NCERT Exemplar Solutions: Squares & Square Roots & Cubes & Cube Roots
Exercise Page: 88
In each of the questions, 1 to 24, write the correct answer from the given four options.
Q.1. 196 is the square of
(a) 11
(b) 12
(c) 14
(d) 16
Ans: (c)
Explanation: Square of 196 is,
= 14 × 14 = 196
Q.2. Which of the following is a square of an even number?
(a) 144
(b) 169
(c) 441
(d) 625
Ans: (a)
Explanation: 144 = 12 × 12, and 12 is even. Hence 144 is the square of an even number.
Q.3. A number ending in 9 will have the units place of its square as
(a) 3
(b) 9
(c) 1
(d) 6
Ans: (c)
Explanation: A number ending in 9 (or 1) gives a square that ends in 1. Example: 11 × 11 = 121.
Q.4. Which of the following will have 4 at the units place?
(a) 142
(b) 622
(c) 272
(d) 352
Ans: (b)
Explanation: A number ending with 2 or 8 has its square ending in 4. 62 × 62 = 3,844, so units digit is 4.
Q.5. How many natural numbers lie between 52 and 62?
(a) 9
(b) 10
(c) 11
(d) 12
Ans: (b)
Explanation: Numbers between 25 and 36 are 26, 27, 28, 29, 30, 31, 32, 33, 34 and 35 - ten numbers.
Q.6. Which of the following cannot be a perfect square?
(a) 841
(b) 529
(c) 198
(d) All of the above
Ans: (c)
Explanation: Perfect squares never end with the digits 2, 3, 7 or 8. Since 198 ends with 8, it cannot be a perfect square.
Q.7. The one's digit of the cube of 23 is
(a) 6
(b) 7
(c) 3
(d) 9
Ans: (b)
Explanation: 23 × 23 × 23 = 12,167. The unit digit is 7.
Q.8. A square board has an area of 144 square units. How long is each side of the board?
(a) 11 units
(b) 12 units
(c) 13 units
(d) 14 units
Ans: (b)
Explanation: Area = side × side, so side = √144 = 12 units.
Q.9. Which letter best represents the location of √25 on a number line?
(a) A
(b) B
(c) C
(d) D
Ans: (c)
Explanation: √25 = 5 since 5 × 5 = 25. The point labelled C corresponds to 5 on the number line.
Q.10. If one member of a Pythagorean triplet is 2m, then the other two members are
(a) m, m2 + 1
(b) m2 + 1, m2 - 1
(c) m2, m2 - 1
(d) m2, m + 1
Ans: (b)
Explanation: For any natural m > 1, the numbers 2m, m2 - 1 and m2 + 1 form a Pythagorean triplet (their squares satisfy a2 + b2 = c2).
Q.11. The sum of successive odd numbers 1, 3, 5, 7, 9, 11, 13 and 15 is
(a) 81
(b) 64
(c) 49
(d) 36
Ans: (b)
Explanation: = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
= 64
Q.12. The sum of first n odd natural numbers is
(a) 2n + 1
(b) n2
(c) n2 - 1
(d) n2 + 1
Ans: (b)
Explanation: The sum of the first n odd natural numbers equals n2.
Q.13. Which of the following numbers is a perfect cube?
(a) 243
(b) 216
(c) 392
(d) 8640
Ans: (b)
Explanation: A perfect cube can be expressed as product of triplets of the same prime. 216 = 6 × 6 × 6 = 63.
Q.14. The hypotenuse of a right triangle with its legs of lengths 3x × 4x is
(a) 5x
(b) 7x
(c) 16x
(d) 25x
Ans: (a)
Explanation: Using Pythagoras theorem,
a = √((3x)2 + (4x)2)
= √(9x2 + 16x2)
= √(25x2)
= 5x.
Hence the hypotenuse is 5x.
Q.15. The next two numbers in the number pattern 1, 4, 9, 16, 25 ... are
(a)35, 48
(b) 36, 49
(c) 36, 48
(d) 35,49
Ans: (b)
Explanation: These are perfect squares: 12, 22, 32, 42, 52. Next are 62=36 and 72=49.
