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Chap 4 : Quadratic Equation, , www.cbse.online, , CHAPTER 4, Quadratic Equation, TOPIC 1 :, , S, E, , Q, , kx2 - 14x + 8 = 0, , We have, , Let one root be a and other root be 6a ., Sum of roots, a + 6a = 14, k, , VERY SHORT ANSWER TYPE QUESTIONS, 1., , 7a = 14 or a = 2, k, k, , 2, , Find the positive root of 3x + 6 = 9 ., Ans :, [Board Term-2, 2015, Set UDICCY2], We have, , Product of roots, , 3x2 + 6 = 9, 3x + 6 = 81, , 6a2 = 8, k, , 6 # 42 = 8, k, k, , Thus, x =! 5, Hence 5 is positive root., , 3 =1, k, k2, , If x =- 12 , is a solution of the quadratic equation, 3x2 + 2kx - 3 = 0 , find the value of k ., [Board Term-2, 2015, Delhi CBSE (Set, I, II, III)], Ans :, [CBSE Marking Scheme, 2015], , 3k = k2, 3k - k2 = 0, k 63 - k @ = 0, , 3x2 + 2kx - 3 = 0, , We have, , Putting x =- 1 we get, 2, , k = 0 or k = 3, Since k = 0 is not possible, therefore k = 3 ., , 2, 3 b- 1 l + 2k b- 1 l - 3 = 0, 2, 2, , 5., , 3 -k-3 = 0, 4, k = 3 -3, 4, , of, , the, , 2, 6b 2 l - 2 - k = 0, 3, 3, , 6# 4 -2-k = 0, 3 3, k = 6 # 4 - 2 = 24 - 6 = 2, 9 3, 9, Thus k = 2 ., , equation, , [Board Term-2, 2012, (35)2011 (A1)], 3 x2 - 2x -, , 3 =0, , 3 x - 3x + x -, , 3 =0, , We have, 2, , 3 x ^x -, , 4., , quadratic, , 3 h + 1 ^x -, , 6x2 - x - k = 0, , Substituting x = 2 , we get, 3, , Hence k = - 9, 4, Find the roots, 3 x2 - 2x - 3 ., Ans :, , If one root of the quadratic equation 6x2 - x - k = 0 is, 2 , then find the value of k ., 3, Ans :, [Board Term-II foreign-2, 2017], We have, , = 3 - 12 = - 9, 4, 4, , 3., , ...(2), , Solving (1) and (2), we obtain, 2, 6b 2 l = 8, k, k, , 3x2 = 81 - 6 = 75, x2 = 75 = 25, 3, , 2., , a (6a) = 8, k, , or,, , 2, , ...(1), , 3h = 0, , ^x - 3 h^ 3 + 1h = 0, Thus, x = 3, -1, 3, Find the value of k , for which one root of the quadratic, equation kx2 - 14x + 8 = 0 is six times the other., Ans :, [Board Term-2, 201], [Board Sample Paper 2016], , 6., , Find the value (s) of k if the quadratic equation, 3x2 - k 3 x + 4 = 0 has real roots., Ans :, [Sample Question Paper 2017], If discriminant of quadratic equation is equal to zero,, or more than zero, then roots are real., We have, , 3x2 - k 3 x + 4 = 0, ax2 + bx + c = 0, , Compare with, , D = b2 - 4ac, b2 - 4ac $ 0, , For real roots, , ^- k 3 h - 4 # 3 # 4 $ 0, 2, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 54

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Chap 4 : Quadratic Equation, , www.rava.org.in, =0, , 2, , 3k - 48 $ 0, , ^ 3 x + 2h^4x - 3 h, , 2, , k - 16 $ 0, , ^k - 4h^k + 4h $ 0, Thus k #- 4 and k $ 4, , =0, , NO NEED TO PURCHASE ANY BOOKS, , For session 2019-2020 free pdf will be available at, www.cbse.online for, 1. Previous 15 Years Exams Chapter-wise Question, Bank, 2. Previous Ten Years Exam Paper (Paper-wise)., 3. 20 Model Paper (All Solved)., 4. NCERT Solutions, All material will be solved and free pdf. It will be, provided by 30 September and will be updated regularly., , Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education,, New Delhi in any manner. www.cbse.online is a private organization which provide free, study material pdfs to students. At www.cbse.online CBSE stands for Canny Books, For School Education, , 4., , We have x2 - ^ 3 + 1h x +, 2, , x x ^x Thus x =, 5., , SHORT ANSWER TYPE QUESTIONS - I, 1., , Thus x = - 2 , 3, 3 4, Solve for x : x2 - ^ 3 + 1h x +, Ans :, , 2, , Find the roots the quadratic equation 6x - x - 2 = 0 ., Ans :, [Board Term-2, 2012, Set (13)], , 3 =0, 3 =0, , 3 x - 1x +, 3 h - 1 ^x -, , 3h = 0, , ^x - 3 h^x - 1h = 0, 3,x = 1, , Find the roots of the following quadratic equation :, 1, ^x + 3h^x - 1h = 3 bx - 3 l, Ans :, , [Board Term-2, 2012, Set (52), 2011 Set (A1)], , We have, , 1, ^x + 3h^x - 1h = 3 bx - 3 l, x2 + 2x - 3 = 3x - 1, , 6x2 - x - 2 = 0, , x2 - x - 2 = 0, , 6x2 + 3x - 4x - 2 = 0, , x2 - 2x + x - 2 = 0, , We have, , x ^x - 2h + 1 ^x - 2h = 0, , 3x ^2x + 1h - 2 ^2x + 1h = 0, , ^2x + 1h^3x - 2h = 0, 3x - 2 = 0 or 2x + 1 = 0, x = 2 or x =- 1, 2, 3, Hence roots of equation are 2 and - 1 ., 3, 2, , ^x - 2h^x + 1h = 0, Thus x = 2, - 1, 6., , Find the roots of the following quadratic equation :, 2 x2 - x - 3 = 0, 5, 5, [Board Term-2, 2012 Set (40)], , Ans :, 2., , Find the roots of the following quadratic equation :, 15x2 - 10 6 x + 10 = 0, Ans :, [Board Term-2, 2012 Set (1)], , 2 x2 - x - 3 = 0, 5, 5, , We have, , 15x - 10 6 x + 10 = 0, , 2x2 - 5x - 3 = 0, 5, , 3x2 - 2 6 x + 2, , 2x2 - 5x - 3 = 0, , 2, , We have, =0, , 2x2 - 6x + x - 3 = 0, 3x2 -, , 6x-, , 2x ^x - 3h + 1 ^x - 3h = 0, , 6x+2, , ^2x + 1h^x - 3h = 0, , =0, 3 x^ 3 x -, , 2h-, , 2^ 3 x -, , 2h, , =0, , 3., , 3 =0, [Foreign Set, II, III, 2015, , ^ 3 x - 2 h^ 3 x - 2 h = 0, Thus x = 2 , 2, 3, 3, Solve the following quadratic equation for x :, 4 3 x 2 + 5x - 2 3 = 0, Ans :, [Board term-2, 2013, 2012, Set (22)], We have, , 4 3 x2 + 5x - 2 3 = 0, 4 3 x2 + 8x - 3x - 2 3, , =0, 4x ^ 3 x + 2h -, , 3 ^ 3 x + 2h, , Thus x = - 1 , 3, 2, 7., , Solve the following quadratic equation for x :, 4x2 - 4a2 x + ^a 4 - b 4h = 0, [Delhi CBSE Term-2, 2015 (Set I, II)], , Ans :, , 4x2 - 4a2 x + ^a 4 - b 4h = 0, , We have, , Compare with Ax2 + Bx + C = 0 we have, A = 4, B =- 4a2, C = ^a 4 - b 4h, 2, x = - B ! B - 4AC, 2A, , =, , 4a 2 !, , 2 2, 4, 4, ^- 4a h - 4 # 4 ^a - b h, 2#4, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 55

