Class 8 Maths Chapter 7 Cubes and Cube roots
NCERT Solutions For Class 8 Maths Chapter 7 Cubes And Cube Roots, (Maths) exam are Students are taught thru NCERT books in some of state board and CBSE Schools. As the chapter involves an end, there is an exercise provided to assist students prepare for evaluation. Students need to clear up those exercises very well because the questions withinside the very last asked from those.
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NCERT Solution For Class 8 Maths Chapter 7 Cubes and Cube roots
Class 8 Maths Chapter 7 Cubes and Cube roots
Exercise 7.1
Page: 114
1. Which of the following numbers are not perfect cubes?
(i) 216
Solution:
By resolving 216 into prime factor,
216 = 2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
216 = (2×2×2)×(3×3×3)
Here, 216 can be grouped into triplets of equal factors,
∴ 216 = (2×3) = 6
Hence, 216 is cube of 6.
(ii) 128
Solution:
By resolving 128 into prime factor,
128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
128 = (2×2×2)×(2×2×2)×2
Here, 128 cannot be grouped into triplets of equal factors, we are left of with one factors 2 .
∴ 128 is not a perfect cube.
(iii) 1000
Solution:
By resolving 1000 into prime factor,
1000 = 2×2×2×5×5×5
By grouping the factors in triplets of equal factors,
1000 = (2×2×2)×(5×5×5)
Here, 1000 can be grouped into triplets of equal factors,
∴ 1000 = (2×5) = 10
Hence, 1000 is cube of 10.
(iv) 100
Solution:
By resolving 100 into prime factor,
100 = 2×2×5×5
Here, 100 cannot be grouped into triplets of equal factors.
∴ 100 is not a perfect cube.
(v) 46656
Solution:
By resolving 46656 into prime factor,
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 46656 is cube of 36.
2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
Solution:
By resolving 243 into prime factor,
243 = 3×3×3×3×3
By grouping the factors in triplets of equal factors, 243 = (3×3×3)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 243 by 3 to get perfect cube.
(ii) 256
Solution:
By resolving 256 into prime factor,
256 = 2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
256 = (2×2×2)×(2×2×2)×2×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will multiply 256 by 2 to get perfect cube.
(iii) 72
Solution:
By resolving 72 into prime factor,
72 = 2×2×2×3×3
By grouping the factors in triplets of equal factors,
72 = (2×2×2)×3×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will multiply 72 by 3 to get perfect cube.
(iv) 675
Solution:
By resolving 675 into prime factor,
675 = 3×3×3×5×5
By grouping the factors in triplets of equal factors,
675 = (3×3×3)×5×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 675 by 5 to get perfect cube.
(v) 100
Solution:
By resolving 100 into prime factor,
100 = 2×2×5×5
Here, 2 and 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 100 by (2×5) 10 to get perfect cube.
3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81
Solution:
By resolving 81 into prime factor,
81 = 3×3×3×3
By grouping the factors in triplets of equal factors, 81 = (3×3×3)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 81 by 3 to get perfect cube.
(ii) 128
Solution:
By resolving 128 into prime factor,
128 = 2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
128 = (2×2×2)×(2×2×2)×2
Here, 2 cannot be grouped into triplets of equal factors.
∴ We will divide 128 by 2 to get perfect cube.
(iii) 135
Solution:
By resolving 135 into prime factor,
135 = 3×3×3×5
By grouping the factors in triplets of equal factors,
135 = (3×3×3)×5
Here, 5 cannot be grouped into triplets of equal factors.
∴ We will divide 135 by 5 to get perfect cube.
(iv) 192
Solution:
By resolving 192 into prime factor,
192 = 2×2×2×2×2×2×3
By grouping the factors in triplets of equal factors,
192 = (2×2×2)×(2×2×2)×3
Here, 3 cannot be grouped into triplets of equal factors.
∴ We will divide 192 by 3 to get perfect cube.
(v) 704
Solution:
By resolving 704 into prime factor,
704 = 2×2×2×2×2×2×11
By grouping the factors in triplets of equal factors,
704 = (2×2×2)×(2×2×2)×11
Here, 11 cannot be grouped into triplets of equal factors.