Q.16. Which among 432, 672, 522, 592 would end with digit 1?
(a) 432
(b) 672
(c) 522
(d) 592
Ans: (d)
Explanation: Squares of numbers ending in 1 or 9 end in 1. Among the options, 59 ends with 9, so 592 ends with 1.
Q.17. A perfect square can never have the following digit in its ones place.
(a) 1
(b) 8
(c) 0
(d) 6
Ans: (b)
Explanation: No integer squared ends with the digit 8, so a perfect square cannot have 8 in the units place.
Q.18. Which of the following numbers is not a perfect cube?
(a) 216
(b) 567
(c) 125
(d) 343
Ans: (b)
Explanation: 216 = 6 × 6 × 6, 125 = 5 × 5 × 5, 343 = 7 × 7 × 7. But 567 = 9 × 9 × 7, not a product of three equal factors, so it is not a perfect cube.
Q.19. 3√1000 is equal to
(a) 10
(b) 100
(c) 1
(d) none of these
Ans: (a)
Explanation: 3√1000 = 3√(10 × 10 × 10) = 3√(103) = 10.
Q.20. If m is the square of a natural number n, then n is
(a) the square of m
(b) greater than m
(c) equal to m
(d) √m
Ans: (d)
Explanation: Given m = n2. Solving for n gives n = √m.
Q.21. A perfect square number having n digits where n is even will have square root with
(a) n + 1 digit
(b) n/2 digit
(c) n/3 digit
(d) (n + 1)/2 digit
Ans: (b) n/2 digit
Q.22. If m is the cube root of n, then n is
(a) m3
(b) √m
(c) m/3
(d) 3√m
Ans: (a)
Explanation: Given 3√n = m, so n = m3.
Q.23. The value of √(248 + √(52 + √144)) is
(a) 14
(b) 12
(c) 16
(d) 13
Ans: (c)
Explanation: √(248 + √(52 + √144))
= √(248 + √(52 + 12))
= √(248 + √64)
= √(248 + 8)
= √256 = 16.
Q.24. Given that √4096 = 64, the value of √4096 + √40.96 is
(a) 74
(b) 60.4
(c) 64.4
(d) 70.4
Ans: (d)
Explanation: √4096 + √40.96 = 64 + 6.4 = 70.4.
In questions 25 to 48, fill in the blanks to make the statements true.
Q.25. There are _________ perfect squares between 1 and 100.
Solution: There are 8 perfect squares between 1 and 100.
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
6 × 6 = 36
7 × 7 = 49
8 × 8 = 64
9 × 9 = 81
Q.26. There are _________ perfect cubes between 1 and 1000.
Solution: There are 8 perfect cubes between 1 and 1000.
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
5 × 5 × 5 = 125
6 × 6 × 6 = 216
7 × 7 × 7 = 343
8 × 8 × 8 = 512
9 × 9 × 9 = 729
Q.27. The units digit in the square of 1294 is _________.
Solution: The units digit in the square of 1294 is 6.
Squares of numbers ending in 4 or 6 end in 6.
Q.28. The square of 500 will have _________ zeroes.
Solution: The square of 500 will have 4 zeroes.
5002 = 500 × 500 = 250000
Q.29. There are _________ natural numbers between n2 and (n + 1)2.
Solution: There are 2n natural numbers between n2 and (n + 1)2.
Q.30. The square root of 24025 will have _________ digits.
Solution: The square root of 24025 will have 3 digits.
The number has 5 digits. For an odd number n of digits, the square root has (n + 1)/2 digits,
= (5 + 1)/2 = 3.
Q.31. The square of 5.5 is _________.
Solution: The square of 5.5 is 30.25.
= 5.52 = 5.5 × 5.5 = 30.25
Q.32. The square root of 5.3 × 5.3 is _________.
Solution: The square root of 5.3 × 5.3 is 5.3.
= √(5.3 × 5.3) = √(28.09) = 5.3
Q.33. The cube of 100 will have _________ zeroes.
Solution: The cube of 100 will have 6 zeroes.
= 1003 = 100 × 100 × 100 = 1,000,000
Q.34. 1m2 = _________ cm2.