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Chap 4 : Quadratic Equation, , www.rava.org.in, , 14. Solve for x (in ferms of a and b ) :, , completing square., Ans :, , a + b = 2, x ! a , b, x-b x-a, [Board Term-2 Foreign Set II, 2016], , Ans :, We have, , [Board Term-2, 2015], , We have, x2 - 4x - 8 = 0, Squaring both side we have, , a ^x - a h + b ^x - b h, =2, ^x - b h^x - a h, , ^x - 2h2 - 8 - 4 = 0, ^x - 2h2 - 12 = 0, , ^x - 2h2 = 12, 2, ^x - 2h2 = ^2 3 h, x - 2 =! 2 3, , a ^x - a h + b ^x - b h = 2 8x2 - ^a + b h x + abB, , ax - a2 + bx - b2 = 2x2 - 2 ^a + b h x + 2ab, 2x2 - 3 ^a + b h x + ^a + b h2 = 0, , 2x2 - 2 ^a + b h x - ^a - b h x + ^a + b h2 = 0, , x = 2!2 3, , 82x - ^a + b hB8x - ^a + b hB = 0, x = a + b, a + b, 2, , Thus, 15. Solve for x :, , Ans :, , Thus x = 2 + 2 3 , 2 - 2 3, NO NEED TO PURCHASE ANY BOOKS, , 3 x2 - 2 2 x - 2 3 = 0, [Board Term-2 Foreign Set II, 2016], , We have 3 x2 - 3 2 x +, 3 x 6x -, , 6@ +, , 2x-2 3 = 0, 2 6x -, , 6@ = 0, , ^x - 6 h^ 3 x + 2 h = 0, x 6, - 2, 3, , Thus, , 2 and x =- 3 are roots of the quadratic, 3 2, equation ax + 7x + b = 0 , find the values of a and b ., Ans :, [Board Term-2 Delhi Set I, II, III, 2016], , 16. If x =, , Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education,, New Delhi in any manner. www.cbse.online is a private organization which provide free, study material pdfs to students. At www.cbse.online CBSE stands for Canny Books, For School Education, , 19. A two digit number is four times the sum of the digits., , ax2 + 7x + b = 0, , We have, , For session 2019-2020 free pdf will be available at, www.cbse.online for, 1. Previous 15 Years Exams Chapter-wise Question, Bank, 2. Previous Ten Years Exam Paper (Paper-wise)., 3. 20 Model Paper (All Solved)., 4. NCERT Solutions, All material will be solved and free pdf. It will be, provided by 30 September and will be updated regularly., , It also equal to 3 times the product of digits. Find the, number., Ans :, [Board Term-2, Foreign Set I, 2016], , Substituting x = 2 in above equation we obtain, 3, 4 a + 14 + b = 0, 9, 3, 4a + 42 + 9b = 0, 4a + 9b =- 42, , (1), , Let units digit and tens digit of the two digit number, be x and y respectively., Thus number is 10y + x, According to question, we have, 10y + x = 4 ^y + x h, 10y + x = 4y + 4x, , and substituting x =- 3 we obtain, 9a - 21 + b = 0, 9a + b = 21, , (2), , 10y - 4y = 4x - x, , Solving (1) and (2), we get a = 3 and b =- 6, 17., , Solve for x :, Ans :, We have, , 2y = x, , 6x + 7 - ^2x - 7h = 0, [O. D. Set III, 2016], , 6x + 7 - ^2x - 7h = 0, , 6x + 7 = ^2x - 7h, Squaring both sides we get, or,, , 6x + 7 = ^2x - 7h2, 6x + 7 = 4x2 - 28x + 49, 4x2 - 34x + 42 = 0, 2x2 - 17x + 21 = 0, 2x2 - 14x - 3x + 21 = 0, 2x ^x - 7h - 3 ^x - 7h = 0, , 6y = 3x, , ^x - 7h^2x - 3h = 0, Thus x = 7 and x = 2, 3, 2, 18. Find the roots of x - 4x - 8 = 0 by the method of, , Also,, , 10y + x = 3xy, 10y + 2y = 3 ^2y h y, 12y = 6y2, 6y2 - 12y = 0, 6y ^y - 2h = 0, y = 0 or y = 2, , As the number can not be zero x = 4 and x = 2y = 4 ., Thus required number is 24., 20. In a cricket match, Harbhajan took three wickets less, , than twice the number of wickets taken by Zahir. The, Product of the number of wickets taken by these two, is 20. Represent the above situation in the form of, quadratic equation., [Board Term-2, 2015], Ans :, [CBSE Marking Scheme, 2015], , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 57

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Chap 4 : Quadratic Equation, , www.rava.org.in, , ^x + 5 3 h^x - 2 3 h = 0, Thus x =- 5 3 , 2 3, , [Board Term-2, Delhi Set III, 2016], , Ans :, , x2 +, , Solve for x : x2 + 5x - ^a2 + a - 6h = 0, Ans :, [Board Term-2 Foreign Set I, 2015], , 8., , x2 + b a + a + b l x + 1 = 0, a, a+b, , We have, , a x+a+bx+1 = 0, a, a+b, , x3 + 5x - ^a2 + a - 6h = 0, , x ax + a k + a + b ax + a k = 0, a, a+b, a+b, , x =, , -5 !, , - b ! b2 - 4ac, ;` x =, E, 2a, , a, a+b, ax + a + b kbx + a l = 0, - ^a + b h, x = -a ,, a, a+b, , Thus, 5., , =, , Solve for x :, 1, 1, = 2 ; x ! 1, 2 , 3, +, ^x - 1h^x - 2h ^x - 2h^x - 3h 3, 1, 1, =2, +, 3, ^x - 1h^x - 2h ^x - 2h^x - 3h, , We have, , x = a - 2, x =- ^a + 3h, Solve for x : x2 - ^2b - 1h x + ^b2 - b - 20h = 0, Ans :, [Board Term-2 Foreign Set II, 2015], , 9., , x-3+x-1, =2, 3, ^x - 1h^x - 2h^x - 3h, , We have x2 - ^2b - 1h x + ^b2 - b - 20h = 0, Compare with Ax2 + Bx + C = 0 we have, , 2x = 4, =2, 3, ^x - 1h^x - 2h^x - 3h, , A = 1, B =- (2b - 1), C = ^b2 - b - 20h, 2, x = - B ! B - 4AC, 2A, , 2 ^x - 2h, =2, 3, x, 1, ^, h^x - 2h^x - 3h, 2, =2, 3, ^x - 1h^x - 3h, , x =, , 3 = ^x - 1h^x - 3h, , =, , 2, , x - 4x + 3 = 3, x ^x - 4h = 0, Solve for x :, 3 x2 - 2 2 x - 2 3 = 0, Ans :, [Board Term-2, Outside Delhi CBSE, 2015 Set I, III,, Foreign Set I, II, II, 2014], 3 x2 - 2 2 x - 2 3 = 0, , 3 x2 -, , 3, , 3 x 6x -, , 3 x 6x Thus x =, 7., , 3 x2 - 63 2 -, , 2 @x - 2 3 = 0, , 3 x2 - 3 2 x +, , 2x-2 3 = 0, , 3, , 2x+, 3 . 2@ +, , 6@ +, , 2x-, , 2, , 2 6x -, , 2 6x -, , 2, 2, , 6@ = 0, , Thus x = b + 4 and x = b - 5, Add 8905629969 in Your Class Whatsapp Group to Get All PDFs, 10. Solve for x :, , Ans :, We have, , 3 =0, , 16x + 16 - 15x = x2 + x, x + 16 = x2 + x, x2 - 16 = 0, x2 = 16, x =! 4, , 2x2 + 6 3 x - 60 = 0, Thus x = - 4 and x =+ 4, , x2 + x 3 x - 30 = 0, x ^x + 5 3 h - 2 3 ^x + 5 3 h = 0, , 16 - 1 = 15, x, x+1, , 16 ^x + 1h - 15x = x2 + x, , ^x - 6 h^ 3 x + 2 h = 0, 6 =- 2, 3, , x2 + 5 3 x - 2 3 x - 30 = 0, , 16 - 1 = 15 ; x ! 0, - 1, x, x+1, [Board Term-2, OD 2014], , 16 - 15 = 1, x, x+1, , 3@ = 0, , Solve for x : 2x2 + 6 3 x - 60 = 0, Ans :, [Board Term-2, O.D. CBSE, 2015, Set II], We have, , ^2b - 1h ! 9, 2, , = b + 4, b - 5, , Thus x = 0 or x = 4, , We have, , 2, ^2b - 1h ! ^2b - 1h2 - 4 ^b - b - 20h, 2, , = 2b + 8 , 2b - 10, 2, 2, , x2 - 4x = 0, , 6., , - 5 ! ^2a + 1h, 2, , = 2a - 4 , - 2a - 6, 2, 2, , [Board Term-2, O.D. Set I, 2016 ], , Ans :, , 25 + 4 ^a2 + a - 6h, 2, , 11., , Solve the quadratic equation ^x - 1h2 - 5 ^x - 1h - 6 = 0, Ans :, [Board Term-2, 2015], , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 59

