 4x + x = 200 ⇒ 20x = 200, ⇒ x = 10, 3 x 3 × 10, =, ∴ Number of `5 coins =, = 15, 2, 2, Hence, the correct option is (a)., Given 5 ×, , 1, 1, <0 ⇒ x − 3, >0 ⇒, 12. Options (d): −, x−3, x−3, , 15. Options (b): 6x + 6 < 7x + 5 ⇒ x > 1, , Assessment Test III, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 13, , 5y −, , x − 2y = 1, 6x + 2y = 20 → (1) × 2, x − 2y = 1 → (2), 7x = 21, (1) × 2 + (2):, x=3, Substituting x = 3 in Eq. (1), we get, 3 (3) + y = 10 ⇒ y = 1, ∴ 3x − 2y + 5z = 17, ⇒ 3 (3) − 2 (1) + 5z = 17, 9 − 2 + 5z = 17, 5z =17 − 7 ⇒ 5z = 10 ⇒ z = 2, Hence, the correct option is (a)., , 14. Options (a): 8x − 8 > 5x − 2 ⇒ 3x > 6 ⇒ x > 2, , 2x + 5y = 1, 7x + 7y = 14, , 5y − 8, , 5. 3x + y = 10, , ⇒x>4, , 2. 5x + 2y = 13, , 9, , ⇒ 6x − 10 + 6x − 42 = 7x + 35 − 15, ⇒ 12x − 7x = 20 + 52, ⇒ 5x = 72, ⇒ x = 14.4, Hence, the correct option is (c)., , 15, < 5 ⇒ 15 < 5x − 5 ⇒ 5x > 20, x −1, , 1. The following are the steps involved in solving, the problem:, (B) Multiply 4x − 3y = 5 by 2 and 5x + 2y = 12, by 3., (A) (4x − 3y) × 2 = 5 × 2 ⇒ 8x − 6y = 10, (5x + 2y) × 3 = 12 × 3 ⇒ 15x + 6y = 36., (C) 8x − 6y = 10 should be added to 15x + 6y = 36., (D) Substitute the value of x in 4x − 3y = 5,, then find the value of y., ∴ BACD is the required sequential order., Hence, the correct option is (b)., , =, , 4. Given, 2 (3x − 5) + 6 (x − 7) = 7 (x + 5) − 15, , ⇒x<3, 13. Options (c):   , , 4, , 11, 2, ⇒ 20 y − 22 = 45y − 72, ⇒ 50 = 25 y, ⇒y=2, Hence, the correct option is (d)., , 11. Let the number of `2 coins be x., 3x, and, The number of `5 coins =, 2, The number of `1 coins =, , A.13, , (1), (2), , 6. 3x − 5 ≤ 5x − 9, 3x − 5 + 9 ≤ 5x, ⇒ 4 ≤ 5x − 3x, ∴ 4 ≤ 2x, ⇒2≤x, ∴x≥2, Hence, the correct option is (a)., 7., , 5x, <1, 5x + 7, ⇒, , 5x, −1<0, 5x + 7, , 4/11/2018 3:28:21 PM

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A.14, , Chapter 2 Linear Equations and Inequations, 5x − (5x + 7 ), <0, 5x + 7, −7, ⇒, <0, 5x + 7, ⇒, , ⇒ 5x + 7 > 0, −7, ⇒x>, 5, Hence, the correct option is (d)., 8. x + 7 > − 3 + 7x, x − 7x > − 3 − 7, − 6x > − 10, 6x < 10, 10, x<, 6, 5, x<, 3, Hence, the correct option is (a)., 9. Let the present ages of the person and his son, be x and y, respectively., x − y = 28, (1), x + 10 3, =, y + 10 1, x + 10 3, =, y + 10 1, ⇒ x + 10 = 3y + 30, x − 3y = 20, (2), Eq. (1) – Eq. (2), x − y = 28, x − 3y = 20, (−) (+) (−), 2y = 8, ⇒y=4, Since x − y = 28, x − 4 = 28, x = 32, The sum of their present ages = 32 + 4 = 36, Hence, the correct option is (b)., 10. The system of equations 3x − 2 (k − 1) y = 20 and, 5x − 3y = 25 is consistent., a, b, a, b, c, ⇒ 1 ≠ 1 or 1 = 1 = 1, a2 b2, a2 b2 c2, As, , a1, a2, , ≠, , c1, c2, , ,, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 14, , 3 −2 ( k − 1), ≠, 5, −3, 3 2k − 2, ≠, 5, 3, 9 ≠ 10k − 10, ⇒ 10k ≠ 19, k≠, , 19, 10, , Hence, the correct option is (b)., 11. Let the number of correctly answered and that, of wrongly answered questions be x and y,, respectively., x + y = 100 , , (1), , 1, y = 61, 2, , (2), , x−, , Eq. (1) – Eq. (2): y +, , y, = 39, 2, , 3y, = 39, 2, ⇒ y = 26, , ⇒, , ∴ x + 26 = 100 [from Eq. (1)], ⇒ x = 74, Hence, the correct option is (c)., 12. y ≤ x + 2, (0, 0); ⇒ 0 ≤ 0 + 2 (True), (1, 5); ⇒ 5 ≤ 1 + 2 (False), (4, 2); ⇒ 2 ≤ 4 + 2 (True), (6, 3); ⇒ 3 ≤ 6 + 2 (True), Hence, the correct option is (b)., a1 b1 c1, =, ≠ , then a1x + b1 y + c1 = 0 and, a2 b2 c2, a2x + b2 y + c2 = 0 have no solution., , 13. If, , Option (a):, , a1, a2, , =, , a, b, 2 b1, 3, =, ;, ⇒ 1 ≠ 1, 3 b2 −2, a2 b2, , (unique solution), Option (b):, , a1, a2, c1, c2, , =, , 5 1 b1 −2 1, =, = ;, = ;, 10 2 b2 −4 2, , =, , 9 1, =, 18 2, , 4/11/2018 3:28:26 PM

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Hints and Explanation, , A.17, , CHAPTER 3, Mensuration, Assessment Test I, 264 = 2p r, 264 × 7, r=, 2 × 22, r = 42 cm, ∴Area of the circle = p r2, 22, × 42 × 42 = 5544 sq. cm, =, 7, Hence, the correct option is (a)., , 1. BACD is the required sequential order., Hence, the correct option is (c)., 2. b = 10 cm, a = 8 cm, b, A=, 4 a2 − b2, 4, 10, =, 4(8)2 − 102, 4, 10, × 156, 4, 5, = × 2 39 = 5 39 sq.cm, 2, Hence, the correct option is (c)., =, , 3. Area of a square =, =, , (32 × 7, , ), , 2, × 24.5, , d2, 2, , 2, 34 × 72 × 29, = 34 × 72 × 28, =, 2, Hence, the correct option is (a)., 4. Inner dimensions of the box are 28 cm × 16 cm, × 12 cm., Number of cakes of soaps is maximum., \ Number of cakes of soaps along length =, 28, =7, 4, Number of cakes of soaps along breadth =,  16 ,   = 2, 8, 12, Number of cakes of soaps along height =, 6, =2, \ Total number of cakes of soaps = 7 × 2 × 2, = 28, Hence, the correct option is (c)., 5. Perimeter of a parallelogram., = 2(65 + 67) = 264 cm, Circumference of the circle, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 17, , 6., 0, 120°, , A, , r, , B, , 120°, 22, ×2×, × r = 924 cm, 360°, 7, 924 × 3 × 7, ⇒r=, 2 × 22, \ r = 21 × 21 = 441 cm, Hence, the correct option is (a)., l=, , 7. p r2 = 154 sq. cm, 7, r2 = 154 ×, 22, r = 7 cm, CSA = 2p rh = 440, 22, 2×, × 7 × h = 440, 7, h = 10 cm, ∴ Volume of the cylinder, v = p r2h, = 154 × 10 = 1540 cm3, Hence, the correct option is (c)., 4, 8. Volume of a sphere (v) = p r3, 3, 4 22 3, × × r = 38,808 cm3, 3 7, r3 =, , 38 , 808 × 3 × 7, ⇒ r = 21, 4 × 22, , 4/11/2018 3:28:39 PM

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A.18, , Chapter 3 Mensuration, ∴ Surface area of the sphere, 22, 4p r2 = 4 × × (21)2, 7, = 4 × 22 × 3 × 21 = 88 × 63 = 5544 sq. cm, Hence, the correct option is (c)., , 2., C, 13, , 9., 4 cm, , 4 cm, , 4 cm, , A, , 4 cm, 4 cm, 4 cm, , ∴ T.S.A. of a cuboid = 2 (lb + bh + hl), = 2 (16 × 4 + 4 × 4 + 4 × 16), = 2 × 16 (4 + 1 + 4), = 32 × 9 = 288 sq. cm, Hence, the correct option is (d)., , 13, 5, , 12, , D, , 12, , B, , Let ABC be the triangle in which AC = BC = 13 cm., Let CD ⊥ AB., ⇒ ∆CDB is a right triangle and BC is the, hypotenuse, BD = 132 − 52 = 12 cm, AB = 2(BD) = 24 cm, 24 × 5, \ Area =, = 60 cm2, 2, Hence, the correct option is (c)., , 10. CSA = 2p r2 = 416, ⇒ p r2 = 208, ⇒ TSA = 3p r2 = 624 sq. cm, Hence, the correct option is (c)., , 3. Area of a square = 24 × 36 × 510 sq. cm, Side of the square = 22 × 33 × 55 cm, \ Diagonal of the square = 2 × 22 × 33 × 55, , 11. Let the length the breadth and the height of the, cuboid be l, b, and h, respectively., (lb) (lh) (bh) = 576, i.e.,, (lbh)2 = 242, ∴, \ lbh = 24 ( lbh > 0), Hence, the correct option is (a)., , 4. Radius of each small lead ball = 2 cm, , 12. (c): Area of the ring = pR2 − p r2, = p(R + r) (R − r), lr Rr, 13. (e): Area of the sector =, =, 2, 2, 14. (a): CSA of a cone = p rl, = p rR = pRr, 15. (b): CSA of a cylinder = 2p rh, = 2pRr, , Assessment Test II, 1. BAECD is the required sequential order, Hence, the correct option is (b)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 18, , Hence, the correct option is (c)., , 4, ×p × 8 × 8 × 8, The required number of balls = 3, 4, ×p × 2 × 2 × 2, 3, = 64, Hence, the correct option is (a)., 5. p r2 = 2p r, r=2, ∴ Circumference of the circle = 2p r = 4p cm, Hence, the correct option is (c)., 6. Length of the arc, l =, ⇒, , x°, × 2p r = 22, 360, , x°, 22, ×2×, × 12 = 22, 360, 7, , x, 22, ×, = 22, 15, 7, ⇒ x = 105°, , 4/11/2018 3:28:42 PM

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Hints and Explanation, \ The required angle of the sector = 360° −, 3(105°) = 45°, Hence, the correct option is (b)., 7. CSA of a cylinder = 2p rh, 22, =2×, × 2016 × 20, 7, = 44 × 288 × 20, = 253440 sq. cm, Hence, the correct option is (a)., 8. Total surface area of a sphere = 4p r2 = 5544, 22, 4×, × r2 = 5544, 7, 7 × 9 × 88 × 7, r2 =, ⇒ r = 21, 4 × 22, 4, Volume of the sphere = pr3, 3, 4, 22, =, ×, × 21 × 21 × 21, 3, 7, = 88 × 441 = 38808 cm3, Hence, the correct option is (a)., 9. Edge of new cube formed = 3(5) = 15 cm, ∴LSA = 4(15)2 = 900 sq. cm, , A.19, , 12. (b): Volume of a cylinder = p r2h, = pR2 × (2r) = 2pR3, 1, 13. (a): Volume of a cone = p r2h, 3, 1, 2, 3, = pR 3R = pR, 3, 4, 14. (c): Volume of a sphere = p r3, 3, 3, 4 3 , = p  R, 3 2 , 4, 27 R3, p×, 3, 8, 9, = pR3, 2, , =, , 15. (e): Volume of a hemisphere =, =, , 2, p (3R)3, 3, , =, , 2, ×p 27R3 = 18pR3, 3, , 2 3, pr, 3, , Assessment Test III, 1. BDAC is the required sequential order., Hence, the correct option is (a)., 2. Given, perimeter of a regular hexagon (6a) =, 36 cm, a = 6 cm, , Hence, the correct option is (c)., 10. Given, p r2 = 670 sq. cm, TSA of solid hemisphere = 3p r2 = 3 × 670 sq. cm, = 2010 sq. cm, Hence, the correct option is (a)., 11. Let the length, breadth, and the height of the, cuboid be l, b, and h, respectively. The surface, areas of three of its adjacent faces are lh, bh,, and lb., (lb) (bh) (lh) = 3600, (lbh)2 = 602, ∴, \ lbh = 60 ( lbh > 0), Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 19, , Area of a regular hexagon =, 3 3, 3, 2, 6×, × a2 =, × (6 ), 4, 2, = 54 3 sq. cm, Hence, the correct option is (b)., 3. Let ‘h’ be the hypotenuse of a right isosceles, triangle., h2, Given, area of a right isosceles triangle =, 4, = 54 sq. cm, ⇒ h2 = 54 × 4, ⇒h=, , 9×6×4, , ⇒ h = 6 6 cm, Hence, the correct option is (d)., , 4/11/2018 3:28:46 PM