Solution: 1 m = 100 cm. Thus 1 m2 = 100 × 100 = 10,000 cm2.
Q.35. 1m3 = _________ cm3.
Solution: 1 m = 100 cm. Thus 1 m3 = 100 × 100 × 100 = 1,000,000 cm3.
Q.36. Ones digit in the cube of 38 is _________.
Solution: Ones digit in the cube of 38 is 2, since 38 × 38 × 38 = 54,872.
Q.37. The square of 0.7 is _________.
Solution: The square of 0.7 is 0.49.
= 0.72 = 0.7 × 0.7 = 0.49
Q.38. The sum of first six odd natural numbers is _________.
Solution: The sum of first six odd natural numbers is 36.
1 + 3 + 5 + 7 + 9 + 11 = 36
Q.39. The digit at the ones place of 572 is _________.
Solution: The digit at the ones place of 572 is 9.
Digits 3 and 7 when squared give unit digit 9. 57 × 57 = 3,249.
Q.40. The sides of a right triangle whose hypotenuse is 17cm are _________ and _________.
Solution: For m > 1, the numbers 2m, m2 - 1 and m2 + 1 form a Pythagorean triplet with hypotenuse m2 + 1. If the hypotenuse is 17, then m2 + 1 = 17 so m2 = 16 and m = 4. The other two sides are 2m = 8 and m2 - 1 = 15. Hence the sides are 8 cm and 15 cm.
Q.41. √(1.96) = _________.
Solution: √1.96 = √(196/100) = √(142/102) = 14/10 = 1.4.
Q.42. (1.2)3 = _________.
Solution: (1.2)3 = (12/10)3 = 1,728/1,000 = 1.728.
Q.43. The cube of an odd number is always an _________ number.
Solution: The cube of an odd number is always an odd number.
Q.44. The cube root of a number x is denoted by _________.
Solution: The cube root of x is denoted by 3√x or x1/3.
Q.45. The least number by which 125 be multiplied to make it a perfect square is _____________.
Solution: 125 = 5 × 5 × 5 = 53. To make all prime powers even, multiply by one more 5. So multiply by 5 to get 625 = 252. The least number is 5.
Q.46. The least number by which 72 be multiplied to make it a perfect cube is _____________.
Solution: 72 = 23 × 32. To make each prime exponent a multiple of 3, multiply by one more 3. So multiply by 3; the least number is 3.
Q.47. The least number by which 72 be divided to make it a perfect cube is _____________.
Solution: Divide 72 by 9 to get 8, which is 23. So the least number is 9.
Q.48. Cube of a number ending in 7 will end in the digit _______________.
Solution: The cube of a number ending in 7 ends with 3. (For example, 73=343.)
In questions 49 to 86, state whether the statements are true (T) or false (F).
Q.49. The square of 86 will have 6 at the units place.
Solution: True.
Numbers ending with 4 or 6 have squares ending in 6.
Q.50. The sum of two perfect squares is a perfect square.
Solution: False.
Example: 25 + 49 = 74, which is not a perfect square.
Q.51. The product of two perfect squares is a perfect square.
Solution: True.
Example: 4 × 9 = 36, which is a perfect square.
Q.52. There is no square number between 50 and 60.
Solution: True.
Q.53. The square root of 1521 is 31.
Solution: False.
√1521 = 39.
Q.54. Each prime factor appears 3 times in its cube.
Solution: True.
A perfect cube's prime factorisation has each prime exponent divisible by 3.
Q.55. The square of 2.8 is 78.4.
Solution: False.
2.82 = (28/10)2 = 784/100 = 7.84.
Q.56. The cube of 0.4 is 0.064.
Solution: True.
0.43 = (4/10)3 = 64/1000 = 0.064.
Q.57. The square root of 0.9 is 0.3.
Solution: False.
√0.9 ≈ 0.9487. Note that 0.32 = 0.09, not 0.9.
Q.58. The square of every natural number is always greater than the number itself.
Solution: False.
For n = 1, n2 = 1 which is equal to n, not greater.
Q.59. The cube root of 8000 is 200.Solution: False.
3√8000 = 20.