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Chap 4 : Quadratic Equation, , www.cbse.online, , As x can’t be negative, x = 6 ., 21. Three consecutive natural number are such that the, , squre of the middle number exceeds the difference of, the squares of the other two by 60. Find the number., Ans :, [Board Term-2, O.D. Set III, 2016], , 3x - 8 x - 60 = 0, x = y , then we have, , Let, , 3y2 - 8y - 60 = 0, 3y2 - 18y + 10y - 60 = 0, 3y ^y - 6h + 10 ^y - 6h = 0, , Let the three consecutive natural numbers be x, x + 1, and x + 2 ., , ^x + 1h2 = ^x + 2h2 - ^x h2 + 60, x2 + 2x + 1 = x2 + 4x + 4 - x2 + 60, , Now, , ^3y + 10h^y - 6h = 0, , y = 6 or y =- 10, 3, 10, Here y = is not possible., 3, Thus y = 6 or y2 = 36 ,, x = y2 = 36, , x2 - 2x - 63 = 0, x2 - 9x + 7x - 63 = 0, x ^x - 9h + 7 ^x - 9h = 0, , Hence the number of camels is 36., , ^x - 9h^x + 7h = 0, x = 9 or x =- 7, , 25. The sum of the squares of two consecutive naturals is, , 421. Find the numbers., , As x can’t be negative, x = 9 ., , Ans :, , Hence three numbers are 9, 10, 11., x, , Let the first natural number be x . The second, , y, , 22. If ^x2 + y2h^a2 + b2h = ^ax + by h2 . Prove that, =, a, b, , [Board Term-2, 2014], , Ans :, , consecutive natural will be .x + 1, According to the question,, , ^x + y h^a + b h = ^ax + by h, 2 2, x a + x2 b2 + y2 a2 + y2 b2 = a2 x2 + b2 y2 + 2abxy, , We have, , 2, , 2, , 2, , [Board Term-2, 2012 Set (12)], [CBSE Marking Scheme, 2012], , 2, , x2 + ^x + 1h2 = 421, , 2, , x2 + x2 + 2x + 1 = 421, , x2 b2 + y2 a2 - 2abxy = 0, , x2 + x - 210 = 0, , ^xb - ya h2 = 0, xb = ya, x =y, Thus, Hence Proved., a, b, 23. The sum of ages (in years) of a son and his father is, 35 years and product of theirs ages is 150 years, find, their ages., Ans :, [ Delhi Term-2, 2014, Term-2, 2012 Set (40)], Let the age of father be x years and age of son be y, years, x + y = 35, , Now, , xy = 150, , and, , (1), (2), , Putting the value of y , from (1) we have, x ^35 - x h = 150, 2, , x - 35x + 150 = 0, , ^x - 30h^x - 5h = 0, x = 30, x = 5 (Rejected), Age of father cant be 5 years, so we reject x = 5 and, take x = 30 ., Now, y =5, Hence the age of father is 30 years and the age of son, is 5 years., 24. One fourth of a herd of camels was seen in forest., , Twice of square root of the herd had gone to mountain, and remaining 15 camels were seen on the bank of a, river, find the total number of camels., Ans :, [Board Term-2, 2012 Set (1)], Let the total number of camels be x ., According to the question,, x + 2 x + 15 = x, 4, , x2 + 15x - 14x - 210 = 0, x ^x + 15h - 14 ^x + 15h = 0, , ^x + 15h^x - 14h = 0, x + 15 = 0 or x - 14 = 0, x =- 15 or x = 14, , Rejecting negative value x = 14 ., Therefore first number is 14 and consecutive number, is 15., 26. In a class test, the sum of the marks obtained by a, , student in mathematics and science is 28. Had he got, 3 marks more in mathematics and 4 marks less in, science, the product of the marks would have been, 180. Find his marks in two subjects., Ans :, [Board Term-2 2012, Set (21)], Let marks obtained in maths be x , the marks obtained, in science will be, Now, , 28 - x, , ^x + 3h^28 - x - 4h = 180, ^x + 3h^24 - x h = 180, 24x - x2 + 72 - 3x = 180, x2 - 21x + 108 = 0, , ^x - 9h^x - 12h = 0, x = 9 or x = 12, Case I : x = 9, Marks obtained in maths = 9, Marks obtained in science = 28 - 9 = 19, Case II : x = 12, Marks obtained in maths = 12, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 62

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Chap 4 : Quadratic Equation, , www.cbse.online, , 7x + 1 = 29, 4x - 1, x2 - 1, , 2, , x + 10x - 600 = 0, , ^x + 30h^x - 20h = 0, x = 20, , 2, ^7x + 1h^4x - 1h = 29x - 29, 28x2 - 7x + 4x - 1 = 29x2 - 29, , Speed while going is 20 km/h and speed while, , - 3x = x2 - 28, x2 + 3x - 28 = 0, x2 + 7x - 4x - 28 = 0, , returning will be = 20 + 10 = 30 km/h, 7., , x ^x + 7h - 4 ^x + 7h = 0, , ^x + 7h^x - 4h = 0, Hence, x = 4, - 7, 5., , Let numerator be x then fraction will be, As per the question we have, x + 2x + 1 = 2 16 = 58, x, 2x + 1, 21 21, , Two pipes running together can fill a tank in 11 19, minutes. If one pipe takes 5 minutes more than the, other to fill the tank, find the time in which each pipe, would fill the tank separately., Ans :, [O. D. Set III, 2016], , x, 2x + 1, , 21 8x2 + ^2x + 1h2B = 58 ^2x2 + x h, or,, , Let time taken by pipe A be x minutes and time, taken by pipe B be x + 5 minutes., In one minute pipe A will fill 1 tank., x, 1 tank., x+5, tank in one, Thus pipes A + B will fill 1 + 1, x x+5, minute., As per question, two pipes running together can fill, 9, a tank in 11 19 = 100, minutes, in one minute 100, tank, 9, will be filled., Now according to the question we have, 1+ 1 = 9, x x+5, 100, , The denominator of a fraction is one more than twice, its numerator. If the sum of the fraction and its, reciprocal is 2 16, 21 , find the fraction., Ans :, [Foreign Set III, 2016], , 11x2 - 26x - 21 = 0, 11x2 + 33x + 7x - 20 = 0, , x = 3, - 7 (rejected), 11, We reject x =- 73 , thus x = 3 and fraction will be, 3 =3, 6+1 7, , In one minute pipe B will fill, , 8., , The perimeter of a right triangle is 60 cm. Its, hypotenuse is 25 cm. Find the area of the triangle., Ans :, [Delhi Set II, 2016], As per question statement figure is given below., , x+5+x = 9, x ^x + 5h, 100, 100 ^2x + 5h = 9x ^x + 5h, 200x + 500 = 9x2 + 45x, 9x2 - 155x - 500 = 0, 9x2 - 180x + 25x - 500 = 0, 9x ^x - 20h + 25 ^x - 20h = 0, , Here, , ^x - 20h^9x + 25h = 0, , a + b = 60 - c = 60 - 25 = 35, , x = 20, - 25, 9, As time can’t be negative we take x = 20 minutes, and, , Using Pythagoras theorem, a2 + b2 = 252 = 625, Substituting the values in ^a + b h2 = a2 + b2 + 2ab ,, 352 = 625 + 2ab, , x + 5 = 25 minutes, , Hence pipe A will fill the tank in 20 minutes and pipe, , 1225 - 625 = 2ab, , B will fill it in 25 minutes., 6., , The time taken by a person to cover 150 km was 2 12, hours more than the time taken in the return journey., If he returned at a speed of 10 km/hour more than, the speed while going, find the speed per hour in each, direction., Ans :, [Delhi Set III, 2016], Let the speed while going be x km/h, Speed while returning = ^x + 10h km/h, According to question we have, 150 - 150 = 5, x, 2, x + 10, , a + b + c = 60, c = 25, , or,, , ab = 300, , Hence, Area of TABC, 1 ab = 150 cm2 ., 2, 9., , Two water taps together can fill a tank in 9 hours 36, minutes. The tap of larger diameter takes 8 hours less, than the smaller one to fill the tank. Find the time in, which each tap can separately fill the tank., Ans :, [Foreign Set III, 2016], Let the tap with smaller diameter fills the tank in, x hours, then the other tap fills the tank in ^x - 8h, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 64

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Chap 4 : Quadratic Equation, hours, , In one hour small tap will fill 1 tank., x, , www.rava.org.in, 2 # 2x ^x - 2h + 3x ^x + 1h, = 23, 2 ^x + 1h^x - 2h, 5, 4x2 - 8x + 3x2 + 3x = 23, 5, 2 ^x2 - x - 2h, , 1 tank., x-8, Thus both tap will fill 1 + 1 tank in one hour., x x-8, 9 hours 36 minutes = 9 + 36 = 9 + 3 = 48, 60, 3, 5, Since two water taps together can fill a tank in 483, 1 = 5 ., hour, tank fill by both pipe in one hour is 48, 48, 5, 1, 1, 5, Thus, =, +, x x-8, 48, In one hour large tap will fill, , 7x2 - 5x, = 23, 5, 2 ^x2 - x - 2h, 35x2 - 25x = 46x2 - 46x - 92, 11x2 - 21x - 92 = 0, 2, x = - b ! b - 4ac, 2a, , =, , x-8+x = 5, x ^x - 8h, 48, , = 21 ! 441 + 4048, 22, , 5x ^x - 8h = ^2x - 8h 48, 2, 5x - 136 + 384 = 0, , = 21 ! 67, 22, , 136 ! ^136h2 - 4 # 5 # 384, x =, 2.5, = 136 ! 18496 - 7680, 10, x = 136 ! 104 = 24, 16, 10, 5, There is no possibility of x = 165 because it is less than, 8 Hours., Thus smaller tap can fill the tank in 24 hours and, larger tap can fill in 24 hrs., For more files visit www.cbse.online, , 10. The denominator of a fraction is two more than its, , numerator. If the sum of the fraction and its reciprocal, is 34 , find the fraction., 15, Ans :, [Board Term-2, 2012 Set (1)], Let numerator be x , then denominator will be x + 2 ., and, fraction = x, x+2, Now, , x = 21 + 67 or x = 21 - 67, 22, 22, Thus, , denominator. If 2 is added to both the numerator and, the denominator, then the sum of the new fraction, and original fraction is 29 . Find the original fraction., 20, Ans :, [Board Term-2, Delhi, 2015 Set I, III], Let the denominator be x , then numerator will x - 3, So the fraction will be x - 3, x, By the given condition, new fraction will, x-3+2 = x-1, x+2, x+2, Now, , 20 ^x2 - x - 6 + x2 - x h = 29x2 + 58x, 20 ^2x2 - 2x - 6h = 29x2 + 58x, , 40x2 - 40x - 240 = 29x2 + 58x, , 30x2 + 60x + 60 = 34x2 + 68x, , 11x2 - 98x - 120 = 0, , 4x2 + 8x - 60 = 0, , x ^x + 5h - 3 ^x + 5h = 0, , ^x + 5h^x - 3h = 0, , We reject the x =- 5 . Thus x = 3 and fracrtion = 3, 5, 2 +, 3, 11. Solve for x :, = 33 ; x ! 0, - 1, 2, x + 1 2 ^x - 2h 5x, Ans :, We have, , [Board Term-2, Delhi 2015, Set I, II], 2 +, 3, = 23, x + 1 2 ^x - 2h, 5x, , x - 3 + x - 1 = 29, x, 20, x+2, 20 6^x - 3h^x + 2h + x ^x - 1h@ = 29 ^x2 + 2x h, , 15 ^x2 + x2 + 4x + 4h = 34 ^x2 + 2x h, , x2 + 5x - 3x - 15 = 0, , x = 4, - 23, 11, , 12. The numerator of a fraction is 3 less than its, , x + x + 2 = 34, x, x+2, 15, , x2 + 2x - 15 = 0, , 21 ! ^- 21h2 - 4 ^11h^- 92h, 2 # 11, , 11x2 - 110x + 12x - 120 = 0, , ^11x + 20h^x - 10h = 0 ,, We take x = 10 and fraction will be is 10 - 3 = 7 ., 10, 10, 13. The diagonal of a rectangular field is 16 metre more, than the shorter side. If the longer side is 14 metre, more than shorter side, then find the length of the of, the field., Ans :, [Board Term, O.D., 2015 Set I, II, III], Let the length of shorter side be x m., Length of diagonal = ^x + 16h m, , and,, , Length of longer side = ^x + 14h m, , 2x + 3x, = 23, x + 1 2 ^x - 2h, 5, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 65