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A.20, , Chapter 3 Mensuration, , 4. Given, angle of the sector (q )= 84°, Circumference of the circle = 2p r = 330 cm, 22, × r = 330, ⇒ 2×, 7, 15 × 7, r=, cm., 2, q, Area of the sector =, × p r2, 360°, 84° 22 15 × 7 15 × 7, =, ×, ×, ×, 360° 7, 2, 2, 8085, =, = 2021.25 sq. cm, 4, Hence, the correct option is (a)., 1, 5. Given that the area of a trapezium = × h( a + b), 2, = 49 sq. cm, 1, × 7( 4 + b) = 49 ⇒ b = 14 − 4 = 10 cm, 2, Hence, the correct option is (c)., ⇒, , 6. Side of the base (a) = 60 cm, height (h) = 1.95 m, = 195 cm, Total length of the edges = 2 (perimeter of, base) + The number of sides of the base ×, height, = 2(3a) + 3 × h = 3(2a + h), = 3(2 × 60 + 195) = 3(120 + 195), = 3 × 315 cm = 945 cm = 9.45 m, Hence, the correct option is (d)., 14, = 7 cm, 2, Outer radius (R) = 7 + 7 = 14 cm, Outer surface area of a hemispherical shell, = 2p R2 + 2p r2(p R2 − p r2), = 3p R2 + p r2 = p (3R2 + r2), = p [3(14)2 + 72] = p (72)[3(2)2 + 1], = 49p [13] ∴ = 637p, Hence, the correct option is (b)., q, 8., × 2p r = 2pR, 360°, (Length of the arc of a sector = Perimeter of, the base of the cone), 120, × 2p × 42 = 2p R, R = 14 cm, 360, , 7. Inner radius (r) =, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 20, , A, 42, , 120°, , 0, B, , C, , 14, , ⇒ From the figure, AO2 = AC2 − OC2, ⇒ AO2 = (42)2 − (14)2, ⇒ AO2 = (14)2 [32 − 1], ⇒ AO2 = (14)2 (8), ⇒ AO = 14 × 2 2, \ AO = 28 2 cm, Hence, the correct option is (b)., 9. Height of the cylinder = 14 cm, Diameter of the base of the cylinder = 60 cm, ∴ Volume of the cylinder = p r2h, =, , 2, , 22  6 , ×   × 14 = 22 × 9 × 2,  2, 7, , = 396 cm3, Hence, the correct option is (b)., 10., Q, , D, , 60°, , C, , 30°, , A, , P, , B, , ∠QAB = 30°, ∠DAQ = 60°, In ∆ADQ, AQ2 = DA2 + DQ2, 142 = 72 + DQ2, ⇒ DQ2 = 147, ⇒ DQ = 7 3 cm, Required area = Area of the sector AQB + Area, of ∆ADQ − Area of the sector ADP, 30°, 1, 90°, × p × (14)2 + × 7 × 7 3 −, × p × (7 )2, =, 360, 2, 360, =, , 196p 49 3 49p, +, −, 12, 2, 4, , =, , 49 22 49 3, × +, 12 7, 2, , 4/11/2018 3:29:48 PM

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Hints and Explanation, , (, , ), , 7, 11 + 21 3 cm2, 6, Hence, the correct option is (b)., =, , w, , 11. No change in the total surface area., Hence, the correct option is (d)., 12., B, , C, , 6 cm, 60°, D, , P, 8 cm, , A, , Draw BP perpendicular to AD., Sides opposite to 30°, 60°, and 90° are in the, ratio = 1 : 3 : 2, Side opposite to 90° is 6 cm., ⇒ Side opposite to 60° is, , 3, × 6 cm = 3 3 cm., 2, , ∴ Area of the parallelogram = 8 × 3 × 3, = 24 3 sq. cm, Hence, the correct option is (c)., 13., , a, , a=5, , b=6, , Area of an isosceles triangle, b, =, 4 a2 − b2, 4, =, , 6, 2, 4 ( 5) − 62, 4, , 6, 6, =, 100 − 36 = × 8 = 12 sq. cm, 4, 4, Hence, the correct option is (c)., 14. Area of the path = Area of outer rectangle −, Area of inner rectangle., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 21, , A.21, , = lb − (l −2w) (b − 2w), = 20 × 16 − (20 − 2 × 1.8) (16 − 2 × 1.8), = 320 − (20 − 3.6) (16 − 3.6), = 320 − (16.4) (12.4), = 320 − 203.36, = 116.64 sq. cm, Hence, the correct option is (b)., 15.     (i) Area of a rhombus, 1, 1, d2, d1 d2 = dd =, 2, 2, 2,   (ii) Area of a rectangle = l × b = d × d = d2, (iii) Area of a trapezium, 1, 3 d2, 1, h( a + b) = d(d + 2d) =, 2, 2, 2, (iv) Area of a quadrilateral, 1, = d( h1 + h2 ), 2, 1 , 3 d2, d, = dd +  =, 2 , 2, 4, =, ,     (a) Area of the square = d2,    (b) Area of the square whose diagonal is d., d2, =, 2,     (c) Area of a parallelogram = bh, d 3 d2, = 3d × =, 4, 4, 1,   (d) Area of a triangle = bh, 2, 1, 3 d2, = × 3d × d =, 2, 2, (i) → (B); (ii) → (A); (iii) → (D); (iv) → (C), Hence, the correct option is (b)., , Assessment Test IV, 1. BADEC is the required sequential order., Hence, the correct option is (b)., 2. Let ‘a’ be the side a hexagon., Given, a = 12 cm, , 4/11/2018 3:29:53 PM

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A.22, , Chapter 3 Mensuration, Area of a regular hexagon = 6 ×, = 6×, , 3 2, ×a, 4, , 3, × 12 × 12, 4, , = 216 3 sq. cm, Hence, the correct option is (d)., 3. Let h be the hypotenuse of a right isosceles, triangle., We know that area of a right isosceles triangle, h2, =, ., 4, , (13 2 ), =, , 2, , 169 × 2, =, = 84.5 sq. cm, 4, 4, Hence, the correct option is (b)., , 4. Given, area of a sector = 770 sq. cm, q, ⇒, × p r2 = 770, 360, 72° 22 2, ×, × r = 770, ⇒, 360° 7, 70 × 5 × 7, ⇒ r2 =, 2, 7×5×2×5×7, ⇒ r2 =, ⇒ r = 35 cm, 2, Circumference of the circle = 2p r, 22, = 2×, × 35 = 220 cm, 7, Hence, the correct option is (a)., 5. a + b = 32 cm, Area of the trapezium = 128 sq. cm, 1, ⇒ h( a + b) = 128, 2, 1, 128, × h(32) = 128 ⇒ h =, 2, 16, h = 8 cm, Hence, the correct option is (b)., 6. Total length of the edges = 2(3a) + 3 × h, = 3(2a + h), where ‘a’ is the side of the base, and h is the height., h, Given that, a = ⇒ h = 2a, 2, ⇒ 3(2a + 2a) = 8.64 m, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 22, , 12a = 8.64 m, 3a = 2.16 m, ∴ Perimeter of the base = 2.16 m, Hence, the correct option is (b)., 7. Outer diameter (2R) = 16.4 cm, Thickness = 1.2 cm, Inner diameter (2r) = 16.4 − 2 × 1.2, = 16.4 − 2.4, 2r = 14 cm, 14, ∴ Inner radius (r) =, = 7 cm, 2, Inner surface area of a hemispherical shell =, 2p r2, 22, 2, = 2×, × (7 ), 7, = 44 × 7 = 308 sq. cm, Hence, the correct option is (c)., 2, 8. Curved surface area of the cone formed = Area, 3, of the circle., 2, 2, p rl = × p × ( 21), 3, 2, r(21) = × 21 × 21, 3, \ r = 14 cm, Hence, the correct option is (c)., 9. Given, volume of a cylinder = p r2h and height, of the cylinder (h) = 2r., p r2h = 2156, 22 2, × r × 2r = 2156, 7, 7, r3 = 2156 ×, 44, r3 = 73, ⇒ r = 7 cm, ∴Height = 14 cm, Hence, the correct option is (a)., 10. Area of the shaded region, = Area of the square − Area of the sector DAB, 2, 2, 90° 22, = 7 3 −, ×, × 7 3, 360° 7, 1, 231, = 49 × 3 − × 22 × 7 × 3 = 147 −, 4, 2, , (, , ), , (, , ), , 4/11/2018 3:29:58 PM

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Hints and Explanation, , bh = 96, where b and h are the base and corresponding, altitude of the triangle., Since bh = 96, only b = 12, h = 8 satisfies the, condition., 2,  12 , The equal side of the triangle is 82 +   =,  2, 10 cm., ∴ Perimeter of the triangle = 10 + 10 + 12 = 32 cm, Another Method:, Let the sides of the triangle be b, b − 2, and, b − 2., ∴Perimeter = 3b − 4, where b is a positive, integer., \ Option (a) follows, since 32 + 4 is divisible, by 3., Hence, the correct option is (a)., , = 147 − 115.5 = 31.5 sq. cm, Hence, the correct option is (a)., 11. Total surface area of each cube = 6 sq. cm, ⇒ Edge of each cube = 1 cm, Length of the cuboid formed = 3 cm, Breadth and height are 1 cm each., Total surface are of the cube, = 2(lb + bh + hl), = 2(3 × 1 + 1 × 1 + 1 × 3), = 2(3 + 1 + 3) = 14 sq. cm, Hence, the correct option is (b)., 12., S, , R, , 10 cm, 30°, P, , A.23, , 14., T, , Q, , Draw ST perpendicular to PQ., Side opposite to 90° is 10 cm., Hence the side opposite to 30° is 5 cm., ∴ ST = 5 cm, Area of the parallelogram = PQ × ST = 55 3, ⇒ PQ × 5 = 55 3, \ PQ = 11 3 cm2, Hence, the correct option is (a)., 13., , 2m, 75 m, 60 m, 2m, , 2m, , 2m, , Perimeter of the outer edge of the path, = 2[(l + 2w) + (b + 2w)], = 2[(75 + 2 × 2) + (60 + 2 × 2)], = 2[79 + 64], = 2[143] = 286 m, Hence, the correct option is (a)., , b−2, , b−2, h, b, , Since the perimeter is an integer and the difference of ‘a’ and ‘b’ is an integer, all its sides, are integers., Also, area is an integer, and altitude drawn on, to the base is also an integer., 1, ∴ × b × h = 48, 2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 23, , 15.    (i) Perimeter of an isosceles triangle, =b+a+a, = b + 2a,    (ii) Perimeter of a rectangle, = 2(l + b), = 2(a + 2b), = 2a + 4b, (iii) Perimeter of a parallelogram, = 2(2a) + 2(b), = 4a + 2b, , 4/11/2018 3:30:00 PM

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A.24, , Chapter 3 Mensuration, (iv) Perimeter of isosceles trapezium, = a + 2a + b + b, = 3a + 2b, (A) Perimeter of the quadrilateral, = a + 2a + b + b, = 3a + 2b,   (B) Perimeter of the pentagon, = a + b + 2a + a + b, = 4a + 2b, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 24, , (C) Perimeter of a hexagon, b b a a a a, = + + + + +, 2 2 2 2 2 2, = b + 2a, (D) Perimeter of a scalene triangle whose, sides are 2a, 3b,and b is 2a + 3b + b., = 2a + 4b, (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A), Hence, the correct option is (c)., , 4/11/2018 3:30:00 PM