Q.60. There are five perfect cubes between 1 and 100.
Solution: False.
There are three perfect cubes between 1 and 100: 8, 27 and 64.
Q.61. There are 200 natural numbers between 1002 and 1012.
Solution: True.
Number of integers between a and b is b - a - 1. Here 1012 - 1002 - 1 = (101 + 100)(101 - 100) - 1 = 201 - 1 = 200.
Q.62. The sum of first n odd natural numbers is n2.
Solution: True. Using the formula gives n2.
Q.63. 1000 is a perfect square.
Solution: False.
1000 = 10 × 10 × 10 is a perfect cube, not a perfect square.
Q.64. A perfect square can have 8 as its unit's digit.
Solution: False.
No perfect square ends with 8.
Q.65. For every natural number m, (2m -1, 2m2 -2m, 2m2 -2m + 1) is a Pythagorean triplet.
Solution: False.
The standard formula is 2m, m2 - 1 and m2 + 1 for m > 1.
Q.66. All numbers of a Pythagorean triplet are odd.
Solution: False.
Example: 5, 12, 13 is a Pythagorean triplet; 12 is even.
Q.67. For an integer a, a3 is always greater than a2.
Solution: False.
For negative integers this is not true. Example: a = -2 gives a3 = -8 and a2 = 4, so a3 >a2.
Q.68. If x and y are integers such that x2 > y2, then x3 > y3.
Solution: False.
Counterexample with negative integers: x = -2, y = -1 gives x2 = 4 > 1 = y2, but x3 = -8 < -1 = y3.
Q.69. Let x and y be natural numbers. If x divides y, then x3 divides y3.
Solution: True.
If y = kx for some integer k, then y3 = k3x3, so x3 divides y3.
Q.70. If a2 ends in 5, then a3 ends in 25.
Solution: False.
Example: a = 15 gives a2 = 225 (ends in 5) but a3 = 3,375 (ends in 75), not 25.
Q.71. If a2 ends in 9, then a3 ends in 7.
Solution: False.
Example: a = 7 gives a2 = 49 (ends in 9) and a3 = 343 (ends in 3), not 7.
Q.72. The square root of a perfect square of n digits will have (n + 1)/2 digits, if n is odd.
Solution: True.
Example: a 9-digit perfect square has a 5-digit square root since (9 + 1)/2 = 5.
Q.73. Square root of a number x is denoted by √x.
Solution: True.
Q.74. A number having 7 at its ones place will have 3 at the units place of its square.
Solution: False.
For numbers ending in 7, the square ends in 9. Example: 472 = 2,209 (units digit 9).
Q.75. A number having 7 at its ones place will have 3 at the ones place of its cube.
Solution: True.
Examples: 373 = 50,653 and 273 = 19,683 - both end with 3.
Q.76. The cube of a one digit number cannot be a two digit number.
Solution: False.
Example: 4 is a one-digit number and 43 = 64, which is a two-digit number.
Q.77. Cube of an even number is odd.
Solution: False.
The cube of an even number is always even. Example: 23=8, 43=64.
Q.78. Cube of an odd number is even.
Solution: False.
The cube of an odd number is always odd. Example: 33=27, 53=125.
Q.79. Cube of an even number is even.
Solution: True.
Example: 83=512, 123=1,728.
Q.80. Cube of an odd number is odd.
Solution: True.
Example: 73=343.
Q.81. 999 is a perfect cube.
Solution: False.
Q.82. 363 × 81 is a perfect cube.
Solution: False.
Prime factors: 363 = 3 × 11 × 11 = 3 × 112, and 81 = 34. So 363 × 81 = 35 × 112. For a perfect cube, every prime exponent must be a multiple of 3; here 11 has exponent 2, so the product is not a perfect cube.
Q.83. Cube roots of 8 are +2 and -2.
Solution: False.
The real cube root of 8 is 2 only, since (-2)3 = -8, not 8.
Q.84. 3√ (8 + 27) = 3√ (8) + 3√ (27).
Solution: False.
3√(8 + 27) = 3√35, while 3√8 + 3√27 = 2 + 3 = 5; these are not equal.