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Chap 4 : Quadratic Equation, , www.cbse.online, 2, , x - 60x + 10x - 600 = 0, x ^x - 60h + 10 ^x - 60h = 0, , ^x - 60h^x + 10h = 0, Negative value is rejected, thus first speed of truck is, 60 km/h., 16. The total cost of a certain length of cloth is Rs 200. If, , Now as per question we have, x2 + ^x + 14h2 = ^x + 16h2, x2 + x2 + 28x + 196 = x2 + 32x + 256, x2 - 4x - 60 = 0, x2 + 6x - 10x - 60 = 0, x ^x + 6h - 10 ^x + 6h = 0, x =- 6, x = 10, As lenght can’t be negative x = 10 m. Therefore, length of sides are 10 m and 24 m., 14. A train travels at a certain average speed for a distance, , of 54 km and then travels a distance of 63 km at an, average speed of 6 km/h more than the first speed. If, it takes 3 hours to complete the total journey, what is, its first speed ?, Ans :, [Board Term-2, O.D., 2015 Set I, II], Let the speed of the train be x km/hr. for first 54 km., and for next 63 km, speed = ^x + 6h km/hr., According to the question, 54 + 63 = 3, x, x+6, 54 ^x + 6h + 63x, =3, x ^x + 6h, 54x + 324 + 63x = 3x ^x + 6h, 117x + 324 = 3x2 + 18x, 3x2 - 99x - 324 = 0, x2 - 33x - 108 = 0, x2 - 36x + 3x - 108 = 0, x ^x - 36h + 3 ^x - 36h = 0, , ^x - 36h^x + 3h = 0, x =- 3, 36, , Negative value is rejected, thus first speed of train is, 36 km/h., 15. A truck covers a distance of 150 km at a certain, , average speed and then covers another 200 km at, average speed which is 20 km per hour more than the, first speed. If the truck covers the total distance in 5, hours, find the first speed of the truck., Ans :, [Board Term-2, O.D., 2015, Set II], Let the average speed of the truck be x km/hr. for, first 150 km and for next 200 km, speed ^x + 20h km/, hr., 150 + 200 = 5, Now, x, x + 20, 150x + 3000 + 200x = 5x ^x + 20h, , the piece was 5 m longer and each metre of cloth coast, Rs 2 less, the cost of the piece would have remained, unchaged. How longer is the piece and what is its, original rate per metre ?, Ans :, [Foreign Set I, II, 2015], Let the length of the cloth be x m., cost per metre = 200, x, New length of the cloth = ^x + 5h m, New cost per metre = b 200 - 2 l, x, Since cost of the piece have remained unchanged,, 200, ^x + 5hb x - 2l = 200, , ^x + 5h^200 - 2x h, = 200, x, 200x - 2x2 + 1000 - 10x = 200x, x2 + 5x - 500 = 0, , ^x + 25h^x - 20h = 0, x = 20, 25, Negative value is rejected, thus length of the piece is, 20 m., Original cost per metre is 200 = 10 Rs., 20, 17. A motorboat whose speed in still water is 18 km/h,, takes 1 hour more to go 24 km upstream than to, return downstream to the same spot. Find the speed, of the stream., Ans :, [CBSE O.D. 2014], Let the speed of stream be x km/h, Then the speed of boat upstream = ^18 - x h km/h, Speed of boat downstream = ^18 + x h km/h, According to the question,, 24 - 24 = 1, 18 - x 18 + x, 24 [(18 + x) - (18 - x)], =1, 182 - x2, 48x = 324 - x2, x2 + 48x - 324 = 0, x2 + 54x - 6x - 324 = 0, x ^x + 54h - 6 ^x + 54h = 0, , ^x + 54h^x - 6h = 0, x + 54 = 0, x - 6 = 0, x =- 54, x = 6, , Since speed cannot be negative, we reject x =- 54 ., The speed of steam is 6 km/h., , 2, , x - 50x - 600 = 0, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 66

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Chap 4 : Quadratic Equation, 18. Solve for x :, , Ans :, We have, , www.rava.org.in, , x - 3 + x - 5 = 10 ; x ! 4, 6, 3, x-4 x-6, , According to question,, , ^32 - x - 2h^x + 4h = 253, ^30 - x h^x + 4h = 253, , x - 3 + x - 5 = 10, 3, x-4 x-6, , 26x - x2 + 120 = 253, , ^x - 3h^x - 6h + ^x - 4h^x - 5h 10, =, 3, ^x - 4h^x - 6h, x2 - 9x + 18 + x2 - 9x + 20 = 10, 3, x2 - 10x + 24, 3 ^2x2 - 18x + 38h = 10x2 - 100x + 240, 6x2 - 54x + 114 = 10x2 - 100x + 240, , x2 - 26x + 133 = 0 ]or,, x2 - 19x - 7x + 133 = 0, x ^x - 19h - 7 ^x - 19h = 0, or,, x = 7 or x = 19, If x = 7 then marks in mathematics = 7 , and marks, in science = 25, If x = 19 , then marks in mathematics = 19 and marks, in science = 13 ., , 4x2 - 46x + 126 = 0, 2x2 - 23x + 63 = 0, 2x2 - 14x - 9x + 63 = 0, 2x ^x - 7h - 9 ^x - 7h = 0, , ^2x - 9h^x - 7h = 0, 2x - 9 = 0, x - 7 = 0, , For more files visit www.cbse.online, , 21. The sum of squares of two consecutive multiples of 7, , is 637. Find the multiples., Ans :, , Let 7x and 7x + 7 be two consecutive multiples of 7., According to question,, , x = 92 , x = 7, , ^7x h2 + ^7x + 7h2 = 637, 49x2 + 49x2 + 49 + 98x = 637, , 19. A motor boat whose speed is 24 km/h in still water, , takes 1 hour more to go 32 km upstream than to, return downstream to the same spot. Find the speed, of the stream., Ans :, [Board Term-2, O.D., Set II 2016], Ans :, Let the speed of stream be x km/h, Then the speed of boat upstream = ^24 - x h km/h, Speed of boat downstream = ^24 + x h km/h, According to the question,, 32 - 32 = 1, 24 - x 24 + x, 32 : 1 - 1 D = 1, 24 - x 24 + x, 32 :24 + x - 242+ x D = 1, 576 - x, 32 ^24 + x - 24 + x h = 576 - x2, , 98x2 + 98x - 588 = 0, x2 + x - 6 = 0, , ^x + 3h^x - 2h = 0, x = - 3, 2, Neglecting negative value, x = 2, Therefore multiples are 14 and 21., 22. Solve for x :, , Ans :, , Let x - 1 be y so 2x + 1 = 1, y, 2x + 1, x-1, Substituting this value we obtain, y+ 1 = 2, y, y2 + 1 = 2y, , x2 + 64x - 576 = 0, x ^x + 72h - 8 ^x + 72h = 0, , ^x - 8h^x + 72h = 0, x = 8, - 72, , y2 - 2y + 1 = 0, , ^y - 1h2 = 0, y =1, x, 1, Putting y =, we have, 2x + 1, , Since speed cannot be negative, we reject x =- 72 ., , x - 1 = 1 or x - 1 = 2x + 1, 2x + 1, , The speed of steam is 8 km/h., 20. A student scored a total of 32 marks in class tests in, , mathematics ans science. Had he scored 2 marks less, in science and 4 more in mathematics, the product of, his marks would have been 253. Find his marks in two, subjects., Ans :, [Board Term-2, 2012 Set (50)], Let marks in Mathematics be x , then marks in science, will be 32 - x, , x - 1 + 2x + 1 = 2 where x ]- 1 , 1, 2, 2x + 1, x-1, [Out Side Delhi Set-III 2017], x - 1 + 2x + 1 = 0, 2x + 1, x-1, , We have, , 64x = 576 - x2, x2 + 72x - 8x - 576 = 0, , [Foreign Set II, 2014], , or, 23. Find for x :, , Ans :, We have, , x =- 2, 1 + 2 = 6 ; x ! 0, 1 , 2, x, x-2 x-1, [Board Outside Delhi Compt. Set-I, II 2017], 1 + 2 =6, x, x-2 x-1, x - 1 + 2x - 4 = 6, x, ^x - 2h^x - 1h, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 67