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Hints and Explanation, 5. P = x3 − 7x + 6, R = x2 − 3x + 2, x+3, x2 − 3 x + 2 x3 + 0 x2 − 7 x + 6, , ), , x3 − 3 x2 + 2 x, − + −, 3 x2 − 9 x + 6, 3 x2 − 9 x + 6, − + −, 0, , Q=x+3, ⇒ P + Q = x3 − 7x + 6 + x + 3, = x3 − 6x + 9, Hence, the correct option is (c)., , ⇒ k = 15., Hence, the correct option is (c)., 9. f(x) = (x + 3) (x2 − 4x + p), g(x) = (x − 3) (x2 + x + q), HCF = x2 − 9 = (x + 3)(x − 3), HCF should be the factor of f(x) and g(x)., \ (x + 3) and (x − 3) are the factors of f(x) and g(x)., Clearly, (x + 3) is a factor of f(x)., \ (x − 3) should be the factor of x2 − 4x + p., x −1, x − 3 x2 − 4 x + p, , ), , = x2 + xy + y2, Hence, the correct option is (a)., 2, 2, 7. ( x + y ) + ( x − y ), 2, ( x + y ) − ( x − y )2, , =, , 2( x2 + y2 ), 4 xy, , =, , x2 + y 2, 2 xy, , =, , y, x, +, 2 y 2x, , 0, ⇒p=3, Clearly, x − 3 is a factor of g(x)., \ (x + 3) should be the factor of x2 + x + q., x−2, x + 3 x2 + x + q, , ), , x2 + 3 x, − −, − 2x + q, − 2x − 6, + +, 0, , Hence, the correct option is (a)., 8. If (x + 3) is a factor of x2 − 2x – k., x−5, x + 3 x2 − 2 x + k, , ), , x2 − 3 x, − +, −x−p, −x+3, + −, , 3, 3, 6. ( x − y )( x + y ), ( x2 − y 2 ), , ( x − y )( x2 + xy + y2 )( x + y ), =, ( x + y )( x − y ), , A.27, , x2 + 3 x, − −, − 5x − k, − 5 x − 15, + +, 0, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 27, , ⇒ q = −6, \ p − q = 3 − (−6) = 9, Hence, the correct option is (a)., 10. Degree of p(x) = 2, Degree of q(x) = 3, Degree of p(x) × q(x), = degree of p(x) + degree of q(x), =2+3=5, Hence, the correct option is (d)., 11. (4x2 − 1) (16x4 + 4x2 + 1), = (4x2 − 1) [(4x2) + 4x2 · 1 + 12], , 4/11/2018 3:30:05 PM

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A.28, , Chapter 4 Polynomials and Their LCM and HCF, = (4x2)3 − 13 = 64x6 − 1, Hence, the correct option is (a)., , 12. Option (b):  3 x2 + 7 x − 1 is a polynomial in one, variable., 1, is not a polynomial., x, 14. Option (a): 100x° = 100 is a constant polynomial., 13. Option (c): x +, , 15. Option (d): ax2 + bx + c, where a = b = c = 0, Therefore, it is a zero polynomial., , Assessment Test III, 1. BDACE is the required sequential order., Hence, the correct option is (b)., 1, 2. x − = 4, x, 2, , 3. We know that (a − b), , (a2, , + ab +, , b2), , 8. x6 + 64y6 = (x2)3 + (4y2)3, =, , a3, , −, , b3, , \(2x − 3y), + 6xy +, 3, 3, = (2x) − (3y) = 8x3 − 27y3, Hence, the correct option is (a)., (4x2, , 9y2), , 4. x2 − 3x − 28 = (x − A) (x − B), x2 − 7x + 4x − 28 = (x − A) (x − B), x(x − 7) + 4 (x − 7) = (x − A) (x − B), (x − 7) (x + 4) = (x − A) (x − B) ⇒ A = 7 and, B = −4, \A+B=7−4=3, Hence, the correct option is (a)., 5., , 3x2, , + 13x + 4 = (Ax + B) (Cx + D), , + 12x + 1x + 4 = (Ax + B) (Cx + D), 3x(x + 4) + 1(x + 4) = (Ax + B) (Cx + D), 3x2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 28, , 6. abc + ab + bc + ac + a + b + c = 692, (abc + bc) + (ab + b) + (ac + c) + (a + 1) = 693, bc(a + 1) + b(a + 1) + c(a + 1) + (a + 1) = 7 × 9 × 11, (a + 1) (bc + b + c + 1) = 7 × 9 × 11, (a + 1) (b + 1) (c + 1) = 7 × 9 × 11, = (6 + 1) (8 + 1) (10 + 1), ∴ a + b + c = 6 + 8 + 10 = 24, Hence, the correct option is (d)., 7. 5x2 + 12x + 7, = 5x2 + 5x + 7x + 7, = 5x(x + 1) + 7(x + 1), = (x + 1)(5x + 7), ∴ (5x2 + 12x + 7) ÷ (5x + 7), (5 x + 7 )( x + 1), =x+1, =, (5x + 7 ), Hence, the correct option is (d)., , 1, 2,  =4, x,  1, 1, x2 − 2x   +, = 16,  x, x2, 1, x2 +, = 16 + 2, x2, 1, \ x2 +, = 18, x2, Hence, the correct option is (c)., ,  x −, , (x + 4) (3x + 1) = (Ax + B) (Cx + D), ⇒ A = 1, B = 4, C = 3, and D = 1, AC + BD = 1 × 3 + 4 × 1 = 7, Hence, the correct option is (c)., Shortcut:, By comparing like terms, AC = 3 and BD = 4, ⇒ AC + BD = 3 + 4 = 7, , = (x2 + 4y2) (x4 − 4x2y2 + 16y4), On comparing with (Ax2 + By2) (Cx4 + Dx2y2 +, Ey4), A = 1, B = 4, C = 1, D = −4, and E = 16, ∴ A + B + C + D + E = 1 + 4 + 1 − 4 + 16 = 18, Hence, the correct option is (b)., 9. a2 + b2 − c2 − 2ab, = a2+ b2 − 2ab − c2, = (a − b)2 − c2, = (a − b + c) (a − b − c), Hence, the correct option is (d)., 10. (x + m)2 = (x + m) (x + m), x2 − m2 = (x + m) (x − m), \ LCM = (x + m)2 (x − m) or (x2 − m2)(x + m), Hence, the correct option is (d)., , 4/11/2018 3:30:07 PM

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A.30, , Chapter 4 Polynomials and Their LCM and HCF, x(x + 2) + 1(x + 2) = (x − A) (x − B), (x + 2) (x + 1) = (x − A) (x − B), ⇒ A = −2 and B = −1, A + B = − 2 + (−1) = −3, Hence, the correct option is (d)., , 5. 6x2 + 11x − 10 = (Px + Q) (Ax + B), 6x2 + 15x − 4x − 10 = (Px + Q) (Ax + B), 3x (2x + 5) − 2(2x + 5) = (Px + Q) (Ax + B), (2x + 5) (3x − 2) = (Px + Q) (Ax + B), ⇒ P = 2, Q = 5, A = 3, and B = −2, AP + BQ = 3(2) + (−2) (5) = 6 − 10 = −4, Shortcut:, On comparing like terms, we get, AP = 6 and BQ = −10, ∴ AP + BQ = 6 + (−10) = −4, Hence, the correct option is (a)., 6. xyz + xy + yz + zx + x + y + z = 959, ⇒ (xyz + yz) + (xy + y) + (zx + z) + (x + 1) = 960, ⇒ yz(x + 1) + y(x + 1) + z(x + 1) + (x + 1) = 960, ⇒ (x +1) [yz + y + z + 1] = 960, ⇒ (x + 1) (y + 1) (z + 1) = 8 × 10 × 12, = (7 + 1) (9 + 1) (11 + 1), ⇒ xyz = 7(9) (11) = 693, Hence, the correct option is (c)., 7. 6x2 + x − 15 = 6x2 + 10x − 9x − 15, = 2x(3x + 5) − 3(3x + 5), = (3x + 5) (2x − 3), 6 x2 + x − 15 (3 x + 5)(2 x − 3), = 2x − 3, =, (3 x + 5), 3x + 5, Hence, the correct option is (a)., , ∴, , = (x + y)2 − z2, = (x + y + z) (x + y − z)., Hence, the correct option is (c)., 10. (x − n)2 = (x − n) (x − n), x2 + n2 = (x2 + n2), ∴ LCM = (x − n)2 (x2 + n2), Hence, the correct option is (c)., 11. (x + 1)2 = (x + 1)2, x2 − 1 = (x + 1) (x − 1), x + 1 = (x + 1), ∴ HCF = (x + 1), Hence, the correct option is (a)., 12. x2 − 9 = (x + 3) (x − 3), x2 − 6x + 9 = (x − 3)2, ∴ HCF = (x − 3), Hence, the correct option is (c)., 13. Let f(x) = (x + 2)2 (x − 2)(x2 − 1), g(x) = (x − 2)2 (x + 2) (x + 1), ∴ LCM = (x − 2)2 (x + 2)2 (x2 − 1), Hence, the correct option is (d)., 14. f(x) = (x − 2) (2x2 + 7x + P), g(x) = (x + 2) (2x2 − 7x + Q), ∴ HCF is x2 − 4 = (x + 2) (x − 2)., ⇒ x + 2 should be a factor of (2x2 + 7x + P)., 2x + 3, x + 2 2 x2 + 7 x + p, , ), , 8. a6 − 729b6 = (a2)3 − (9b2)3, = (a2 − 9b2)(a4 + 9a2b2 + 81b4) on comparing, with, (Pa2 + QB2) (Ra4 + Sa2b2 + Tb4), P = 1, Q = − 9, R = 1, S = 9, and T = 81, ∴ P + Q + R − S + T = 1 − 9 + 1 − 9 + 81 = 65, Hence, the correct option is (d)., 9. x2 + y2 − z2 + 2xy, = x2 + y2 + 2xy − z2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 30, , 2 x2 + 4 x, ( −) ( −), 3x + p, 3x + 6, ( −) ( −), p−6 = 0, , , , ∴P=6, , (x + 2)(x − 2) is a factor of (x + 2) (2x2 − 7x + Q)., ∴ x − 2 is the factor of (2x2 − 7x + Q)., , 4/11/2018 3:30:10 PM

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A.32, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , CHAPTER 5, Ratio, Proportion, and Variations:, Commercial Mathematics, Assessment Test I, 1. CBAD is the required sequential order., Hence, the correct option is (b)., 2. 2.7p = 0.09q ⇒ 30p = q, q + 6p, 36 p, 3, =, \, = ., 2, q − 6p, 24 p, Hence, the correct option is (b)., e, =k, 3. Let, f, a, c, e, =k, =, =, b, d, f, a = bk, c = dk, and e = fk, a+c, bk + dk, k (b + d ), =, =, =k, b+d, b+d, b+d, a+e, c+e, and, equal k., Similarly,, b+ f, d+ f, Option (d) follows., Hence, the correct option is (d)., A, 4. For A to vary directly with B,, must be a, B, constant., Option (a), A, = a constant, B, A, \, = (a constant)3 = a constant, B, \ A varies directly with B., Option (b), 3, , 3, ,  A,   = a constant, B, A, = 3 a constant = a constant, B, \ A varies directly with B., Option (d) follows., Hence, the correct option is (d)., 5. Let the income of P and Q be `2a and `3a, respectively., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 32, , Let the expenditure of P and Q be `5b and `6b,, respectively., Method 1, Savings of P = ` (2a − 5b) saving of Q = ` (3a − 6b), 2 a − 5b, Part of his income that P saves =, =, 2a, 2.5b, 1−, a, 3 a − 6b, Part of his income that Q saves =, =, 3a, 2b, 1−, a, b, b, 2.5, >2, a, a, b, b, \ − 2.5 < −2, a, a, b, b, \ 1 − 2.5 < 1 − 2, a, a, \ Q saves a greater part of his income., Method 2, The person who saves a greater part of his income will spend a lesser part of his income., Parts of their incomes spent by P and Q are, 5b, 6b, b, b, and, , respectively, i.e., 2.5 > 2 ., 2a, 3a, a, a, \ Q spends a lower part of his income and, hence, saves a greater part of his income., Hence, the correct option is (b)., 6. Let the total number of stamps be x. Ratio of, the number of stamps with P, Q, and R, 1 1 1, 3 2 1, = : : = : : = 3:2:1. The total is 6 parts., 2 3 6, 6 6 6, If this ratio had been 2:3:6, the total is 11 parts., 3:2:1 = 33:22:11, 2:3:6 = 12:18:36, \ P and Q would have less and R would have, more., Hence, the correct option is (d)., , 4/11/2018 3:30:18 PM