Q.85. There is no cube root of a negative integer.
Solution: False.
Negative integers can have real cube roots. Example: 3√(-27) = -3.
Q.86. Square of a number is positive, so the cube of that number will also be positive.
Solution: False.
Example: number -3 has square 9 (positive) but cube -27 (negative).
Solve the following questions.
Q.87. Write the first five square numbers.
Solution: The first five square numbers are,
1 × 1 = 1
2 × 2 = 4
3 × 3 = 9
4 × 4 = 16
5 × 5 = 25
Q.88. Write cubes of first three multiples of 3.
Solution: The first three multiples of 3 are 3, 6 and 9.
Cubes:
33 = 27
63 = 216
93 = 729
Q.89. Show that 500 is not a perfect square.
Solution: Prime factors: 500 = 2 × 2 × 5 × 5 × 5 = 22 × 53. Since the exponent of 5 is odd, prime factors do not all occur in pairs. Hence 500 is not a perfect square.
Q.90. Express 81 as the sum of first nine consecutive odd numbers.
Solution: First nine odd numbers: 1, 3, 5, 7, 9, 11, 13, 15, 17. Their sum is 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 = 81. Also 81 = 92.
Q.91. Using prime factorisation, find which of the following are perfect squares.
(a) 484
Solution: Prime factors: 484 = 2 × 2 × 11 × 11 = 22 × 112. All primes occur in pairs, so 484 is a perfect square.
(b) 11250
Solution: Prime factors: 11250 = 2 × 3 × 3 × 5 × 5 × 5 × 5 = 2 × 32 × 54. The prime 2 occurs only once (no pair), so 11250 is not a perfect square.
(c) 841
Solution: 841 = 29 × 29 = 292. All factors pair up, so 841 is a perfect square.
(d) 729
Solution: 729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3) × (3 × 3) × (3 × 3) = 36 = (33)2. Thus 729 = 272, so it is a perfect square.
Q.92. Using prime factorisation, find which of the following are perfect cubes.
(a)128
Solution: 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27. Grouping in triplets leaves one 2 ungrouped, so 128 is not a perfect cube.
(b) 343
Solution: 343 = 7 × 7 × 7 = 73. All factors group into one triplet, so 343 is a perfect cube.
(c) 729
Solution: 729 = 3 × 3 × 3 × 3 × 3 × 3 = (3 × 3 × 3) × (3 × 3 × 3) = 33 × 33. Thus 729 is a perfect cube (93 = 729).
(d) 1331
Solution: 1331 = 11 × 11 × 11 = 113, so 1331 is a perfect cube.
Q.93. Using distributive law, find the squares of
(a)101
Solution: 101 = 100 + 1, so 1012 = (100 + 1)2 = 10000 + 100 + 100 + 1 = 10,201.
(b) 72
Solution: 72 = 70 + 2, so 722 = (70 + 2)2 = 4,900 + 140 + 140 + 4 = 5,184.
Q.94. Can a right triangle with sides 6cm, 10cm and 8cm be formed? Give reason.
Solution: Check Pythagoras: If 10 is hypotenuse, 102 = 100 and 82 + 62 = 64 + 36 = 100. Equality holds, so a right triangle with sides 6 cm, 8 cm and hypotenuse 10 cm can be formed.
Q.95. Write the Pythagorean triplet whose one of the numbers is 4.
Solution: If 2m = 4 then m = 2. The triplet is 2m = 4, m2 - 1 = 3 and m2 + 1 = 5. Thus 3, 4, 5 is the Pythagorean triplet.
Q.96. Using prime factorisation, find the square roots of
(a) 11025
Solution: 11025 = 3 × 3 × 5 × 5 × 7 × 7 = 32 × 52 × 72. √11025 = 3 × 5 × 7 = 105.
(b) 4761
Solution: 4761 = 3 × 3 × 23 × 23 = 32 × 232. √4761 = 3 × 23 = 69.
Q.97. Using prime factorisation, find the cube roots of
(a) 512
Solution: 512 = 29 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2). Grouping into triplets gives 23 × 23 × 23, so 3√512 = 2 × 2 × 2 = 8.