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Chap 4 : Quadratic Equation, , www.cbse.online, Ans :, , 3x2 - 5x = 6x2 - 18x + 12, 2, , 3x - 13x + 12 = 0, , Let the number be x and x + 2, , ^x h2 + ^x + 2h2 = 340, x + x2 + 4 + 4x = 340, , Now, , 2, , 3x - 4x - 9x + 12 = 0, , 2, , x ^3x - 4h - 3 ^3x - 4h = 0, , ^3x - 4h^x - 3h = 0, x = 4 and 3, 3, 4, Hence, x = 3,, 3, 24. Solve, for x :, , Ans :, We have, , 3 x2 + 10x + 7 3 = 0, [Board Foreign II, III 2017], 3 x2 + 10x + 7 3 = 0, 3 x2 + 3x + 7x = 7 3 = 0, , ^x + 3 h^ 3 x + 7h = 0, ^x + 3 h^ 3 x + 7h = 0, , 2x2 + 4x - 336 = 0, x2 + 2x - 168 = 0, , ^x + 14h^x - 12h = 0, x = 12, Thus numbers are 12 and 14., 27. The sum of the squares of two consecutive odd, , numbers is 394. Find the numbers., Ans :, [Foreign Set I, 2014] [Board Term-2, 2012 Set(12)], Let the odd number be 2x + 1, then consecutive odd, number will be 2x + 1 + 2 = 2x + 3, Now, according to question, , ^2x + 1h2 + ^2x + 3h2 = 394, 4x + 4x + 1 + 4x2 + 12x + 9 = 394, , x =- 3 and x = - 7, 3, Hence roots x =- 3 or x = - 7, 3, , 2, , 8x2 + 16x - 384 = 0, x2 + 2x - 48 = 0, , 25. The difference of two numbers is 5 and the difference, , x2 + 8x - 6x - 48 = 0, , of their reciprocals is 1 . Find the numbers, 10, Ans :, [Board Term-2, 2014, Delhi], , Let the first number be x , then second number will, be x + 5, Now according to the question, 1- 1 = 1, x x+5, 10, , x ^x + 8h - 6 ^x + 8h = 0, x =- 8, 6, Rejecting the negative value,, Ist number = 2 # 6 + 1 = 13, and second odd number = 15, 28. Sum of the areas of two squares is 400 cm2. If the, , x+5-x = 1, x ^x + 5h, 10, 50 = x2 + 5x, x2 + 5x - 50 = 0, , difference of their perimeters is 16 cm, find the sides, of the two squares., Ans :, [Board Term-2, 2013], Let the sides of two squares be a and b ,, , x2 + 10x - 5x - 50 = 0, , then, , x ^x + 10h - 5 ^x + 10h = 0, , ^x + 10h^x - 5h = 0, x = 5, - 10, , Rejecting the negative value, numbers are 5 and 10., , and, , is 340. Find the numbers., Ans :, , [Foreign Set I, 2014], , 4 ^a - b h = 16, a = 4+b, , From equations (1) and (2), we obtain, 2, ^4 + b h2 + b = 400, 16 + b2 + 8b = b2 = 400, , For session 2019-2020 free pdf will be available at, www.cbse.online for, 1. Previous 15 Years Exams Chapter-wise Question, Bank, 2. Previous Ten Years Exam Paper (Paper-wise)., 3. 20 Model Paper (All Solved)., 4. NCERT Solutions, All material will be solved and free pdf. It will be, provided by 30 September and will be updated regularly., , 26. The sum of squares of two consecutive even numbers, , (1), , a-b = 4, , NO NEED TO PURCHASE ANY BOOKS, , Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education,, New Delhi in any manner. www.cbse.online is a private organization which provide free, study material pdfs to students. At www.cbse.online CBSE stands for Canny Books, For School Education, , a2 + b2 = 400, , 2b2 + 8b - 384 = 0, b2 = 4b - 192 = 0, b2 + 16b - 12b - 192 = 0, b ^b + 16h - 12 ^b + 16h = 0, , ^b + 16h^b - 12h = 0, b =- 16, 12, , Rejecting the negative value, b = 12 cm, then, , a = 4 + 12 = 16 cm., , 29. A train takes 2 hours less for a journey of 300 km if, , its speed is increased by 5 km/hr from its usual speed., Find the usual speed of the train., Ans :, [Board Term-2, 2012, Set(22)], , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 68

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Chap 4 : Quadratic Equation, , www.rava.org.in, , Let the usual speed of train be x km/hr., According to question we have, 300 - 300 = 2, x, x+5, x2 + 5x - 750 = 0, x2 = 30x - 25x - 750 = 0, , ^x = 30h^x - 25h = 0, x = - 30 or x = 25, , A’ one day work =, , and (A+B)’s one day work = 1, 4, According to the question,, 1+ 1 =1, x x-6, 4, x2 - 14x + 24 = 0, x2 - 12x - 2x + 24 = 0, , Since, speed cannot be negative x !- 30 ., , x ^x - 12h - 2 ^x - 12h = 0, , Thus speed of train is 25 km/hr., 30. The length of the sides forming right angle of a right, , triangle are 5x cm and ^3x - 1h cm. If the area of the, triangle is 60 cm2. Find its hypotenuse., Ans :, [Board Term-2, 2012 Set (44)], , According to the question we have drawn figure below., , 1, x-6, , ^x - 12h^x - 2h = 0, x = 12 or x = 2, , ,, , But x cannot be less than 6. So x = 12, Hence B can finish the work in 12 days., 32. The length of the hypotenuse of a right triangle, , exceeds the length of its base by 2 cm and exceeds, twice the length of altitude by 1 cm. Find the length, of each side of the triangle., Ans :, [Board Term-2, 2012 Set (12)], et altitude of triangle be x ., Hypotenuse of triangle = 2x + 1, and, , Now, , base of triangle = 2x - 1, , Area of triangle = 1 # base # height, 2, 60 = 1 # 5x # ^3x - 1h, 2, 15x2 - 5x = 120, 3x2 - x - 24 = 0, 3x2 - 9x + 8x - 24 = 0, 3x ^x - 3h + 8 ^x - 3h = 0, , ^x - 3h^3x + 8h = 0, , Using Pythagoras theorem,, 2, ^2x + 1h2 = x + ^2x - 1h2, 4x2 + 1 + 4x = x2 + 4x2 + 1 - 4x, , x = 3, x =- 8, 3, Length can’t be negative, so x = 3, , x2 - 8x = 0, , Now AB = 5 # 3 = 15 cm,, BC = 3x - 1 = 9 - 1 = 8 cm, Now, , AC =, , 152 + 82, , =, , 225 + 64, , =, , 289 = 17 cm., , Hence hypotenuse = 17 cm., 31. A takes 6 days less than the time taken by B to finish, , a piece of work. If both A and B together can finish it, in 4 days, find the time taken by B to finish the work., Ans :, [Board Term-2, 2012 Set (5)], Suppose B alone finish the work in x days and A, alone takes ^x - 6h days., B’s one day work = 1, x, , x ^x - 8h = 0, Rejecting x = 0, we get x = 8, Thus altitude of triangle is 8 cm, Hypotenuse of triangle is 2 # 8 + 1 = 17 cm, and base of triangle is 2 # 8 - 1 = 15 cm, 33. The perimeter of a rectangular field is 82 m and its, , area is 400 square metre. Find the length and breadth, of the rectangle., Ans :, [Board Term-2, 2012 Set (21)], We have, or,, , Perimeter = 2 ^l + b h = 82 m, l + b = 41 m, , Let length be x m, then breadth = ^41 - x h m., , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 69

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Chap 4 : Quadratic Equation, , www.cbse.online, , Area = l # b = 400 m, , 2, , =, , x ^41 - x h = 400, 41x - x2 = 400, x - 41x + 400 = 0, , ^x - 16h^x - 25h = 0, x = 16 or x = 25, If length is 16 m, then breadth will be 25 m., If length is 25 m, then breadth will be 16 m., , = - 4b ! 4a, 8, =-, , 34. The product of Tanay’s age (in years) five years ago, , 1., , ^x - 5h^x + 10h = 16, x2 + 5x - 50 = 16, , 2, ^k + 1h x - 2 ^k + 1h x + 1 = 0, Compare with Ax2 + Bx + C = 0 we get, A = ^k + 1h, B = - 2 ^k + 1h, C = 1, If roots are equal, D = 0 , i.e., , x2 + 11x - 6x - 66 =- 66, x ^x + 11h - 16 ^x - 11h = 0, , ^x + 11h^x - 6h = 0, x =- 11, 6 ., As age cannot be negative, we reject x = - 11. Thus, present age of Tanay is 6 years., , We have, , Find k, so that the quadratic equation, 2, ^k + 1h x - 2 ^k + 1h x + 1 = 0 has equal roots., Ans :, [Board Term-2, 2016 Set HODM40L], We have, , x2 + 5x - 66 = 0, , B2 = 4AC, 4 ^k + 1h2 = 4 ^k + 1h, k2 + 2k + 1 = k + 1, k2 + k = 0, , x + 3 - 1 - x = 17 ; x ! 0.2, x, x-2, 4, , k ^k + 1h = 0, k = 0, - 1, , [Delhi Compt. Set-I 2017], x + 3 - 1 - x = 17, x, x-2, 4, , k = - 1 does not satisfy the equation, thus k = 0, 2., , x ^x + 3h - ^1 - x h^x - 2h, = 17, x ^x - 2h, 4, 2, 2, ^x + 3x h - ^- x + 3x - 2h, = 17, 4, x 2 - 2x, , If 2 is a root or the equation x2 + kx + 12 = 0 and, the equation x2 + kx + q = 0 has equal roots, find the, value of q ., Ans :, [Board Sample Paper 2016], , If 2 is the root of above equation, it must satisfy it., ,, , 8x2 + 8 = 17x2 - 34x, , 2, , Substituting k = - 8 in x + kx + q = 0 we have, , 9x2 - 36x + 2x - 8 = 0, , x2 - 8x + q = 0, , 9x ^x - 4h + 2 ^x - 4h = 0, , For equal roots,, , ^x - 4h^9x + 2h = 0, , ^- 8h2 - 4 ^1 h q = 0, 64 - 4q = 0, , x = 4 or x =- 2, 9, , 4q = 64, , 36. Solve for x : 4x2 + 4bx - ^a2 - b2h = 0, , [Board Foreign Set- III 2017], , Ans :, , 4x + 4bx - ^a - b h = 0, Compare with Ax2 + Bx + C = 0 we get, We have, , ^2h2 + 2k + 12 = 0, 2k + 16 = 0, k =- 8, , 9x2 - 34x - 8 = 0, , 2, , x2 + kx + 12 = 0, , We have, , 2x2 + 2 = 17, 4, x2 - 2x, , Hence, x = 4, - 2, 9, , ^a + b h, ^a - b h, and, 2, 2, , SHORT ANSWER TYPE QUESTIONS - I, , Let the present age of Tanay’s be x years., , Ans :, , ^a + b h ^a - b h, ,, 2, 2, , Hence the roots are -, , and his age ten years later is 16. Determine Tanay’s, present age., Ans :, [Board Term-2, 2012 Set (31)], , 35. Solve for x :, , 2, , 2, 2, 2, = - 4b ! 16b - 16b + 16a, 8, , 2, , According to question we have, , - 4b ! ^4b h - 4.4 ^b - a2h, 2.4, 2, , 2, , 2, , A = 4, B = 4b and C = b2 - a2, 2, x = - B ! B - 4AC, 2A, , q = 16, 3., , Find the values of k for which the quadratic equation, 9x2 - 3kx + k = 0 has euqal roots., Ans :, [CBSE Delhi, O.D. 2014], We have, , 9x2 - 3kx + k = 0, , Compare with ax2 + bx + c = 0 we get, a = 9, b = - 3k, c = k, Since roots of the equation are equal, b2 - 4ac = 0, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 70