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A.33, , Hints and Explanation, 7. Let the maximum marks be x., , It becomes `3x in 12 years., It becomes `4x in 18 years., \ It takes 18 years to become 4 times itself., , Given that 0.4x − 0.3x = 30., ⇒ 0.1x = 30 ⇒ x = 300, , n, , Hence, the correct option is (d)., 8. For every 900 g that the trader sells, he earns a, profit of 100 g., 100, Therefore, percentage of profit =, × 100 =, 900, 1, 11 %., 9, Hence, the correct option is (d)., 9. Let C.P. = `100, M.P. = 1.3 × 100 = `130, S.P. = 0.9 × 130 = `117, Therefore, percentage of profit =, 100 = 17%., Hence, the correct option is (d)., , 117 − 100, ×, 100, , 10. Total discount given by trader B, 25 × 5, = 25 + 5 −, = 30 − 1.25 = 28.75%, which, 100, is less than that given by trader A., Hence, the correct option is (a)., 11. Let the cost price of articles A and B be a and b,, respectively., Given, 1.3a + 0.8b = 1160 , (1), 0.8a + 1.05b = 1160 , (2), 1.3 × Eq. (2) - 0.8 × Eq. (1) gives 0.725b = 580, 580, = 800, ⇒b=, 0.725, Hence, the correct option is (b)., 12. Option (d): Let the number be x., 60 x, = 16, Given x + 60% of x = 16 ⇒ x +, 100, 16 x, ⇒, = 16 ⇒ x = 10, 10, 13. Option (e): Required profit percentage =, 225 − 150, × 100%., 150, 75, =, × 100% = 50%, 150, 14. Option (a): Let the initial sum be `x., It becomes `2x in 6 years., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 33, , 15. Option (c): A = P  1 + R  ⇒ 216 = 125, , , 100 , 3, R, , ,  1 +, , 100 , 3, , 3, , R ,  6, , ⇒, ⇒   = 1+,  5,  100 , 6, 6, R, R, = 1+, ⇒, = −1, 5, 100, 100 5, R, 1, = ⇒ R = 20%, ⇒, 100 5, , Assessment Test II, 1. BADC is the required sequential order., Hence, the correct option is (a)., 2. Let x = 4a and y = 9a., x = 2 a and y = 3 a, ⇒, , 4 x +5 y, 6 x +7 y, , =, , ( ) ( ) = 23, 6 ( 2 a ) + 7 ( 3 a ) 33, 4 2 a +5 3 a, , a, a, , =, , 23, 33, , Hence, the correct option is (d)., , 3. When two or more ratios are equal, each of, them is equal to the sum of all the numerators, divided by the sum of all the denominators, provided the sum of the denominators is not, zero., b, a, c, \ If, =, =, , then, a+b−c, a+c−b, b+c−a, each of them equals, b+a+c, =1, a+b−c+a+c−b+b+c−a, \ When a + b + c ≠ 0, each ratio can be 1., Hence, the correct option is (a)., 4. For P to vary inversely with Q, PQ must be a, constant., Option (a), PQ = a constant, Squaring both sides, we get, PQ = (a constant) = a constant, \ P varies inversely with Q., , 4/11/2018 3:30:23 PM

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A.34, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, 8. Let, the trader use x kg for 1 kg., Now, 1.25x = 1, 1, ⇒x=, = 0.8 kg = 800 g, 1.25, Hence, the correct option is (c)., , Option (b), 2PQ = a constant, a constant, PQ =, = a constant, 2, \ P varies inversely with Q., Option (c), (PQ)3 = a constant, PQ = 3 a constant = a constant, \ P varies inversely with A., Option (d) follows., Hence, the correct option is (d)., , 9. Let the cost price of the product be `100., Therefore, the marked price = 1.6 × 100 = `160, Selling price = 0.7 × 160 = `112, Therefore, the trader earns `12 on every `100., Therefore, the profit per cent = 12%, Hence, the correct option is (d)., , 5. Let the income of X and Y be `3a and `4a,, respectively., Let the expenditure of X and Y be `4b and `5b,, respectively., 4b, b, Part of the income that X spends =, = 1.3, 3a, a, 5b, b, Part of the income that Y spends =, = 1.25, 4a, a, b, b, 1.3 > 1.25, a, a, \ X spends a greater part of his income., Hence, the correct option is (a)., 6. The data is tabulated below., A, , B, , C, , Total, , Intended, , 2, , 3, , 4, , 9, , Actual, , 1, 2, , 1, 3, , 1, 4, , Actual, , 6, , 4, , 3, , 10. Let M.P. = `100, S.P. for traders A = 0.7 × 100 = `70, S.P. for trader B = 0.8 × 0.9 × 100 = `72, S.P. for trader C = 0.85 × 0.9 × 0.95 × 100, = `72.675, Therefore, a customer should buy from trader A., Hence, the correct option is (a)., 11. Let the cost price of the product be `x., Therefore, 1.27x − 1.5x = 24, 24, ⇒ 0.12x = 24 ⇒ x =, = 200, 0.12, Hence, the correct option is (b)., , \A gained, whereas B and C lost., , 12. Option (a): Let the number be x., Given, x - 40% of x = 12, 40 x, = 12, ⇒x100, 60 x, = 12 ⇒ x = 20, ⇒, 100, 150 − 100, 13. Option (f): Required loss% =, × 100%, 150, 50, =, × 100%, 150, = 1/3 × 100%, = 331/3%, , Hence, the correct option is (d)., , 14. Option (c): Let the initial sum be `x., , 13, , Multiplying the intended row by 13 and the, actual by 9, we get, Intended, Actual, , A, 26, 54, , B, 39, 36, , C, 52, 27, , 117, 117, , 7. Let the total marks be x., Therefore, 0.4x − 10 = 0.2x + 20, 30, ⇒ 0.2x = 30 ⇒ x =, = 150, 0.2, Hence, the correct option is (c)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 34, , It becomes `3x in 3 years., i.e., `2x is interest in 3 years., \It becomes `5x in 6 years., n, , R , , ⇒ 1210, 15. Option (d): A = P  1 +,  100 , , 4/11/2018 3:30:27 PM

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Hints and Explanation, R , , , = 1000  1 +,  100 , , 2, , 1210 , R , = 1+, , , 1000, 100 , , 2, , 2, , R ,  11 , , ⇒   = 1+,  10 ,  100 , , 5. Given, successive discounts are 25% and 28%., The required single discount, , 2, , 11, R, = 1+, 10, 100, R, 11, 1, ⇒, =, −1 =, 100 10, 10, ⇒ R = 10%, , ⇒, , Assessment Test III, 1. ACBD is a required sequential order., Hence, the correct option is (d)., 2. Given, A:B = 4:5, B:C = 4:5, LCM of 4 and 5 is 20., A:B = 4:5 = (4 × 4):(5 × 4) = 16:20, B:C = 4:5 = (4 × 5):(5 × 5) = 20:25, A:B = 16:20, B:C = 20:25, ⇒ A:B:C = 16:20:25, Hence, the correct option is (d)., 1 1, 3. Given, A:B :, 3 2, ⇒ A:B = 2:3, Let the incomes of A and B be 2x and 3x,, respectively., 3 x − 2x, Required percentage =, × 100%, 3x, 1, 1, = × 100% = 33 %, 3, 3, Hence, the correct option is (c)., 4. The compound ratio = (x + 5) × 9:25x = 12:25, 9 x + 45 12, =, 25 x, 25, 9x + 45 = 12x, ⇒ 3x = 45, ⇒ x = 15, Hence, the correct option is (b)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 35, , A.35, , 25 × 28 , , =  25 + 28 −, %, , 100 , = (53 − 7)% = 46%, Hence, the correct option is (c)., 6. Given,, P + 3I = 4300, P + 4I = 5100, I = 800, We have P + 3I = 4300, ⇒ P + 2400 = 4300, ⇒ P = `1900, Hence, the correct option is (a)., 9 + 8x + 9 − 8x, , 3, 2, 9 + 8x − 9 − 8x, Applying componendo and dividendo, we get, , 7. Given,, , (, (, , =, , ) (, 9 − 8x ) − (, , =, , ), 9 − 8x ), , 9 + 8x + 9 − 8x +, , 9 + 8x − 9 − 8x, , 9 + 8x +, , 9 + 8x −, , 3+2, 3−2, , 2 9 + 8x, , =, , 5, 1, , 2 9 − 8x, 9 + 8x, 25, =, 9 − 8x, 1, Again applying componendo and dividendo,, we get, 9 + 8x + 9 − 8x, 25 + 1, =, 9 + 8x − 9 + 8x, 25 − 1, 18, 26, =, 16x, 24, 18 × 24, 27, ⇒x=, ⇒=, 16 × 26, 26, Hence, the correct option is (d)., 8. Let x be the required number., 14 − x, 11 − x, 17 – x, and 13 − x are in proportion., (14 − x) (13 − x) = (17 − x) (11 − x), 182 − 14x − 13x + x2 = 187 − 17x − 11x + x2, 182 − 27x = 187 − 28x, ⇒ 28x − 27x = 187 − 182, , 4/11/2018 3:30:33 PM

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A.36, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, ⇒x=5, Hence, the correct option is (c)., , 9. Let the number of days be x., M1D1, MD, = 2 2, W1, W2, 150 × D2, 250 × 60, =, 720, 2250, 250 × 60 × 720, ⇒ D2 =, 2250 × 150, ⇒ D2 = 32 days, Hence, the correct option is (b)., 10. Given, y is mean proportion of x and z., y2 = xz ⇒, ⇒, , y, x−y, x, =, =, y, z, y−z, , y4, ( x − y )4, x4, =, =, y4, z4, ( y − z)4, , Hence, the correct option is (b)., 11. The ratio of wages of X, Y, and Z = 3 × 6:4 × 7:5, ×8, = 18:28:40 = 9:14:20, 9, Amount received by X =, × 86,000, 9 + 14 + 20, 9, × 86,000 = `18,000, 43, Hence, the correct option is (d)., 2,  15 , 12. Loss percentage =   %,  10 , 225, =, % = 2.25%, 100, Hence, the correct option is (c)., =, , 13. Given, S.P. of 50 articles = C.P. of 75 articles, S.P. × 50 = C.P.× 75, S.P., 75, =, C.P., 50, S.P., 3, =, C.P., 2, ∴, Profit is made ( S.P. > C.P.)., Profit% =, =, , S.P. − C.P., × 100%, C.P., 3−2, × 100%, 2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 36, , 1, × 100% = 50%, 2, Hence, the correct option is (a)., =, , 14. Given, total number students in the school =, 1320., Let b and g be the number of boys and the, number of girls, respectively., Given, b:g = 5:6, 5, ⇒b=, × 1320 = 600, 11, ⇒ g = 1320 − 600 = 720, 18, 18% of the number of boys =, × 600 = 108, 100, Strength of the boys = 600 + 108 = 708, Required ratio = 708:720, = 59:60, Hence, the correct option is (c)., 15.   (i) C.P. < S.P., ∴ Profit% =,    (ii) C.P. > S.P., ∴ Loss% =, (iii) C.P. < S.P., ∴ Profit% =, (iv) C.P. > S.P., , 50, × 100% = 25% → ( b), 200, 50, × 100% = 25% → (c), 200, 30, × 100% = 20% → (a), 150, , 30, × 100% = 20% → (d), 150,   (i) → (B); (ii) → (C); (iii) → (A); (iv) → (D), ∴ Loss% =, , Hence, the correct option is (b)., , Assessment Test IV, 1. CADB is a required sequential order., Hence, the correct option is (c)., 2. Given, A:B = 2:5, B:C = 4:3, LCM of 4 and 5 is 20., A:B = 2:5 = (2 × 4):(5 × 4) = 8:20, B:C = 4:3 = (4 × 5):(3 × 5) = 20:15, \ A:B:C = 8:20:15, A:C = 8:15., Hence, the correct option is (c)., , 4/11/2018 3:30:38 PM