(b) 2197
Solution: 2197 = 13 × 13 × 13 = 133. So 3√2197 = 13.
Q.98. Is 176 a perfect square? If not, find the smallest number by which it should be multiplied to get a perfect square.
Solution: Prime factors: 176 = 2 × 2 × 2 × 2 × 11 = 24 × 11. The factor 11 is unpaired. Multiply by 11 to pair it: 176 × 11 = 1,936 = 442. So 176 is not a perfect square; the smallest multiplier is 11 and √1936 = 44.
Q.99. Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.
Solution: Prime factors: 9720 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 = 23 × 35 × 5. To make exponents multiples of 3 in the quotient, divide out the extra factors 32 × 5 = 9 × 5 = 45. So divide by 45 to get 9720 ÷ 45 = 216, and 3√216 = 6.
Q.100. Write two Pythagorean triplets each having one of the numbers as 5.
Solution: One triplet with 5 is 3, 4, 5. Another is 5, 12, 13.
Q.101. By what smallest number should 216 be divided so that the quotient is a perfect square. Also find the square root of the quotient.
Solution:
Prime factors: 216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33. To get a perfect square after division, reduce each exponent to the nearest even number. Dividing by 2 × 3 = 6 leaves exponents 2 and 2, so quotient = 216 ÷ 6 = 36. √36 = 6. Thus divide by 6; the quotient is 36 and its square root is 6.
Q.102. By what smallest number should 3600 be multiplied so that the quotient is a perfect cube. Also find the cube root of the quotient.
Solution:
Prime factors: 3600 = 24 × 32 × 52. To make each exponent a multiple of 3, multiply by 22 × 3 × 5 = 4 × 3 × 5 = 60. Then 3600 × 60 = 216,000 = 26 × 33 × 53 = (22)3 × 33 × 53. So the cube root is 22 × 3 × 5 = 4 × 3 × 5 = 60. The smallest multiplier is 60 and the cube root of the product is 60.
Q.103. Find the square root of the following by long division method.
(a) 1369
Solution: Using long division method, √1369 = 37.
(b) 5625
Solution: Using long division method, √5625 = 75.
Q.104. Find the square root of the following by long division method.
(a) 27.04
Solution: By long division, √27.04 = 5.2.
(b) 1.44
Solution: By long division, √1.44 = 1.2.
Q.105. What is the least number that should be subtracted from 1385 to get a perfect square? Also find the square root of the perfect square.
Solution:
From long division, the nearest lower perfect square is 1,369. So subtract 1385 - 1369 = 16. The resulting perfect square is 1369 and √1369 = 37.
Q.106. What is the least number that should be added to 6200 to make it a perfect square?
Solution:
782 = 6,084 and 792 = 6,241. The next perfect square after 6,200 is 6,241, so add 6,241 - 6,200 = 41. The required perfect square is 6,241 whose square root is 79.
Q.107. Find the least number of four digits that is a perfect square.
Solution: The smallest four-digit number is 1,000. The least four-digit perfect square is 1,024 = 322 (found by long division), so 1,024 is the required number.
Q.108. Find the greatest number of three digits that is a perfect square.
Solution: The greatest three-digit number is 999. The greatest three-digit perfect square is 961 = 312. Thus 961 is required.
Q.109. Find the least square number which is exactly divisible by 3, 4, 5, 6 and 8.
Solution: Find LCM of 3, 4, 5, 6 and 8: LCM = 2 × 2 × 2 × 3 × 5 = 120. To make it a perfect square multiply by 2 × 3 × 5 = 30, giving 120 × 30 = 3,600. Hence 3,600 is the least square number divisible by all given numbers.
Q.110. Find the length of the side of a square if the length of its diagonal is 10cm.
Solution:
In right triangle formed by half the square, diagonal2 = side2 + side2. So 102 = 2y2, giving y2 = 50 and y = √50 = 5√2 cm. Thus each side is √50 cm or 5√2 cm.
Q.111. A decimal number is multiplied by itself. If the product is 51.84, find the number.
Solution: Let the number be a. Then a2 = 51.84 so a = √51.84. Using long division, √51.84 = 7.2. Thus the number is 7.2.