∴ We will divide 704 by 11 to get perfect cube.
4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution:
Given, side of cube is 5 cm, 2 cm and 5 cm.
∴ Volume of cube = 5×2×5 = 50
50 = 2×5×5
Here, 2 , 5 and 5 cannot be grouped into triplets of equal factors.
∴ We will multiply 50 by (2×2×5) 20 to get perfect cube. Hence, 20 cuboid is needed.
Exercise 7.2
Page: 116
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64
Solution:
64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
64 = (2×2×2)×(2×2×2)
Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.
(ii) 512
Solution:
512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors,
512 = (2×2×2)×(2×2×2)×(2×2×2)
Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.
(iii) 10648
Solution:
10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors,
10648 = (2×2×2)×(11×11×11)
Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.
(iv) 27000
Solution:
27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors,
27000 = (2×2×2)×(3×3×3)×(5×5×5)
Here, 27000 can be grouped into triplets of equal factors,
∴ 27000 = (2×3×5) = 30
Hence, 30 is cube root of 27000.
(v) 15625
Solution:
15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors,
15625 = (5×5×5)×(5×5×5)
Here, 15625 can be grouped into triplets of equal factors,
∴ 15625 = (5×5) = 25
Hence, 25 is cube root of 15625.
(vi) 13824
Solution:
13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 13824 can be grouped into triplets of equal factors,
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is cube root of 13824.
(vii) 110592
Solution:
110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)
Here, 110592 can be grouped into triplets of equal factors,
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is cube root of 110592.
(viii) 46656
Solution:
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)
Here, 46656 can be grouped into triplets of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 36 is cube root of 46656.
(ix) 175616
Solution:
175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)
Here, 175616 can be grouped into triplets of equal factors,
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is cube root of 175616.
(x) 91125
Solution:
91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors,
91125 = (3×3×3)×(3×3×3)×(5×5×5)
Here, 91125 can be grouped into triplets of equal factors,
∴ 91125 = (3×3×5) = 45
Hence, 45 is cube root of 91125.
2. State true or false.
(i) Cube of any odd number is even.
Solution:
False
(ii) A perfect cube does not end with two zeros.
Solution:
True
(iii) If cube of a number ends with 5, then its cube ends with 25.
Solution:
False
(iv) There is no perfect cube which ends with 8.
Solution:
False
(v) The cube of a two digit number may be a three digit number.
Solution:
False
(vi) The cube of a two digit number may have seven or more digits.
Solution:
False
(vii) The cube of a single digit number may be a single digit number.
Solution:
True
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution:
(i) By grouping the digits, we get 1 and 331
We know that, since, the unit digit of cube is 1, the unit digit of cube root is 1.
∴ We get 1 as unit digit of the cube root of 1331.
The cube of 1 matches with the number of second group.
∴ The ten’s digit of our cube root is taken as the unit place of smallest number.
We know that, the unit’s digit of the cube of a number having digit as unit’s place 1 is 1.
∴ ∛1331 = 11
(ii) By grouping the digits, we get 4 and 913
We know that, since, the unit digit of cube is 3, the unit digit of cube root is 7.
∴ we get 7 as unit digit of the cube root of 4913. We know 13 = 1 and 23 = 8 , 1 > 4 > 8.
Thus, 1 is taken as ten digit of cube root.
∴ ∛4913 = 17
(iii) By grouping the digits, we get 12 and 167.
We know that, since, the unit digit of cube is 7, the unit digit of cube root is 3.
∴ 3 is the unit digit of the cube root of 12167 .We know 23 = 8 and 33 = 27 , 8 > 12 > 27.
Thus, 2 is taken as ten digit of cube root.
∴ ∛12167= 23
(iv) By grouping the digits, we get 32 and 768.
We know that, since, the unit digit of cube is 8, the unit digit of cube root is 2.
∴ 2 is the unit digit of the cube root of 32768. We know 33 = 27 and 43 = 64 , 27 > 32 > 64.
Thus, 3 is taken as ten digit of cube root.
∴ ∛32768= 32
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