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Hints and Explanation, 3. Given, P:Q =, , 1 1, :, 3 4, , 7 + 3x, , ⇒ P:Q = 4:3, Let the incomes of P and Q be 4x and 3x,, respectively., 4x − 3x, The required percentage =, × 100%, 4x, x, =, × 100% = 25%, 4x, Hence, the correct option is (b)., 4. The compound ratio = (x + 6) × 3:x × 7 = 12:7, (3x + 18):7x = 12:7, 3 x + 18, 12, =, 7x, 7, 3x + 18 = 12x, ⇒ 9x = 18, ⇒x=2, Hence, the correct option is (b)., 5. Given, successive discounts are 20% and 25%., The required single discount, 20 × 25 , , =  20 + 25 −, %, , 100 , = (45 − 5)% = 40%, Hence, the correct option is (d)., 6. Given, , 7 − 3x, , =, , 5, 4, , 7 + 3x, 25, =, 16, 7 − 3x, Again applying componendo and dividendo,, we get, 7 + 3 x + 7 − 3 x 25 + 16, =, 7 + 3 x − 7 + 3 x 25 − 16, 14, 41, =, 6x, 9, 9 × 14, 21, ⇒x=, 6 × 41, 41, Hence, the correct option is (a)., ⇒x=, , 8. Let x be the required number., 5 + x, 2 + x, 12 + x, and 6 + x are in proportion., (5 + x) (6 + x) = (12 + x) (2 + x), 30 + 5x + 6x + x2 = 24 + 12x + 2x + x2, 30 + 11x = 24 + 14x, ⇒ 14x − 11x = 30 − 24, ⇒ 3x = 6 ⇒ x = 2, Hence, the correct option is (d)., 9. Given, M1 = 30, D1 = 24, H2 = 6 h, M2 = 18, D2 = ?, H2 = 8 h, , P + 4I = `4300, , M1D1H1 = M2D2H2, , P + 6I = `5200, , 30 × 24 × 6 = 18 × D2 × 8, 30 × 24 × 6, ⇒ D2 =, 18 × 8, D2 = 30 days, Hence, the correct option is (d)., , 2I = `900 ⇒ I = `450, ∴ 3I = `1350, Hence, the correct option is (a)., 7 + 3x + 7 − 3x, , 9, 1, 7 + 3x − 7 − 3x, Applying componendo and dividendo rule,, we get, , 7. Given,, , (, (, , ) (, 7 − 3x ) − (, , =, , )=, 7 − 3x ), , 7 + 3x + 7 − 3x +, , 7 + 3x − 7 − 3x, , 7 + 3x +, , 7 + 3x −, , 9+1, 9−1, 2 7 + 3x, 2 7 − 3x, , =, , 10, 8, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 37, , A.37, , 10. Given, b is the mean propositional of a and c., a+b, a, b, b2 = ac ⇒ =, =, b, c, b+c, ( a + b)3, a3, b3, =, =, b3, c3, (b + c)3, Hence, the correct option is (d)., ⇒, , 11. Ratios of wages = 6 × 8:7 × 6:9 × 5, = 48:42:45 = 16:14:15, 14, Amount received by B =, × 90,000, 16 + 14 + 15, , 4/11/2018 3:30:43 PM

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A.38, , Chapter 5 Ratio, Proportion, and Variations: Commercial Mathematics, , 14, × 90,000 = `28,000., 45, Hence, the correct option is (a)., 2,  9, 12. Loss percentage =   %,  10 , 81, =, % = 0.81%, 100, Hence, the correct option is (a)., =, , 13. Given, C.P. of 125 articles = S.P. of 150 articles, 125 × C.P. = S.P. × 150, C.P. 150, =, S.P. 125, C.P. 6, =, S.P. 5, ⇒ C.P. > S.P., loss is made., C.P. − S.P., Loss% =, × 100%, S.P., 6−5, Loss% =, × 100%, 6, 2, ∴ Loss% = 16 %., 3, Hence, the correct option is (c)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 38, , 14. Given, total number of students = 800, Let b and g be the boys and girls in the school,, respectively., Given b:g = 2:3, 2, b = × 800 = 320, 5, g = 480, Let x be number boys added to the school., b+x 3, =, g, 2, 320 + x 3, =, 480, 2, 640 + 2x = 1440, 2x = 800, x = 400, 400, ∴ Required percentage =, × 100, 320, = 125%, Hence, the correct option is (b)., 15. (i) → (D); (ii) → (A); (iii) → (C); (iv) → (B), Hence, the correct option is (d)., , 4/11/2018 3:30:47 PM

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Hints and Explanation, , A.39, , CHAPTER 6, Real Numbers, LCM, and HCF, Assessment Test I, 1. If n = pa × qb × rc, then the number of factors of n, = (a + 1) (b + 1) (c + 1), 360 = 23 × 32 × 51, The number of factors of 360, = (3 + 1) (2 + 1) (1 + 1), = 4 × 3 × 2 = 24, \ The required sequential order is BCAD., Hence, the correct option is (b)., 2. 1315 × 1517 × 1719, The units digit of the product, = The units digit of 7 × 5 × 3 = 5., Alternative method: The product of an odd number and a number ending with 5 ends in 5., Hence, the correct option is (b)., 2648 − 2 2646 98, =, =, 999, 999 37, Hence, the correct option is (b)., , 3. 2.648 =, , 4. The GCD of 145 and 256 is 1., LCM of 144 and 1728 is 1728., ∴ AB = 1 × 1728 = 1728, Alternative method: The product of A and B, is a multiple of 1728., ∴ Only option (D) follows., Hence, the correct option is (d).,  2 2 3  HCF ( 2, 2, 3 ) 1, 5. HCF of  , ,  =,  3 5 5  LCM ( 3 , 5, 5) 15,  2 2 3  LCM ( 2, 2, 3 ) 6, =, LCM of  , ,  =,  3 5 5  HCF ( 3 , 5, 5) 1, 1, 2, ×6 =, 15, 5, Hence, the correct option is (a)., \ HCF × LCM =, , 6. Required number is (LCM of 8, 18, and 28) k + 7., 2 8 , 18 , 28, 2 4 , 9, 14, 2, 9, 7, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 39, , LCM = 504, \ The required number is (504) k + 7., When k = 1, the number is 511., Hence, the correct option is (d)., 7. 15 − 7 = 8, 25 − 17 = 8, 35 − 27 = 8, \ Required number is LCM of (15, 25, 35) k − 8., 5 15, 25, 35, 3,, , 5,, , 7, , \ LCM of 5, 25, and 35 is 525., \ Required number is 525 k − 8., When k = 1, the number is 517., Hence, the correct option is (d)., 8. HCF = 18 and LCM = 144, The numbers A and B should be factors of, 144. From the options, 72 or 144 is possible for, B. However, the greatest possible one is 144., Hence, the correct option is (b)., 9. 7700 = 7 × 11 × 100, = 7 × 11 × 2 × 2 × 5 × 5, = 2 × 2 × 5 × 5 × 7 × 11, (5, 7) is the only pair of twin primes., 3080 = 2 × 154 × 10, = 2 × 2 × 77 × 2 × 5, = 2 × 2 × 2 × 5 × 7 × 11, (5, 7) is the only pair of twin primes., 5005 = 5 × 1001, = 5 × 7 × 143, = 5 × 7 × 11 × 13, (5, 7) and (11, 13) are the two pairs of twin primes., 3003 = 3 × 1001, = 3 × 7 × 11 × 13, (11, 13) only a pair of twin primes., Hence, the correct option is (c)., , 4/11/2018 3:30:50 PM

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A.40, , Chapter 6 Real Numbers, LCM, and HCF, , 10. 984x6y is a six-digit number, where x ≠ 0 and y, ≠ 0., 9 + 8 + 4 + x + 6 + y = 27 + x + y, As the given number is a multiple of 9, x + y, should be a multiple of 9., i.e., x = 1, y = 8 ⇒ xy = 8, x = 2, y = 7 ⇒ xy = 14, x = 3, y = 6 ⇒ xy = 18, x = 4, y = 5 ⇒ xy = 20, ∴ The minimum value is 8., Hence, the correct option is (c)., 11. If 2k − 1 is a prime number, then 2k − 1 (2k − 1) is, a perfect number., 25 − 1 = 31 is a prime number., \ 25 −1 (25 − 1) is a perfect number., Hence, the correct option is (c)., 12. → Option (b): 48 = 24 × 31 ⇒ The number of, factors = 5 × 2 = 10., 13. → Option (d): The sum of the factors of 6 = 1 +, 2 + 3 + 6 = 12, = 2 × 6., 14. → Option (f): 887 ends in 2., 15. → Option (e): The least composite number is 4., , Assessment Test II, 1. CADBE is the required sequential order., Hence, the correct option is (c)., 2. Unit’s digit of 415 × 516 + 617 × 718 = the unit’s, digit of (4 × 5 + 6 × 9) = the unit’s digits of (20 +, 54) = 4, Hence, the correct option is (a)., 3345 − 3, 999, 3342 1114, =, =, ., 999, 333, Hence, the correct option is (d)., , 3. 3.345 =, , 4. A × B = P × Q, A Q, ⇒ =, P B, Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 40, , 5. LCM =, LCM =, HCF =, , LCM ( 4 , 6 , 7 ), HCF ( 5, 7 , 9), , 84, = 84, 1, , HCF ( 4 , 6 , 7 ), , LCM ( 5, 7 , 9), , 1, =, 315, \ LCM × HCF = 84 ×, , 1, 315, , 4, 15, Hence, the correct option is (c)., , =, , 6. The required number is LCM of (8, 12, and 15), k + 6., i.e., 120 k + 6, \ When k = 1, the number is 126., Hence, the correct option is (d)., 7. 12 − 6 = 6, 14 − 8 = 6, 16 − 10 = 6, \ Required number is LCM of (12, 14, and 16), k − 6., i.e., 336 k − 6, \ When k = 1, the number is 330., Hence, the correct option is (c)., 8. If HCF of two numbers is one of the numbers,, then the LCM is the other number., As P ≠ 1 and LCM = 94, the numbers are (2, 94), and (47, 94)., \ PQ = 2 × 94 or 47 × 94, Hence, the correct option is (b)., 9. 1, 2, 3, 4, 6, 8, 12, and 24 are the factors of 24., Therefore, the sum is 60 ≠ 2 × 24., \ 24 is not a perfect number., Hence, the correct option is (c)., 10. 2k − 1 (2k − 1) is a prime number., ⇒ 2k − 1 is a prime number., ⇒ k can be 3., Hence, the correct option is (d)., , 4/11/2018 3:31:33 PM

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Hints and Explanation, 11. Given that 684m7n is divisible by 11., The sum of digits in odd places is n + m + 8., The sum of digits in even places is 7 + 4 + 6 = 17., m + n + 8 − 17 is a multiple of 11 or zero., m + n − 9 cannot be multiple of 11., \ m + n − 9 should be 0., For maximum of mn, m = 4, n = 5 or m = 5, n = 4, \ m × n = 20, Hence, the correct option is (d)., , =, , (23 − 1), (33 − 1), ×, 1, 2, , = (8 − 1) ×, =7×, , 26, 2, , 6. 64 = 26, \ The number of factors of 64 = 6 + 1 = 7., Hence, the correct option is (c)., 7., 2, , 32, 48, , 14. → Option (b): 9 is odd but not prime., , 2, , 16, 24, , 15. → Option (g): 62 = 21 × 311, , 2, , 8, 12, , 2, , 4, 6, , The number of factors = 2 × 2 = 4., , Assessment Test III, 1. The required sequential order is BCDA., Hence, the correct option is (c)., 6, 2. 0.6 =, 9, 5, 0.5 =, 9, , 11, 6 5, +, =, = 1. 2, 9 9, 9, Hence, the correct option is (b)., x + 6 = 5 × 125, , = 5 × 25 × 5 = 5 × 5 5, =5×5, = 25, ⇒ x = 19 ⇒ x = 361, Hence, the correct option is (b)., 4. If the sum of two integers is negative, then at, least one of the integers must be negative., Hence, the correct option is (c)., 5. 36 = ×, (22 +1 − 1) (32 +1 − 1), ×, \ The sum of the factors =, (2 − 1), (3 − 1), 22, , 2, 3, LCM = 2 × 2 × 2 × 2 × 2 × 3, = 96, Hence, the correct option is (b)., 8. LCM of (a andb ) = a, \ LCM of (k a and k b ) = ka, Hence, the correct option is (c)., , 0. 6 + 0. 5 =, , 3., , (27 − 1), 2, , = 91, Hence, the correct option is (b)., , 12. → Option (a): The number of primes between, 50 and 100 is 10., 13. → Option (d): The unit’s digit of 777 is same as, the units digit of 71., , A.41, , 32, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 41, , 9. 28 = 22 × 71, Sum of the factors =, , (22 + 1 − 1) (71 + 1 − 1), ×, (2 − 1), (7 − 1), , (23 − 1) (72 − 1), ×, 1, 6, 48, =7×, 6, = 56 = 2 × 28, \ 28 is a perfect number., Hence, the correct option is (c)., =, , 7 7 2 HCF of (7 , 7 , 2), 10. HCF =  , ,  =,  3 9 5  LCM of (3 , 9, 5), 1, 45, Hence, the correct option is (b)., =, , 4/11/2018 3:31:36 PM