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Chap 4 : Quadratic Equation, , www.rava.org.in, , ^- 3k h - ^4 # 9 # k h = 0, 9k2 - 36k = 0, , Since 2 is a root of above equation, it must satisfy it., Substituting x = 2 in 3x2 + px - 8 = 0 we have, , 2, , 12 + 2p - 8 = 0, , k2 - 4k = 0, , p =- 2, , k ^k - 4h = 0, Hence,, 4., , 2, , k = 0 or k = 4, , Since 4x - 2px + k = 0 has equal roots,, , k = 4., , or, , D = b2 - 4ac = 0, , If the equation kx2 - 2kx + 6 = 0 has equal roots, then, find the value of k ., Ans :, [Board Term-2, 2012 Set (22)], We have, , 42 - 4 ^4 h^k h = 0, 16 - 16k = 0, 16k = 16, , kx2 - 2kx + 6 = 0, , Thus, , Compare with ax2 + bx + c = 0 we get, a = k, b = - 2k, c = 6, Since roots of the equation are equal, b2 - 4ac = 0, , 2., , ^- 2k h2 - 4 ^k h^6 h = 0, 4k2 - 24k = 0, 4k ^k - 6h = 0, k = 0, 6, , We have, , kx ^x - 2 5 h + 10 = 0, , or,, , kx2 - 2 5 kx + 10 = 0, , Compare with ax2 + bx + c = 0 we get, a = k, b = - 2 5 , c = 10, , k =6, , Since, roots are equal, D = b2 - 4ac = 0, , ^- 2 5 k h - 4 # k # 10 = 0, 20k2 - 40k = 0, 2, , Find the values of p for which the quadratic equation, 4x2 + px + 3 = 0 has equal roots., Ans :, [Board Term-2, 2014], We have, , 20k ^k - 2h = 0, k ^k - 2h = 0, , 4x2 + px + 3 = 0, Since k ! 0 , we get k = 2, , Compare with ax2 + bx + c = 0 we get, a = 4, b = p, c = 3, Since roots of the equation are equal, b2 - 4ac = 0, , 3., , p2 - 4 # 4 # 3 = 0, p2 - 48 = 0, , 6., , We have, , p =! 4 3, , Compare with ax2 + bx + c = 0 we get, , 13 3 x2 + 10x +, , b2 - 4ac = ^- 4 3 h - 4 ^3 h^4 h, 2, , = 48 - 48 = 0, Thus roots are real and equal., , 3 =0, , a = 13 3 , b = 10, c =, , Roots are b- b l, b- b l or 2 3 , 2 3, 2a, 2a, 3, 3, 3, , b2 - 4ac = ^10h2 - 4 ^13 3 h^ 3 h, = 100 - 156, =- 56, As D 1 0 , the equation has not real roots., , SHORT ANSWER TYPE QUESTIONS - II, If 2 is a root of the quadratic equation 3x2 + px - 8 = 0, and the quadratic equation 4x2 - 2px + k = 0 has, equal roots, find k ., Ans :, [Foreign Set II, 2014], We have, , 3x2 - 4 3 x + 4 = 0, a = 3k, b = - 4 3 , c = 4, , Compare with ax2 + bx + c = 0 we get, , 1., , Find the nature of the roots of the following, quadratic equation. If the real roots exist, find them :, 3x2 - 4 3 x + 4 = 0, Ans :, [Board Term-2, 2012 Set (5)], , p2 = 48, , Find the nature of the roots of the quadratic equation, :, 13 3 x2 + 10x + 3 = 0, Ans :, [Board Term-2, 2012, (12)], We have, , k =1, , For what value of k , the roots of the quadratic, equation kx ^x - 2 5 h + 10 = 0 are equal ?, Ans :, [Delhi CBSE, Term-2, 2014] [Delhi 2013], , But k ! 0 , as coefficient of x2 can not be zero, , 5., , 4x2 + 4x + k = 0 has equal roots,, , 4., , Determine the positive value of 'k' for which the, equation x2 + kx + 64 = 0 and x2 - 8x + k = 0 will, both have real and equal roots., Ans : [Board Term-2, 2012 Set (44)] [Delhi CBSE Term-2, 2014], We have, , x2 + kx + 64 = 0, , Compare with ax2 + bx + c = 0 we get, a = 1, b = k, c = 64, For real and equal roots, b2 - 4ac = 0, Thus, , k2 - 4 # 1 # 64 = 0, k2 - 256 = 0, k = ! 16, , (1), , 3x2 + px - 8 = 0, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 71