Q.112. Find the decimal fraction which when multiplied by itself gives 84.64.
Solution: Let the decimal be a. a2 = 84.64, so a = √84.64. By long division, √84.64 = 9.2. Hence the decimal fraction is 9.2.
Q.113. A farmer wants to plough his square field of side 150m. How much area will he have to plough?
Solution: Area = side × side = 150 × 150 = 22,500 m2.
Q.114. What will be the number of unit squares on each side of a square graph paper if the total number of unit squares is 256?
Solution: If a × a = 256 then a = √256 = 16. So there are 16 unit squares on each side.
Q.115. If one side of a cube is 15m in length, find its volume.
Solution: Volume = side3 = 153 = 15 × 15 × 15 = 3,375 m3.
Q.116. The dimensions of a rectangular field are 80m and 18m. Find the length of its diagonal.
Solution: Diagonal = √(802 + 182) = √(6400 + 324) = √6724 = 82 m.
Q.117. Find the area of a square field if its perimeter is 96m.
Solution: Side = 96/4 = 24 m. Area = 242 = 576 m2.
Q.118. Find the length of each side of a cube if its volume is 512 cm3.
Solution: Side = 3√512 = 8 cm, since 8 × 8 × 8 = 512.
Q.119. Three numbers are in the ratio 1:2:3 and the sum of their cubes is 4500. Find the numbers.
Solution: Let numbers be a, 2a, 3a. Then a3 + 8a3 + 27a3 = 36a3 = 4,500. Thus a3 = 125 and a = 5. Numbers are 5, 10 and 15.
Q.120. How many square metres of carpet will be required for a square room of side 6.5m to be carpeted.
Solution: Area = 6.5 × 6.5 = 42.25 m2.
Q.121. Find the side of a square whose area is equal to the area of a rectangle with sides 6.4m and 2.5m.
Solution: Area of rectangle = 6.4 × 2.5 = 16 m2. For a square with area 16, side = √16 = 4 m.
Q.122. Difference of two perfect cubes is 189. If the cube root of the smaller of the two numbers is 3, find the cube root of the larger number.
Solution: Smaller number = 33 = 27. Let larger cube be a3. Then a3 - 27 = 189, so a3 = 216 and a = 3√216 = 6. Thus the cube root of the larger number is 6.
Q.123. Find the number of plants in each row if 1024 plants are arranged so that number of plants in a row is the same as the number of rows.
Solution: If a × a = 1024 then a = √1024 = 32. So there are 32 plants in each row.
Q.124. A hall has a capacity of 2704 seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.
Solution: If a × a = 2704 then a = √2704 = 52. So there are 52 seats in each row.
Q.125. A General wishes to draw up his 7500 soldiers in the form of a square. After arranging, he found out that some of them are left out. How many soldiers were left out?
Solution: Largest square less than or equal to 7,500 is 862 = 7,396. Leftover soldiers = 7,500 - 7,396 = 104.
Q.126. 8649 students were sitting in a lecture room in such a manner that there were as many students in the row as there were rows in the lecture room. How many students were there in each row of the lecture room?
Solution: If a × a = 8,649 then a = √8,649 = 93. So there were 93 students in each row.
Q.127. Rahul walks 12m north from his house and turns west to walk 35m to reach his friend's house. While returning, he walks diagonally from his friend's house to reach back to his house. What distance did he walk while returning?
Solution:
Consider right triangle with legs 12 m and 35 m. By Pythagoras, diagonal = √(122 + 352) = √(144 + 1,225) = √1,369 = 37 m. Rahul walked 37 m while returning.
Q.128. A 5.5m long ladder is leaned against a wall. The ladder reaches the wall to a height of 4.4m. Find the distance between the wall and the foot of the ladder.
Solution: Consider right triangle with hypotenuse 5.5 m and vertical side 4.4 m. Using Pythagoras,
distance2 = 5.52 - 4.42 = 30.25 - 19.36 = 10.89. Distance = √10.89 = 3.3 m.