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A.42, , Chapter 6 Real Numbers, LCM, and HCF, , 11. Let the numbers be 9x and 9y., 9x + 9y = 72 (Given), 9(x + y) = 72, x+y=8, \ (x, y) = (1, 7) or (3, 5), \ Number of pairs = 2, Hence, the correct option is (b)., , 2. 0.4 =, , 23 − 2 21, =, 9, 9, 21 4, −, 9, 9, 17, =, 9, 2.3 =, , 12. We know that the product of the numbers =, HCF × LCM of the numbers., 28 × x = 896, x = 32, \ The required number is 32., Hence, the correct option is (b)., , = 1.8, Hence, the correct option is (b)., , 14. A number having exactly 3 factors must be the, square of a prime number., , 6 × 216 = 6 × 36 × 6, = 6 ×6 6, =6×6, = 36, y = 40 ⇒ y = 1600, Hence, the correct option is (a)., 4. All the given choices can be true., Hence, the correct option is (d)., 5. 56 = 23 × 71, \ The sum of the factor =, , Prime numbers whose squares are less than, 1000 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and 31., The square of any prime number will have 3, factors., N = p2; the factors are 1, p, and p2., \ There are 11 such numbers which are less, than 1000., Hence, the correct option is (a)., , Assessment Test IV, 1. The required sequential order is ACBD., Hence, the correct option is (a)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 42, , (23 + 1 − 1), (71 + 1 − 1), ×, (2 − 1), (7 − 1), , (24 − 1), (72 − 1), ×, 1, 6, ( 49 − 1), = (16 − 1) ×, 6, =, , = 15 ×, , 48, 6, , = 120, Hence, the correct option is (c)., , 15. (i) Units digit of (725), = 74 × 6 + 1 = 7, (ii) Units digit of (239) = 24 × 9 + 3 = 8, (iii) Units digit of 6100 = 6, (iv) Units digit of 99 = 94 × 2 + 1 = 9, Option (b) follows., Hence, the correct option is (b)., , y − 4 = 6 × 216, , 3., , 13. Let the greatest four-digit number be N. Each, of the divisors exceeds the remainder it leaves, by the same value which is 3., \ N = k LCM (13, 11, 15) - 3 = k (2145) - 3 (k < 5), = 4(2145) - 3 = 8577, Hence, the correct option is (d)., , 4−0, 4, =, 1−, 0, 10 10, 9, , 6. 512 = 29, \ The number of factors of 512 = 9 + 1 = 10., Hence, the correct option is (c)., 7., , 2 128, 142, 64, 71, LCM = 2 × 64 × 71, = 9088, Hence, the correct option is (a)., , 4/11/2018 3:31:38 PM

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Hints and Explanation, 8. The HCF of (a and b ) = k, b  HCF (a , b ), k, a, The HCF of  and  =, = =1, k, k  LCM of k , k k, , \ Number of pairs = 9, Hence, the correct option is (d)., 12. Let a and b be the required numbers., , Hence, the correct option is (a)., 9. 496 =, , 24, , × 31, , Sum of the factors =, =, , Let a = 12, Given HCF = 12 and a; LCM = 72, a × b = 12 × 72, 24 × b = 12 × 72, b = 36, Hence, the correct option is (b)., , (24 +1 − 1) (311+1 − 1), ×, (2 − 1), (31 − 1), , (25 − 1) (312 − 1), ×, 1, 30, , (961 − 1), 30, 960, = 31 ×, 30, = 31 × 32, = 992, = 2 × 496, \ 496 is a perfect number., Hence, the correct option is (d)., = 31 ×, , 13. LCM of (7, 8, 11) = 616, The greatest number in 5 digits which is divisible by 616 is 9856., The difference between divisor and corresponding remainder in all the three cases is 3., The required number = 9856 - 3 = 9853., Hence, the correct option is (d)., 14. Only perfect squares have odd number factors., We have to find the number of perfect squares, between 50 and 500., The perfect squares in the given range are 82,, 92, 102, …, 222. So there are 22 − 7 = 15 perfect, squares in the given range which have an odd, number of factors., Hence, the correct option is (c)., , HCF of (5, 4 , 2), 5 4 2, 10. HCF =  ,,  7 11 , 7  = LCM of (7 , 11, 7 ), 1, 77, Hence, the correct option is (b)., =, , 11. Let the number be 11x and 11y., 11x + 11y = 209 ( Given), 11(x + y) = 209, x + y = 19, \ (x, y) = (1, 18), (2,17), (3, 16), (4, 15), (5, 14),, (6, 13), (7, 12), (8, 11), or (9, 10)., \, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 43, , A.43, , 15., , (i) Units digit of 489 = 44 × 22 + 1 = 4, (ii) Units digit of 536 = 5, (iii) Units digit of 1418 = 144 × 4 + 2 = 6, (iv) Units digit of (33)89 = (33)4 × 22 + 1 = 3, Hence, the correct option is (d)., , 4/11/2018 3:31:40 PM

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A.44, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, , CHAPTER 7, Squares and Square Roots, Cubes, and Cube Roots, and Indices, Assessment Test I, = −17 + 12, = −5, = 3 −125, Hence, the correct option is (a)., , 1. BDCA is the sequential order., Hence, the correct option is (d)., 2. 10,584 = 22 × 21 × 31 × 32 × 72, \ 10,584 should be divided by 6., i.e., (2 × 3), So that the quotient is a perfect square., Hence, the correct option is (d)., 3. A is a natural number., The sum of the first B consecutive odd numbers required to give the sum equal to A., \ B2 = A (successive subtraction method), \ A − B2 = 0, Hence, the correct option is (d)., , 8., , 50.41 + 34.81, , 4., , 5041, 3481, +, 100, 100, , =, , 5., , 7. 153 < 3388 < 163, 153 = 3375, \ 3388 − 3375 = 13, \ 13 should be subtracted form 3388 so that, the difference is a perfect cube., Hence, the correct option is (c)., x+, , 3, , 3, , 3, , 5 2, 4, x − 8 = x2 + 34, 8, 7, , 36, 4, = 4 + 9 = 13 = RHS, , =, , −4913 + 3 1728, 3, , ( −17 )3 + 3 123, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 44, , x+, , 36, 3, , x, , = 13, , LHS = 4 +, , If, , 35 x2 − 32 x2, = 42, 56, 3x2 = 42 × 56, x2 = 14 × 56, x2 = 14 × 14 × 4, ∴ x = ± 28, Hence, the correct option is (a)., 3, , x is, , x =4, , Let, 3, , 3, , x should be less than 13 and greater than 3., , 71 59 130, =, = 13, +, =, 10 10 10, Hence, the correct option is (c)., , 5 2 4 2, x − x = 34 + 8, 8, 7, , 6., , 36, , = 13, x, 3, x should be a factor of 36. Hence,, either 1, 2, 3, 4, 6, 9, 12, 18, or 36., Since 3 x + some number is 13., , 3, , 3, , x = 6, then 3 x +, , 36, 3, , = 6+, , x, , 36, 6, , = 6 + 6 = 12, \, , 3, , x≠ 6, , If, , 3, , x = 9, then, , = 9 + 4 = 13, , 3, , x+, , 36, 3, , x, , = 9+, , 36, 9, , \ 3 x = 4 or 9, Hence, the correct option is (b)., 9., , x6 × 3 x6, = x3 × x6/3, , 4/11/2018 3:31:45 PM

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Hints and Explanation, = x3 × x2, = x5, = x10, Hence, the correct option is (a)., 10. (729)x + 3 = (27) 5 + x, (36)x + 3 = (33)5 + x, 36x + 18 = 315 + 3x, ⇒ 6x + 18 = 15 + 3x, ⇒ 6x − 3x = 15 − 18, 3x = −3, x = −1, 1, = −1, \ xx = (−1)(−1) =, −1, Hence, the correct option is (c)., 11. a = 21458 = (26)243 = (64)243, b = 3972 = (34)243 = (81)243, c = 5729 = (53)243 = (125)243, d = 6486 = (62)243 = (36)243, \ c = 5729 is the greatest., Hence, the correct option is (b)., 12. Option (d): (7x)2 > 435, 202 = 400, 212 = 441, Let (7x)2 = (21)2 = 441, ⇒ 7x = 21 ⇒ x = 3, \ x can be 3., 13. Option (c): (3x)3 = 729, (3x)3 = 93, ⇒ 3x = 9 ⇒ x = 2, , Assessment Test II, 1. CADB is the sequential order of steps., Hence, the correct option is (c)., 2., 2, 2, 3, 3, 3, 3, 3, , \ 2916 should be divided by 22, i.e., 4., So that the quotient is a perfect cube., Hence, the correct option is (c)., 3., , 3844 = 62, \ The sum of the first 62 odd numbers is, required to give a sum of 3844., Hence, the correct option is (b)., , 4., , 3, 3, , 5., , 2, , 14x2 = 210 × 15, , 23, , x2 = 15 × 15, ∴ x = ± 15, Hence, the correct option is (c)., , (7x)2, , x=, , 1, 2, , \ x can be, , 1, ., 2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 45, , 7 2, 7, x − 10 = x2 + 200, 3, 5, , 35 x2 − 21x2, = 210, 15, , 2, , 5x = 51, ⇒ x2 = 1 ⇒ x = ± 1, <8, 72x < 8, Let 72x = 71 ⇒ 2x = 1, , (16)3 − 3 113, , 7 2 7 2, x − x = 200 + 10, 3, 5, , 5x = 3 53, , <, , 4096 − 3 1331, , = 16 − 11 = 5 = 25, Hence, the correct option is (b)., , 2, , 15. Option (a):, , 2916, 1458, 729, 243, 81, 27, 9, 3, , 2916 = 22 × 33 × 33, , 14. Option (b): 5x = 3 125, , (7x)2, , A.45, , 6., , 3, , −27 + 729 − 3 216 = x, −3 + 27 − 6 = x, ⇒ x = 18, , 4/11/2018 3:31:49 PM

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A.46, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, a = (2401)7 = (74)7 = 728, \ b = 730 is the greatest., Hence, the correct option is (b)., , 32 x = 32 × 18 = 16 × 2 × 2 × 29, = 4 × 2 × 3 = 24., Hence, the correct option is (a)., 7. 163 < 4096 < 173, 173 = 4913 > 4904, And 4913 − 4904 = 9, \ 9 should be added to 4904 so that the sum, is a perfect cube., Hence, the correct option is (b)., 8., , x+, , 42, x, 42, , x+, , x, , = 289, = 17, , x should be a factor of 42., i.e., 1, 2, 3, 6, 7, 14, 21, and 42, Either x = 3 or 14 satisfies., ⇒ x = 9 or x = 196, Hence, the correct option is (a)., 9., , 3, , = 7 × 28, = 7 × 2 = 14, Hence, the correct option is (b)., , x, , 14. Option (c): (216)x = 83 × 39, (63)x = 29 × 39, 63x = (2 × 3)9, ⇒ 3x = 9, x=3, 15. Option (b): (14x)x = 42 × (49)2, (14x)x = (22)2 × (72)2, (14x)x = (142)2, ⇒x=2, , 1. CBAD is the required sequential order., Hence, the correct option is (b)., , = 7 ×7 ×4, , x, , 13. Option (d): (125)x = (625)3, 53x = (54)3, ⇒ 3x = 4 × 3, ⇒x=4, , Assessment Test III, , 343 × 784, , 5, 2 , , 10.  3 −  .  1 + , , , , 7, 14 , , 12. Option (e): 83x = (3x)2, for x = 0, , −x, , 2. 4(x + 9)2 = 676, 676, ⇒ (x + 9)2 =, 4, , = (8 ), , 3, ,  26 , ⇒ (x + 9)2 =  ,  2, , x, ,  19   14 , 3,     = (8 ), 7, 19, , 26 ∴, ( x is positive.), 2, ⇒ x + 9 = 13, x=4, Hence, the correct option is (c)., , ⇒x+9=, , (19)x (14)x, 3, ×, = (8 ), x, x, (7 ) (19), (2)x = (8)3, 2x = 2 3 × 3, ⇒x=9, Hence, the correct option is (d)., 11. d =, c = (343)9 = (73)9 = 727, b = (49)15 = (72)15 = 730, 729, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 46, , 2, , 3., , 3, , 46656 + 3 42875, = 36 + 35 = 71, Hence, the correct option is (d)., , 4. (1000)3 = 1,00,00,00,000 has 10 digits., ∴ The maximum number of digits in the cube, of a 3-digit number is 9., Hence, the correct option is (d)., , 4/11/2018 3:31:52 PM