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Chap 4 : Quadratic Equation, 9., , www.rava.org.in, , If the equation ^1 + m h x + 2mcx + ^c - a h = 0 has, equal roots, prove that c2 = a2 ^1 + m2h, Ans :, [Delhi CBSE Board, 2015], 2, , 2, , 2, , 2, , ^1 + m h x + 2mcx + ^c - a h = 0, Compare with Ax2 + Bx + C = 0 we get, 2, , We have, , 2, , 2, , 2, , A = 1 + m , B = 2mc, C = ^c - a h, If roots are equal, B2 - 4AC = 0, 2, , 2, , 2, , 2 2, 2, ^2mc h2 - 4 `1 + m ^c - a hj = 0, 4m2 c2 - 4 ^1 + m2h^c2 - a2h = 0, , m2 c2 - ^c2 - a2 + m2 c2 - m2 a2h = 0, , m2 c2 - c2 + a2 - m2 c2 + m2 a2 = 0, - c2 + a2 + m2 a2 = 0, c2 = a2 ^1 + m2h, , ,, Hence Proved., , ^- 5h is a root of the quadratic equation, 2x + px + 15 = 0 and the quadratic equation, p ^x2 + x h + k = 0 has equal roots, then find the values, of p and k ., Ans :, [Delhi CBSE Board, 2015 (Set II)], , 10. If, , 2, , We have, 2x2 + px - 15 = 0, Since x =- 5 is the root of above equation. It must, satisfy it., 2 ^- 5h2 + p ^- 5h - 15 = 0, 50 - 5p - 15 = 0, 5p = 35 & p = 7, p ^x + x h + k = 0 has equal roots, 7x2 + 7x + k = 0, , 2, ^a + b - 2ab h + ^b + c - 2bc h + ^c + a - 2ac hB = 0, 1, 2, 2, 2, 2 8^a - b h + ^b - c h + ^c - a h B = 0, , 1, 28, , 2, , 2, , or, , 2, , 2, , ^a - b h2 + ^b - c h2 + ^c - a h2 = 0, If a ! b ! c, or,, , ^a - b h2 2 0, ^b - c h2 2 0, ^c - a h2 2 0, If, ^a - b h2 = 0 & a = b, ^a - c h2 = 0 & b = c, ^c - a h2 = 0 & c = a, , Thus a = b = c, Hence Proved, 12. If, , the, roots, of, the, quadratic, equation, 2, 2, 2, 2, in, are, equal, c, ab, x, 2, a, bc, x, b, ac, 0, x, =, +, ^, h, ^, h, then show that either a = 0 or a3 + b3 + c3 = 3abc, Ans :, [Board Outside Delhi Set II, III 2017], Ans :, , 2, 2, 2, 2, ^c - ab h x - 2 ^a - bc h x + b - ac = 0, Compare with Ax2 + Bx + C = 0 we get, , We have, , A = (c2 - ab), B = (a2 - bc), C = (b2 - ac), If roots are equal, B2 - 4AC = 0, 2, 2, 2, 82 ^a - bc hB - 4 ^c - ab h^b - ac h = 0, 2, , 4 6a 4 + b2 c2 - 2a2 bc@ - 4 ^b2 c2 - c3 a - ab3 - a2 bc h, , =0, , 4 6a + b c - 2a bc - b c + c a + ab - a bc@ = 0, 4, , 2 2, , 2, , 2 2, , 3, , 3, , 2, , 4 6a 4 + ac3 + ab3 - 3a2 bc@ = 0, a ^a3 + c3 + b3 - 3abc h = 0, , 2, , Now, , 2, , a = 0 or a3 + b3 + c3 = 3abc, , 2, , Taking b - 4ac = 0 we have, 72 - 4 # 7 # k = 0, 7 - 4k = 0, k =7, 4, Hence p = 7 and k = 7 ., 4, 11., , If, the, roots, of, the, quadratic, equation, x, a, x, b, x, b, x, c, x, c, x, a, +, +, ^, h^, h ^, h^, h ^, h^, h = 0 are, equal. Then show that a = b = c., Ans :, [Delhi CBSE Board, 2015 (Set II)], We have, , ^x - a h^x - b h + ^x - b h^x - c h + ^x - c h^x - a h = 0, x2 - ax - bx + ab +, 2, , + x - bx - cx + bc +, + x2 - cx - ax + ac = 0, 3x2 - 2ac - 2bx - 2cx + ab + bc + ca = 0, 2, , For equal roots B - 4AC = 0, 2, $- 2 ^a + b + c h. - 4 # 3 ^ab + bc + ca h = 0, , 4 ^a + b + c h - 12 ^ab + bc + ca h = 0, 2, , a2 + b2 + c2 - 3 ^ab + bc + ca h = 0, , a2 + b2 + c2 + 2ab + 2bc + 2ac - 3ab - 3bc - 3ac = 0, 2, , 2, , 2, , a + b + c - ab - ac - bc = 0, 1, 2, , 2, 2, 62a + 2b + 2c - 2ab - 2ac - 2bc@ = 0, 2, , NO NEED TO PURCHASE ANY BOOKS, , For session 2019-2020 free pdf will be available at, www.cbse.online for, 1. Previous 15 Years Exams Chapter-wise Question, Bank, 2. Previous Ten Years Exam Paper (Paper-wise)., 3. 20 Model Paper (All Solved)., 4. NCERT Solutions, All material will be solved and free pdf. It will be, provided by 30 September and will be updated regularly., , Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education,, New Delhi in any manner. www.cbse.online is a private organization which provide free, study material pdfs to students. At www.cbse.online CBSE stands for Canny Books, For School Education, , 1, for x, :, where, = 1+1+1, a+b+x a b x, a + b + x ! 0 and a, b, x ! 0, Ans :, [Board Foreign Set II, III 2017], , 13. Solve, , We have, , 1, -1 = 1+1, a b, a+b+c x, - ^a + b h, =b+a, ab, x2 + ^a + b h x, x2 + ^a + b h x + ab = 0, , ^x + a h^x + b h = 0, x = - a, x = - b, , Hence x = - a, - b, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 75

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Chap 4 : Quadratic Equation, , www.cbse.online, , 14. The difference between the radii of the smaller circle, , So the equation has real and two distinct roots., , and the larger circle is 7 cm and the difference between, their areas of the two circles in 1078 sq. cm. Find the, radius of the smaller circle., Ans :, [Board Comp. I, II, III 2017], We have, , r2 - r1 = 7 cm, r2 2 r1, , 5x2 - 6x = 2, Dividing both the sides by 5 we get, x2 - 6 x = 2, 5, 5, , (1), , x2 - 2x b 3 l = 2, 5, 5, Adding square of the half of coefficient of x, x2 - 2x b 3 l + 9 = 2 + 9, 5, 25, 5 25, , p ^r - r h = 1078 cm2, p ^r2 - r1h^r2 + r1h = 1078, 22 7 r + r = 1078, 7 # ^ 2 1h, 2, 2, , and, , 2, 1, , r2 + r1 = 1078 = 49, 22, Adding (1) and (2) we get, , 3 2, 19, bx - 5 l = 25, , (2), , x - 3 = ! 19, 5, 5, , 2r2 = 56, r2 = 28 cm, r1 = 49 - 28 = 21, , and, , Verification :, , Hence radii of two circles are 28 cm and 21 cm., , 2, , 5 ;3 + 19 E - 6 ; 3 + 19 E - 2, 5, 5, , 15. A train travelling at a uniform speed for 360 km have, , taken 48 minutes less to travel the same distance if its, speed were 5 km/hour more. Find the original speed, of the train, Ans :, [Sample Paper 2017], , = 9 + 6 19 + 19 - c 18 + 6 19 m - 2, 5, 5, = 28 + 6 19 - 18 + 6 19 - 2, 5, 5, , Ans :, , = 28 + 6 19 - 18 - 6 19 - 10, 5, , Let the original speed of the train be x km/hr., Time taken = Dis tan ce = 360 hours, x, Speed, Time taken at increase speed = 360 hours., x+5, According to the question, 360 - 360 = 48, x, 60, x+5, , =0, Similarly, 2, , 5 ; 3 - 19 E - 6 ; 3 - 19 E - 2 = 0, 5, 5, Hence verified., , 360 : 1 - 1 D = 4, x x+5, 5, 1800 = 4, 5, x + 5x, , x = 3 + 19 or 3 - 19, 5, 5, , HOTS QUESTIONS, , 2, , x2 + 5x - 2250 = 0, , 1., , x2 + ^50 - 45h x - 2250 = 0, , 1, Solve, = 1 + 1 + 1 , a + b ! 0., ^a + b + x h a b x, Ans :, [Sample Paper, 2016], 1, = 1+1+1, a b x, a+b+x, , We have, , x2 - 50x - 45x - 2250 = 0, , ^x + 50h^x - 45h = 0, x = - 50 or x = 45, , 1, -1 = 1+1, a b, a+b+x x, x - ^a + b + x h, =a+b, x ^a + b + x h, ab, , As speed can not be negative, original speed of train, is 45 km/hr., , x-a-b-x =a+b, x ^a + b + x h, ab, , 16. Check whether the equation 5x2 - 6x - 2 = 0 has real, , roots if it has, find them by the method of completing, the square. Also verify that roots obtained satisfy the, given equation., Ans :, [Sample Question Paper 2017], We have, , - ^a + b h, =a+b, x ^a + b + x h, ab, x ^a + b + x h = - ab, , 5x2 - 6x - 2 = 0, , x + ^a + b h x + ab = 0, 2, , Compare with ax2 + bx + c = 0 we get, , ^x + a h^x + b h = 0, x = - a or x =- b, , a = 5, b = ^- 6h and c = ^- 2h, , b2 - 4ac = ^- 6h2 - 4 # 5 # - 2, = 36 + 40 = 36 2 0, , 2., , A pole has to erected at a point on the boundary of, a circular park of diameter 17 m in such a way that, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 76

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Chap 4 : Quadratic Equation, , www.rava.org.in, , the differences of its distances from two diametrically, opposite fixed gates A and B on the boundary is 7, meters. Find the distances from the two gates where, the pole is to be erected., [Foreign Set I, II, 2016], Ans :, [CBSE Marking Scheme, 2011, 2012], , x2 + 10x - 600 = 0, ,, , Since number of books cannot be negative, x = 20, Thus number of books bought is 20., , As per question the figure is shown below., 5., , Let p be the location of the pole such that its distance, from gate B , x metres., , and, , x2 + ^x + 7h2 = ^17h2, x2 + x2 + 14x - 240 = 0, , 192x + 192 # 16 - 192x = 2 ^x2 + 16x h, , = - 2 ! 23 = 8, - 15, 2, , x2 + 16x - 1536 = 0, x2 + 48x - 32x - 1536 = 0, , Thus x = 8 m and x + 7 = 15 m, Hence distance between two gates are 8 m and 15 m., , x ^x + 48h - 32 ^x + 48h = 0, , ^x - 32h^x + 48h = 0, x = 32 or - 48, Since speed can’t be negative, therefore - 48 is not, possible. Speed of passenger train is 32 km/h, , Find the value of k for which the distance between, ^9, 2h and ^3, k h is 10 units., Ans :, [Board Term-2, 2012 set (43); Set 2011 set (B1)], , ^x1 - x2h2 + ^y1 - y2h2, ^3 - 9h2 + ^k - 2h2 = 10, d =, , Thus, , 6., , 2, ^- 6h2 + k - 4k + 4 = 100, k2 - 4k + 40 = 100, , No. of books bought for Rs. 300 = 300, x, Reduced list price of the book = (x - 5), No. of books brough for Rs. 300 = 300, x-5, According to questions, we have, 300 - 300 = 5, x, x-5, , k2 - 10k + 6k - 60 = 0, k (k - 10) + 6 (k - 10) = 0, , ^k - 10h^k + 6h = 0, k = 10, - 6, A shopkeeper buys a number of books for Rs. 1200. If, he had bought 10 more books for the same amount,, each book would have cost him Rs. 20 less. How many, books did he buy?, Ans :, [Board Term-2, 2012 Set (22)], , x2 - 5x - 300 = 0, , ^x - 20h^x + 15h = 0, Since price cannot be negative, x = 20, Thus original list price is 20 rs., , Let the number of books bought be x., As per question we have, 1200 - 1200 = 20, x, x + 10, , If the price of a book is reduced by Rs. 5, a person, can buy 5 more book for Rs.300. Find the original list, price of the book., Ans :, [Board Term-2, 2012 sEt (17)], Let the original list price be Rs. x, , k2 - 4k - 60 = 0, , 4., , 192, ^x + 16h, , 192 ^x + 16h - 192x = 2 ^x2 + 16x h, , x = - 7 ! 49 + 480, 2, , [CBSE Marking Scheme, 2011, 2012], , Tsuperfast =, , As per question we have, 192 - 192 = 2, x, x + 16, , x2 + 7x - 120 = 0, , Ans :, , A journey of 192 km from a town A to town B takes, 2 hours more by an ordinary passenger train than a, super fast train. If the speed of the faster train is, 16 km/h more, find the speed of the faster and the, passanger train., Ans :, [Board Term-2, 2012 Set (28)], , Thus speed of super-fast train = (x + 16) km/h, Now, Tpassenger = 192, x, , Thus, AP = x + 7, Here AP is diameter or, +APB = 90c and AB = 17, m, , 3., , ^x + 30h^x - 20h = 0, x = - 30 or x = 20, , 7., , In a rectangular part of dimension 50m # 40m a, rectangular pond is constructed so that the area of, grass strip of uniform breadth surrounding the pond, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 77