Q.129. A king wanted to reward his advisor, a wise man of the kingdom. So he asked the wiseman to name his own reward. The wiseman thanked the king but said that he would ask only for some gold coins each day for a month. The coins were to be counted out in a pattern of one coin for the first day, 3 coins for the second day, 5 coins for the third day and so on for 30 days. Without making calculations, find how many coins will the advisor get in that month?
Solution: The adviser receives the first 30 odd numbers. Sum of first 30 odd numbers = 302 = 900. So he gets 900 coins.
Q.130. Find three numbers in the ratio 2:3:5, the sum of whose squares is 608.
Solution: Let numbers be 2a, 3a, 5a. Then 4a2 + 9a2 + 25a2 = 38a2 = 608 ⇒ a2 = 16 ⇒ a = 4. Numbers are 8, 12 and 20.
Q.131. Find the smallest square number divisible by each one of the numbers 8, 9 and 10.
Solution: LCM(8,9,10) = 360. To make it a perfect square multiply by 2 × 5 = 10 to balance prime exponents. Thus 360 × 10 = 3,600 is the least square divisible by 8, 9 and 10.
Q.132. The area of a square plot is 101 1/400 m2. Find the length of one side of the plot.
Solution: Convert area to improper fraction: 101 1/400 = (101 × 400 + 1)/400 = 40,401/400. Side = √(40,401/400) = 201/20 = 10.05 m. So one side is 201/20 m (10.05 m).
Q.133. Find the square root of 324 by the method of repeated subtraction.
Solution: Subtract successive odd numbers from 324: 324 - 1 - 3 - 5 - 7 - ... - 35 = 0. It takes 18 subtractions, so √324 = 18.
Q.134. Three numbers are in the ratio 2:3:4. The sum of their cubes is 0.334125. Find the numbers.
Solution: Let numbers be 2a, 3a, 4a. Then 8a3 + 27a3 + 64a3 = 99a3 = 0.334125. So a3 = 0.003375 = 3375/1,000,000. Therefore a = 15/1,000 = 0.015. Numbers are 0.03, 0.045 and 0.06.
Q.135. Evaluate: 3√27 + 3√0.008 + 3√0.064
Solution: = 3√27 + 3√(8/1000) + 3√(64/1000)
= 3 + 2/10 + 4/10 = 3 + 0.2 + 0.4 = 3.6.
Q.136. {(52 + (122)1/2)}3
Solution: {52 + √(122)}3 = (25 + 12)3 = 373 = 50,653.
Q.137. {(62 + (82)1/2)}3
Solution: {36 + √64}3 = (36 + 8)3 = 443 = 85,184.
Q.138. A perfect square number has four digis, none of which is zero. The digits from left to right have values that are: even, even, odd, even. Find the number.
Solution: A four-digit perfect square with digit pattern even, even, odd, even is 8,8,3,6 giving 8,836 = 942. The intended number is 8,836.
Q.139. Put three different numbers in the circles so that when you add the numbers at the end of each line you always get a perfect square.
Solution: Choose numbers 6, 19 and 30 in the three circles. Sums at each end: 6 + 19 = 25, 6 + 30 = 36, 19 + 30 = 49, all perfect squares.
Q.140. The perimeters of two squares are 40 and 96 metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.
Solution: Sides are 40/4 = 10 m and 96/4 = 24 m. Areas are 102 = 100 and 242 = 576. Sum = 676, so new side = √676 = 26 m. Perimeter = 4 × 26 = 104 m.
Q.141. A three digit perfect square is such that if it is viewed upside down, the number seen is also a perfect square. What is the number?
(Hint: The digits 1, 0 and 8 stay the same when viewed upside down, whereas 9 becomes 6 and 6 becomes 9.)
Solution: The three-digit perfect squares 196 and 961 are mirror images under the given digit mapping; both are perfect squares.
Q.142. 13 and 31 is a strange pair of numbers such that their squares 169 and 961 are also mirror images of each other. Can you find two other such pairs?
Solution: Two other pairs are 12 and 21 (122=144, 212=441) and 102 and 201 (1022=10,404, 2012=40,401).
FAQs on NCERT Exemplar Solutions: Squares & Square Roots & Cubes & Cube Roots
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