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Hints and Explanation, 5. 720 = 2 × 360, = 2 × 2 × 180, ⇒ 2 × 2 × 2 × 90, = 2 × 2 × 2 × 2 × 45, = 24 × 32 × 5, ∴ 720 should be multiplied by 5 so that the, product is a perfect square., Hence, the correct option is (a)., x°, × p r2, 360, , 6. A =, , 80°, × p r2, 360, 4 × 36 = r2, ⇒ r = 12, Hence, the correct option is (a)., 32p =, , 63, , 7., 6, , 365, -369, -4, , \ 3965 + 4 = 3669 = (63)2, Hence, the correct option is (a)., 8. 87,880 = 23 × 133 × 51, 23 × 51 × 13x = 23 × 51 × 133, \x=3, Hence, the correct option is (c)., 3, , ( x )3, 144 z2 y, =, , (6 y )3, 3, ,  y, 144 y  ,  12 , , 2, , 63 × y3, =6, 3, y2, 144 y ×, 144, Hence, the correct option is (c)., =, , 10., , (, , 4, , )( 5 − 3)(, = ( 5 ) − ( 3 )  (, , , 5+43, 4, , 5− 3, , )(, , 5+ 3, , ), , =5−3=2, Hence, the correct option is (a)., 11., , ( pa+ b )6 . ( pb+ c )6 ( pc+ a )6, ( pa pb .pc )12, 6, pa + b + b + c + c + a ), (, =, ( pa+ b+ c )12, 6, p2 a + 2b + 2c ), (, =, ( pa+ b+ c )12, ( pa+ b+ c )12 = 1, =, ( pa+ b+ c )12, ,  x,  y, 12.   and   are reciprocals to each other.,  x,  y, , -36, , 9., , (, , Hence, the correct option is (b)., , 3965, , 123, , =, , A.47, , 4, , 2, , 4, , 4, , 2, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 47, , ), 3), , 5+ 3, 5+, , If the sum of the reciprocals is ‘2’, then each, value is 1.,  x  y, \  =  =1,  y  x, 8, , 8, ,  x  11  y  11, ⇒  +  =1+1=2,  x,  y, Hence, the correct option is (a)., 13. 33x − 11 × 27y − 9 = 2592, 33x − 11 × 27y − 9 = 34 × 25, ⇒ 3x − 11 = 4 and 7y − 9 = 5, ⇒ 3x = 15 and 7y = 14, ⇒ x = 5 and y = 2, \ 3x + 2y, = 3(5) + 2(2), = 15 + 4 = 19, Hence, the correct option is (c)., 14. (2401)3x − 5 = (49)3y − 1 = 75x − 6, ⇒ (74)3x − 5 = 75x − 6, ⇒ 12x − 20 = 5x − 6, ⇒ 7x = 14, ⇒x=2, , 4/11/2018 3:31:58 PM

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A.48, , Chapter 7 Squares and Square Roots, Cubes and Cube Roots, and Indices, And 72(3y − 1) = 75x − 6, 76y − 2 = 75x − 6, ⇒ 6y − 2 = 5x − 6, 6y − 2 = 5 × 2 − 6, 6y − 2 = 4, y=1, \x+y=2+1=3, Hence, the correct option is (c)., , 15., , (i) 4 256 = 4 44 = 4,   (ii) The units digit of, , 3, , 1, 40 , 608 is 2., , (iii) The units digit of 5, 22, 729 is 3 or 7, (since the last digit is 9)., (iv) 5° = 1, (i) → (D); (ii) → (B); (iii) → (C); (iv) → (A), Hence, the correct option is (c)., , \ 432 should be multiplied by 3 so that the, product is a perfect square., Hence, the correct option is (b)., 6. l = 12p and r = 15 cm, x, l=, × 2p r, 360, x, × 2 × p × 15, 12p =, 360, 12 × 360, x=, 2 × 15, \ x = 144, Hence, the correct option is (b)., 64, 6 4091, 36, 124 491, 496, -5, , 7., , Assessment Test IV, , \ 4092 + 5 = (64)2, Hence, the correct option is (c)., , 1. BCADFE is the required sequential order., Hence, the correct option is (a)., , 8. 40,824 = 71 × 36 × 23, , 2. 9(x − 5)2 = 1296, 1296, ⇒ (x − 5)2 =, 9,  36 , ⇒ (x − 5)2 =  ,  3, , 23 × 3x × 71 = 23 × 36 × 71, ⇒x=6, Hence, the correct option is (d)., , 2, , ⇒ x − 5 = 12, , 9., , 3, , x = 17, Hence, the correct option is (d)., 3., , 17576 − 3 4913 = 26 − 17 = 9, Hence, the correct option is (c)., , 4. (1000)3 = 1,00,00,00,000 has 10 digits., \ The minimum number of digits in a 4-digit, number is 10., Hence, the correct option is (c)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 48, , 8(7 y )3, , =, , 3, , 5. 432 = 2 × 216, = 2 × 2 × 108, = 2 × 2 × 2 × 54, = 2 × 2 × 2 × 2 × 27, = 24 × 32 × 3, , 8 x3, 49 z2 y, , 3, ,  y, 49 y  ,  7, , 2, , 8(73 ) × y3, , =, , 3, ,  y2 , 49 y  ,  49 , , = 3 8 × 73, = 2 × 7 = 14, Hence, the correct option is (a)., 10., , (, , ) ( 8 + 5) ( 8 + 5), = ( 8 − 5 )( 8 + 5 ) ( 8 + 5 ), 4, , 8−45, 4, , 4, , 4, , 4, , 4, , 4, , 4/11/2018 3:32:03 PM

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Hints and Explanation, , ( 8 ) − ( 5 )   8 +, = ( 8 − 5) ( 8 + 5), =, , , 2, , 4, , 2, , 4, , ⇒ 10x = 10 and 5y = 10, x = 1 and y = 2, \ 2x + 5y = 2 × 1 + 5 × 2, = 2 + 10 = 12, Hence, the correct option is (a)., , 5 , , =8−5=3, Hence, the correct option is (b)., , 11., , (, , )( )(, ( xa xb xc )15, , 8, xa + b, , =, , 8, xb + c, , ), , 14. (81)3x − 5 = (9)3y + 2 = 35x − 6, , 8, xc + a, , ⇒ (81)3x − 5 = 35x − 6 and 93y + 2 = 35x − 6, ⇒ 34(3x − 5) = 35x − 6 and 36y + 4 = 35x − 6, = 12x − 20 = 5x − 6 and 6y + 4 = 5x − 6, ⇒ 12x − 20 = 5x − 6, ⇒ 7x = 14, x=2, And 6y + 4 = 5x − 6, ⇒ 6y = 5(2) − 6 −4, ⇒ 6y = 0, ⇒ y = 10, \x−y=2−0=2, Hence, the correct option is (b)., , ( xa+ b+ b+ c+ c+ a )8, ( xa+ b+ c )15, , x16( a + b + c ), x15( a + b + c ), = xa + b + c, Hence, the correct option is (d)., =, ,  a,  b, 12.   and   are reciprocals to each other.,  b,  a, If the sum of the reciprocals is 2, then each, value is 1., a b, = =1, b a, 7, , 7, ,  a  8  b  10, ⇒  − ,  b  a, =1−1=0, Hence, the correct option is (a)., 13. 210x − 6 × 55y − 5 = 50,000, 210x − 6 × 55y − 5 = 24 × 55, ⇒ 10x − 6 = 4 and 5y − 5 = 5, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 49, , A.49, , 15., , (i) 3−5 × 93 = 3−5 × 36 = 3, (ii) 2−3 × 3−3 × 64, = (6)−3 × 64 = 61, (iii) 4−3 × 9−2 × 363 = 4−1 × 36−2 × 363, 1,   = × 36 = 9, 4, (iv) 52 × 62 × (30)−1 × 3−1,   = (30)2 × (30)−1 × 3−1, 1,   = 30 × = 10, 3, (i) → (D); (ii) → (C); (iii) → (B); (iv) → (A), Hence, the correct option is (b)., , 4/11/2018 3:32:07 PM

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A.50, , Chapter 8 Time and Work, Time and Distance, and Statistics, , CHAPTER 8, Time and Work, Time and, Distance, and Statistics, Assessment Test I, 1. CBAD is the required sequential order., Hence, the correct option is (d)., 2. m1 d1 h1/w1 = m2 d2 h2/w2, 13 × 18 × 8/26 = (x + 13) 6 × 6/30, ⇒ x + 13 = 60 ⇒ x = 47, Hence, the correct option is (b)., 3. As A is twice as good a workman as B, A alone, can complete the work in half the time in which, B alone completes, if A takes x days and B takes, 2x days. As C is thrice as good a workman as B,, C alone completes a work in one-third the time, in which B alone completes., 2x, days., i.e., C takes, 3, 1 1, 3, 1, Then, +, +, =, x 2x 2x 2, 6, 1, ⇒, = ⇒x=6, 2x 2, Hence, the correct option is (d)., 4. Let C alone complete the work in x days., 6 6 4, + +, = 1 ⇒ x = 18 days, 12 x 24, Hence, the correct option is (b)., 5. Let us say the first pipe should be closed after x, minutes. Then,, x 16, +, =1, 18 24, ⇒, , x 2, + =1, 18 3, , x 1, = ⇒x=6, ⇒, 18 3, , 6. Time taken by the first car to reach Pune, 192, = 6h, 32, 1, The second car started 2 h later and reached, 2, 1, an hour earlier than the first car., 2, \ It takes 3 h less than the first car to reach Pune., 192, = 64 kmph, \Speed of the second car =, 6−3, Hence, the correct option is (a)., =, , 7. Let speed be x. After increasing by 25%, the, new speed = 1.25x, Time taken to cover a distance d with speed x., d, = 20, x, or d = 20x, Since distance is same,, d = 20x = t × 1.25x, 20, ⇒ t=, = 16 h, 1.25, Hence, the correct option is (b)., 20, 8. Speed of B =, = 4 m/s, 5, 500, Time taken by B to complete the race =, 4, = 185 s, Time taken by A to complete the race = 125 − 5, = 120 s, 500 25, =, m/s, Speed of A =, 120, 6, Hence, the correct option is (c)., 15, 20, A, B, , 10, 25, C, D, , The first pipe should be closed after 6 min., Hence, the correct option is (a)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 50, , 4/11/2018 3:32:11 PM

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Hints and Explanation, 9. Let the speed of the boat in still water be, x kmph., Let the speed of the stream be y kmph., (x + y) 3 = 48 ⇒ x + y = 16, (1), (x − y) 4 = 40 ⇒ x − y = 10, (2), The speed of the boat in still water is given by, 1, (Its upstream speed + Its downstream speed), 2, 1, =, (10 + 16) = 13 kmph, 2, Hence, the correct option is (a)., , Assessment Test II, 1. BADC is the required sequential order., Hence, the correct option is (b)., 2. m1 d1/w1 = m2 d2/w2, 10, 12, =x×, 900, 660, ⇒ x = 11, Hence, the correct option is (c)., 18 ×, , 3. Let A take x days to complete the work., \ B takes 2x days., , 10. Mode = 3 Median − 2 Mean, 12 = 3 Median − 2 × 3 ⇒ Median = 18/3 = 6, Hence, the correct option is (d)., , 1 1, 1, +, =, x 2 x 14, , x, , f, , fx, , 4, , 3, , 12, , 7, , 10, , 70, , 3, = 21 days, 2, Hence, the correct option is (d)., , 11., , 9, , 15, , 135, , 13, , p, , 13p, , 18, , 12, , 216, , 20, , 2p − 4, , 40p − 80, , Sf = 36 + 3p, , Sfx = 353 + 53p, , Given, AM = 12.95 AM and =, , ∑ fx, ∑f, , 353 + 53 p, 36 + 3 p, 466.20 + 38.85p = 353 + 53p, 113.20 = 14.15 p, 113.20, p=, =8, 14.15, Hence, the correct option is (b)., ⇒ 12.95 =, , 12. Option (h): B is twice as efficient as A., 13. Option (e): 2 men’s capacity is equal to 3, women’s capacity., 14. Option (d): , Q is faster than P since 10 m/s, 18, kmph = 36 km/h, 5, 15. Option (f): E takes less time to fill the tank than F., = 10 ×, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 51, , A.51, , (1 + 2) 1, =, 2x, 14, ⇒ x = 14 ×, , 4. Let the remaining work be completed in x days., 5, 5, x+, +, =1, 10 15, 5, 2, x+, =, 10, 3, 5, 3x + 15 = 20 ⇒ x =, 3, Hence, the correct option is (d)., 5. The time taken by the outlet pipe to empty the, tank is less than the time taken by the inlet pipe, to fill the tank. If the inlet pipe and the outlet, pipe are opened when the tank is partially or, totally full, then the tank will get emptied. As, both pipes are opened when the tank is half, full, the tank will become empty., So the part of tank emptied by both the pipes, working together in one minute, 1 1, 1, −, =, 8 12 24, The time taken to empty half the tank, =, , 1 24, ×, = 12 min, 2 1, Hence, the correct option is (d)., =, , 4/11/2018 3:32:16 PM