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Chap 4 : Quadratic Equation, , www.cbse.online, , 2, , 2, , 9x = 2025 - 1800 = 225, , would be 1184m . Find the length and breadth of the, pond., , x2 = 25, x =! 5, Hence, the speed of the stream = 5 km/hr, 10. A takes 6 days less than B to do a work. If both A, , and B working together can do it in 4 days, how many, days will B take to finish it ?, Ans :, [Board 2016], , [Board Foregin Set-I, III, 2017], , Ans :, , Let width of grass strip be x mts., , Let B complete a work in x days., Then A takes x - 6 days to complete it., Together they complete it in 4 days., According to work done per day,, 1 +1 =1, x-6 x, 4, , Length of pond = ^50 - 2x h mt, and breadth of pond = ^40 - 2x h mt, and area of park – area of pond = area of grass strip, , ^50 # 40h - ^50 - 2x h^40 - 2x h = 1184, 2000 - 2000 + 180x - 4x2 = 1184, , x+x-6 = 1, x ^x - 6h, 4, , x2 - 45x + 296 = 0, , 4 ^2x - 6h = x ^x - 6h, 8x - 24 = x2 - 6x, , x2 - 37x - 8x + 296 = 0, x ^x - 37h - 8 ^x - 37h = 0, , x2 - 14x + 24 = 0, , x = 8, 37, , x2 - 12x - 2x + 24 = 0, , Here 37 is not possible, thus x = 8, , x ^x - 12h - 2 ^x - 12h = 0, , Length of pond = 50 - 2 # 8, , ^x - 2h^x - 12h = 0, x = 2 or 12, Here x = 2 is not possible because x - 6 is ^- 4h which, is not possible., , = 34 m, and breadth of pond = 40 - 2 # 8, = 24 m., 8., , A car covers a distance of 2592 km with a uniform, speed. The number of hours taken for journey is one, half the number representing the speed in km/hour., Find the time taken to ever the distances., Ans :, [Delhi Compt Set-I, III 2017], Let the speed of the car be x km/hr., Therefore time taken = x hour, 2, Now, , Speed = Dis tan ce, Time, x = 2592, x, 2, , 2, , x = 2592 # 2 = 5184, x = 5184 = 72, Hence the time taken 72 = 36 hours., 2, 9., , Speed of a boat in still water is 15 km/hour. It goes, 30 km up stream and returns back at the same point, in 4 hours 30 minutes. Find the speed of the stream., Ans :, [Board Delhi Set-I, III, 2017], Let the speed of the Stream be x km/hr., Speed of boat up stream = 15 - x, and speed of boat down stream = 15 + x, According to the question, 30 + 30 = 4 1, 2, 10 - x 15 + x, 30 ^15 + x h + 30 ^15 - x h, =9, 2, 152 - x2, 900 # 2 = 9 ^152 - x2h, , Thus x = 12 and B takes 12 days to finish the work., 11., , In a class test Raveena got a total of 30 mark in, English and Mathematics. Had she got 2 more marks, in Mathematics and 3 marks less in English then the, product of her marks obtained would have be 210., Find the individual marks obtained in two subjects., Ans :, [Outside Delhi Compt. I, II, III 2017], Let Raveena got marks in English be x ., Marks in Mathematics = ^30 - x h, Marks in English = ^x - 3h, According to problem, , ^x - 3h^30 - x + 2h = 210, 35x - 96 - x2 = 210, x2 - 35x + 306 = 0, x2 - 18x - 17x + 306 = 0, x ^x - 18h - 17 ^x - 18h = 0, , ^x - 18h^x - 17h = 0, x = 18, 17, Hence, if she got 18 marks in English, then she got 12, in mathematics., If she got 17 marks in English, then she got 13 marks, in mathematics., 1, 13, hours. If one tap takes 3 hours more than the other to, fill the tank, then how much time will each tap take, to fill the tank?, Ans :, [Outside Delhi I 2017], , 12. Two taps running together can fill a tank in 3, , Tow tap running together fill the tank in 3 1 hr., 13, , Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in, , Page 78

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Chap 4 : Quadratic Equation, , www.rava.org.in, , = 40 hours, 13, Thus, It will fill in 1 hour = 13 tank, 40, If first tap alone fills the tank in x hrs, then second, tap alone fills it in ^x + 3h hr., 1 + 1 = 13, Now,, x x+3, 40, , alone will finish in ^x - 5h days., 1 +1 =1, Now,, 6, x-5 x, x+x-5 = 1, 6, x ^x - 5h, 12x - 30 = x2 - 5x, x2 - 17x + 30 = 0, , x + 3 + x = 13, x ^x + 3h, 40, , x2 - ^15 + 2h x + 30 = 0, x2 - 15x - 2x + 30 = 0, , 80x + 120 = 13x2 + 39x, or,, , x ^x - 15h - ^x - 15h = 0, , 13x2 - 41x - 120 = 0, , ^x - 15h^x - 2h = 0, x = 15, x = 2, , 13x2 - ^65 - 24h x + 120 = 0, , ^x - 5h^13x + 24h = 0, , x = 5 , x =- 24, 13, Here x =- 24 is not possible. Hence, 1st tap takes 5, 13, hours and 2nd tap takes = 5 + 3 = 8 hours, 8, 11, minutes. If one tap takes 1 minute more than the, other to fill the cistern, find the time in which each, tap separately can fill the cistern., Ans :, [Outside Delhi Set-III 2017], , 13. Two taps running together can fill a cistern in 2, , Two taps together fill the cistern in 2 8 minutes, 11, = 30 minutes, 11, Thus It will fill in 1 minute = 11 cistern, 30, Let first tap fills the same cistern in x minutes, and 2nd tap will take = ^x + 1h minutes, Thus, , Here x = 2 is not possible. Hence, B finishes the work, in 15 days., 15. Ram takes 6 days less than Bhagat to finish a place, , of work. If both of them together can finish the work, in 4 days, in how many days Bhagat alone can finish, the work ?, Ans :, [Delhi Compt. Set-III 2017, Outside Delhi Set-II2017], Ram and Bhagat together do the work in 4 days, Ram and Bhagat will do in one days = 1 work, 4, Let Bhagat alone does the same work in x days., Now, , Ram will take = ^x - 6h days, 1+ 1 =1, x x-6, 4, x-6+x = 1, x ^x - 6h, 4, 8x - 24 = x2 - 6x, x2 - 14x + 24 = 0, , 1+ 1, = 11, x ^x + 1h, 30, , x2 - ^12 + 2h x + 24 = 0, , x + 1 + x = 11, 30, x ^x + 1h, , x ^x - 12h - 2 ^x - 12h = 0, , 60x + 30 = 11x2 + 11x, 11x2 - 49x - 30 = 0, 11x2 - ^55 - 6h x - 30 = 0, 11x2 - 55x + 6x - 30 = 0, 11x ^x - 5h + 6 ^x - 5h = 0, , ^x - 5h^11x + 6h = 0, x = 5, x =- 6, 11, , Here x =- 6 is not possible. Hence, 1st tap takes 5, 11, minute and 2nd tap takes 6 minute., 14. A and B working together can do a work in 6 days. If, , A takes 5 days less than B to finish the work, in how, many days B alone can do it alone?, Ans :, [Outside Delhi Compt. Set-I], Since A + B finish the work in 6 days., They will finish in one day = 1 work, 6, , x2 - 12x - 2x + 24 = 0, , ^x - 12h^x - 12h = 0, x = 12, x = 2, If Bhagat complete the work in 2 days, Ram will take, = 2 - 6 =- 4 days that is impossible. Hence, Bhagat, can finish in 12 days., For more files visit www.cbse.online, , NO NEED TO PURCHASE ANY BOOKS, , For session 2019-2020 free pdf will be available at, www.cbse.online for, 1. Previous 15 Years Exams Chapter-wise Question, Bank, 2. Previous Ten Years Exam Paper (Paper-wise)., 3. 20 Model Paper (All Solved)., 4. NCERT Solutions, All material will be solved and free pdf. It will be, provided by 30 September and will be updated regularly., , Disclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education,, New Delhi in any manner. www.cbse.online is a private organization which provide free, study material pdfs to students. At www.cbse.online CBSE stands for Canny Books, For School Education, , Let B alone does the same work in x days, then A, , Get all GUIDE and Sample Paper PDFs by whatsapp from +91 89056 29969, , Page 79