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A.52, , Chapter 8 Time and Work, Time and Distance, and Statistics, , 6. Distance travelled by the thief from 11:00 am to, 1:00 pm = 2 × 30 = 60 km, Time in which the police will catch the thief, =, , Distance, 60, =, =6h, Relative speed ( 40 − 30), , Time at which the police catches the thief is, 1:00 pm + 6 h, i.e., 7:00 pm, Hence, the correct option is (d)., 7. The distance remains constant. If the usual, speed be x and usual time be t, then, 4, 9, x × t = x  t +  = distance, 7  60 , 4, 3, t+, 7, 35, 3 7 1, × = h, t=, 35 3 5, 1, t = × 60 = 12 min, 5, Alternatively,, t=, , 11. Sf1 = 12 + x + y = 22 ⇒ x + y = 10, and S fixi = 20 + 8x + 18 + 60 + 15y = 98 + 8x + 15y, , ∑ fi xi, ∑ fi, , =, , 98 + 8 x + 15 y, = 10, 22, , On solving 8x + 15y = 122 and x + y = 10, we get, x=4, Hence, the correct option is (b)., , Hence, the correct option is (b)., 8. As Rohan beats Mohan by 60 m or 15 s, in order, to run the final 60 m, Mohan takes 15 s., 60, = 4 m/s, 15, , ∴ The time taken by Mohan to run the race =, 300, = 75 s, 4, The time taken by Rohan to run the race =, 75 − 15 = 60 s, 300, = 5 m/s, 60, Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 52, , We know that, Mode = 3 Median − 2 Mean, Mode − Median = 2(Median − Mean) = 2 × 8 = 16, Hence, the correct option is (a)., , \ 8x + 15y = 122, , 4, of usual speed, he takes, 7, 7, 3, th of usual time, i.e.,, of usual time extra., 4, 4, 3, If, → 9 min, 4, 1→?, 9, ∴, = 12 min, 3/4, , Rohan’s speed =, , 10. Given, Median − Mean = 8, , x=, , As he travels at, , ∴ Mohan’s speed =, , 9. Speed of boat in still water, 28 + 44, =, = 36 kmph, 2, 44 − 28, Speed of stream =, = 8 kmph, 2, Hence, the correct option is (b)., , 12. Option (d): A is more efficient than B., 13. Option (e): 1 man can do the work in 45 days., 18, kmph, 14. Option (a): Car’s speed = 5 m/s = 5 ×, 5, = 18 kmph, Train’s speed = 20 kmph, Train is faster than car., \ Car takes more time to travel a certain distance when compared with train., 15. Option (g): R takes more time to fill the tank, when compared with P., , Assessment Test III, 1. CDAB is the required sequential order., Hence, the correct option is (a)., 2. Let p, q, and r be the capacities of P, Q, and R,, respectively., 1, 1, 1, + +, (p + q + q + r + r + p) =, 24 18 12, 3+4+6, 2(p + q + r) =, 72, , 4/11/2018 3:32:20 PM

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Hints and Explanation, , 5. X and Y are filling pipes, whereas Z is an emptying pipe., , 13, 72, 13, p+q+r=, 144, 2(p + q + r) =, , (p + q + r) − (p + r) =, q=, , 13 − 12, 144, , 13, 1, −, 144 12, , 1, 144, \ Q can complete the work in 144 days., Hence, the correct option is (a)., q=, , 3., , M1 D1 H1, W1, , =, , M2 D2 H2, , W2, 3, M×D×8, M × 12 × 6 4, =, W, 3W, 12 × 6 × 3 × 4, D=, = 36, 3×8, Hence, the correct option is (b)., , 4. Let m, n, and p are the capacities of M, N, and P,, respectively., 1, m+n+p=, 3, 1 1, 1, + +p=, 8 12, 3, 1 1 1, ⇒p= − −, 3 8 12, 1 1, p= −, 6 24, 4 −1 3 1, =, =, =, 24, 24 8, 1 1 1, Ratio of the shares of M, N, and P = 18 : 12 : 8, = 3:2:3, 2, × 1200 = `300, 8, Hence, the correct option is (d)., Share of N =, , Another Method:, N’s work =, , 3 1, =, 12 4, , Share of N =, , 1, (`1200) = `300, 4, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 53, , A.53, , The part of the tank that can be filled in 1 h =, 1 1 1 4+3−2 5, + − =, =, 12, 3 4 6, 12, 12, 2, \ Time required to fill the tank =, =2 h, 5, 5, Hence, the correct option is (b)., 6. Let x and y be the capacities of a man and a, woman, respectively., 1, 3x + 4y = (1), 12, 1, 12x + 4y = (2), 4, 1 1, From Eqs (1) and (2), ⇒ 9x = −, 4 12, 1, 9x =, 6, 1, x=, , (3), 54, 3, 1, + 4y =, From Eqs (1) and (3), ⇒, 54, 12, 1, 1, 4y =, −, 12 18, 1, 4y =, 36, 1, y=, 144, \ A woman can complete the work in 144 days., Hence, the correct option is (a)., 7. Let the speed of the current be x kmph., Speed of the boat upstream = (6 − x) kmph., 25, \, =5, 6−x, ⇒5=6−x, x=1, Hence, the correct option is (a)., 8. Given P:Q =, , P 1000, =, Q 900, , Q 1000, =, R, 920, P Q 1000 1000, × =, ×, Q R 900 920, P 1000, ⇒ =, R 828, , 4/11/2018 3:32:28 PM

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A.54, , Chapter 8 Time and Work, Time and Distance, and Statistics, P:R = 1000:828, \ P beats R by 172 m., Hence, the correct option is (c)., , 9. Speed of the train =, , Length of the train, Time taken to cross the pole, , 300, = 20 m/s, 15, Time taken by the train to cross a platform of, 180 + 300, 480, =, length 180 m =, = 24 s, 20, 20, Hence, the correct option is (b)., , =, , 10. Distance travelled, =, , VU, ( p + q), V −U, , [where V and U are the different speeds, and, p and q are the time early and late in hours,, respectively.], 8 × 6  12 10  48 22, +, ×, , =, 8 − 6  60 60 , 2, 60, 44, =, = 8.8 km, 5, Hence, the correct option is (c)., Alternate Solution:, Let the distance be d km., The difference between the time taken = 22 min, , 13. Sfx = 2 × 2 + 3 × 4 + 6 × 5 + 7 × 7 + P × 4, Sf = 2 + 4 + 5 + 7 + 4 = 22, Σfx 95 + 4 P, =, = 6.5, Σf, 22, ⇒ 95 + 4P = 143, ⇒ 4P = 48, ⇒ P = 12, Hence, the correct option is (a)., , Mean =, , 14. Mode = 3 Median − 2 Mean, 14 = 3 × Median − 2 × 17, 14 + 34 = 3 Median, \ Median = 16, Hence, the correct option is (b)., 15.    (i) Given data 12, 14, 13, 12, 14, 13, 14, 12, Arrange in ascending order., 12, 12, 12, 12, 13, 14, 14, 14, 12 + 13, Median =, = 12.5, 2, Mode is 12., , =, , d d 22, − =, 6 8 60, 22, d, ⇒, =, 24 60, 44, ⇒d=, = 8.8 km, 5, ⇒, , 11. Required time =, =, , Length of the faster train, Relative speed, , 103, 8,    (ii) Given data 14, 13, 12, 12, 13, 13, 14, 14, Arrange in ascending order., 12, 12, 13, 13, 13, 14, 14, 14, Mean =, , 13 + 13, = 13, 2, Mode is 13 and 14., Mean = 13.125, (iii) Given data 11, 12, 13, 14., (iv) Given data 11, 11, 12, 13, 14, 14, 13, 13, Median =, , Mean =, , 120, , (92 − 74) ×, , 5, 18, , 120 18, ×, = 24 s, 18, 5, Hence, the correct option is (a)., =, , 12. 720 units = 6 cm, 1 cm = 120 units, \ K = 120, Hence, the correct option is (b)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 54, , 11 + 11 + 12 + 13 + 14 + 14 + 13 + 13, 8, , 101, 8, (i) → (C); (ii) → (B); (iii) → (A); (iv) → (D), , , , =, , Hence, the correct option is (b)., , Assessment Test IV, 1. BADC is the required sequential order., Hence, the correct option is (b)., , 4/11/2018 3:32:33 PM

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Hints and Explanation, 2. Let x, y, and z be the capacities of X, Y, and Z,, respectively., 1, 1, 1, +, +, (x + y + y + z + z + x) =, 20 24 30, 2(x + y + z) =, x+y+z=, , 6+5+ 4, 120, , 15, 240, , (x + y + z) − (y + z) =, , 15, 1, 15 − 10, −, =, 240 24, 240, , 5, 1, =, 240 48, \ X can do the work in 48 days., , 3., , =, , 6 − 2 −1 1, =, 12, 2, , 1, (`2400) = `1200, 2, 5. M and N are filling pipes, whereas P is an emptying pipe., Share of P =, , The part of the tank that can be filled in one, minute when all the three pipes are turned on, 1, 1, 1, +, −, =, 20 30 24, 6+4−5, 5, 1, =, =, 120, 120 24, , x=, , =, , Hence, the correct option is (b)., , \ In 24 min, the tank will be filled., Hence, the correct option is (d)., , M1 D1 H1, W1, , =, , M2 D2 H2, W2, , 5, M×D×6, M × 15 × 9 4, =, W, 2W, 15 × 9 × 2 × 4, D=, 5×6, D = 36, Hence, the correct option is (c)., 4. Let a, b, and p be the capacities of A, B, and P, respectively., 1, a+b+p=, 2, 1, 1, 1 1, +, + =, 6 12 p 2, 1 1 1 6 − 2−1 3 1, − −, =, =, =, 2 6 12, 12, 12 4, Ratio of the share of A, B, and P, p=, , 1 1 1, :, : = 2:1:3, 6 12 4, 3, Share of P = × 2400 = `1200, 6, Hence, the correct option is (c)., Another method:, 2 2, Work done by P = 1 − −, 6 12, =1−, , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 55, , 1 1, −, 3 6, , A.55, , 6. Let x and y be the capacities of a man and a, woman, respectively., 1, 5x + 3y =, (1), 20, 9x + 8y =, , 1, (2), 8, , Eq. (1) × 8 − Eq. (2) × 3, 13x =, , 16 − 15, 2, 3, −, =, 5, 8, 40, , 1, 520, \ A man can complete the work in 520 days., Hence, the correct option is (b)., ⇒x=, , 7. Let the speed of the boat in still water be x, kmph., The speed of the boat in downstream = x + 5, 24, =8, 3, ⇒8=x+5, =, , ⇒ x = 3 kmph, Hence, the correct option is (c)., 8. Given A:B = 1000:875, B:C = 1000:900, A B 1000 1000, ×, =, ×, B C, 875, 900, , 4/11/2018 3:32:42 PM

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A.56, , Chapter 8 Time and Work, Time and Distance, and Statistics, A 1000, =, C 787.5, \ A:C = 1000:787.5, , 11. Required time =, , \ A beats C by 212.5 m., Hence, the correct option is (a)., 9. The speed of the train, =, , Length of the train + Length of the platfrom, Time taken, , 450 + 150 600, =, 20, 20, = 30 m/s, Time taken by the train to cross a pole, , =, , =, , Length of the train, Speed of the train, , 450, = 15 s, 30, Hence, the correct option is (a)., , 10. Required distance =, 8×5 6, 9, +, =, , , 8 − 5  60 60 , =, , 40 15, ×, 3, 60, , =, , 10, 1, = 3 km, 3, 3, , VU, ( p + q), V −U, , Hence, the correct option is (d)., , M01_IIT-FOUNDATION_XXXX_CH01_HINTS.indd 56, , 360 × 18, =, =9s, 5, 144 × 5, (100 + 44) ×, 18, Hence, the correct option is (d)., , =, , 160 + 200, , 12. 1 cm = 40 units, 580, 580 units =, cm, 40, = 14.5 cm, Hence, the correct option is (c)., 13. Mean =, ⇒, , =, , Length of the trains, Relative speed, , Σfx, =9, Σf, , 60 + 56 + 96 + 90 + 10 K + 180, =9, 55 + K, , 482 + 10K = 9K + 495, K = 13, Hence, the correct option is (c)., 14. Mode = 3 Medians − 2 Mean, 16 = 3 × 15 − 2 Mean, 2 Mean = 45 − 16, Mean = 14.5., Hence, the correct option is (a)., 15. (i) → (C); (ii) → (A); (iii) → (B); (iv) → (D), Hence, the correct option is (c)., , 4/11/2018 3:32:45 PM