Class 8 SELINA Solutions Maths Chapter 4: Cubes and Cube Roots
Cubes and Cube Roots Exercise Ex. 4(A)
Solution 1(i)
Correct option: (b) 0.125
Solution 1(ii)
Correct option: (c) –64
(–4)3 = –4 × –4 × –4 = –(4 × 4 × 4) = –64
Solution 1(iii)
Correct option: (a) 3
72 = 2 × 2 × 2 × 3 × 3
Here, number of 3s is 2. It is not forming a triplet.
Hence, 72 must be multiplied by 3 to obtain a perfect cube.
Solution 1(iv)
Correct option: (b) 3
81 = 3 × 3 × 3 × 3
Here, number of 3s is 4.
So, 81 must be divided by 3 to obtain a perfect cube.
Note: Back answer is incorrect
Solution 2(i)
(7)3 = 7 × 7 × 7 = 343
Solution 2(ii)
(11)3 = 11 × 11 × 11 = 1331
Solution 2(iii)
(16)3 = 16 × 16 × 16 = 4096
Solution 2(iv)
(23)3 = 23 × 23 × 23 = 12167
Solution 2(v)
(31)3 = 31 × 31 × 31 = 29791
Solution 2(vi)
(42)3 = 42 × 42 × 42 = 74088
Solution 2(vii)
(54)3 = 54 × 54 × 54 = 157464
Solution 3(i)
243 = (3 × 3 × 3) × 3 × 3
Since triplet of number 3 is not formed.
Therefore, 243 is not a perfect cube.
Solution 3(ii)
588 = 2 × 2 × 3 × 7 × 7
Here, none of 2, 3 and 7 have formed triplets.
Therefore, 588 is not a perfect cube.
Solution 3(iii)
1331 = 11 × 11 × 11
Number 11 is forming a triplet and no number is left over.
Therefore, 1331 is a perfect cube.
Solution 3(iv)
24000 = (2 × 2 × 2) × (2 × 2 × 2) × 3 × (5 × 5 × 5)
Since triplet of number 3 is not formed.
Therefore, 24000 is not a perfect cube.
Solution 3(v)
1728 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3)
Since all numbers are forming triplets.
Therefore, 1728 is a perfect cube.
Solution 3(vi)
1938 = 2 × 3 × 17 × 19
None of the numbers are forming triplets.
Therefore, 1938 is not a perfect cube.
Solution 4(i)
Solution 4(ii)
Solution 4(iii)
Solution 4(iv)
Solution 4(v)
Solution 4(vi)
Solution 4(vii)
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 5(iv)
Solution 5(v)
Solution 6(i)
(–3)3 = –3 × –3 × –3 = –(3 × 3 × 3) = –27
Solution 6(ii)
(–7)3 = –7 × –7 × –7 = –(7 × 7 × 7) = –343
Solution 6(iii)
(–12)3 = –12 × –12 × –12 = –(12 × 12 × 12) = –1728
Solution 6(iv)
(–18)3 = –18 × –18 × –18 = –(18 × 18 × 18) = –5832
Solution 6(v)
(–25)3 = –25 × –25 × –25 = –(25 × 25 × 25) = –15625
Solution 6(vi)
(–30)3 = –30 × –30 × –30 = –(30 × 30 × 30) = –27000
Solution 6(vii)
(–50)3 = –50 × –50 × –50 = –(50 × 50 × 50) = –125000
Solution 7
Cubes of even natural numbers are even, and those of odd natural numbers are odd.
Therefore, the cubes of
(i) 216, 8000, 4096 are even
(ii) 729, 3375, 125, 343, 92561 are odd
Solution 8
1323 = (3 × 3 × 3) × 7 × 7
Here, the number 7 is not forming a triplet.
To form a triplet, one more 7 is required.
Hence, 1323 must be multiplied by 7 to obtain a perfect cube.
Solution 9
8768 = (2 × 2 × 2) × (2 × 2 × 2) × 137
Here, the number 137 is not forming a triplet.
Hence, 8768 must be divided by 137 to obtain a perfect cube.
Solution 10
27783 = 3 × (3 × 3 × 3) × (7 × 7 × 7)
Here, the number of 3s is 4. They form a triplet, and one 3 is left.
To form a triplet, two more 3s are required.
Hence, 27783 must be multiplied by 3 × 3 = 9 to obtain a perfect cube.
Solution 11
8640 = (2 × 2 × 2) × (2 × 2 × 2) × (3 × 3 × 3) × 5
Here, the number 5 is not forming a triplet.
Hence, 8640 must be divided by 5 to obtain a perfect cube.
Solution 12
77175 = 3 × 3 × 5 × 5 × (7 × 7 × 7)
Here, the number of 3s as well as 5s is 2. They are not forming a triplet.
Hence, 77175 must be multiplied by 3 × 5 = 15 to obtain a perfect cube.
Cubes and Cube Roots Exercise Ex. 4(B)
Solution 1(i)
Correct option: (a) 0.03
Solution 1(ii)
Correct option: (d) –0.4
Solution 2(i)
Solution 2(ii)
Solution 2(iii)
Solution 2(iv)
Solution 2(v)
Solution 2(vi)
Solution 2(vii)
Solution 2(viii)
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 3(iv)
Solution 3(v)
Solution 3(vi)
Solution 3(vii)
Solution 4(i)
Solution 4(ii)
Solution 4(iii)
Solution 4(iv)
Solution 4(v)
Solution 4(vi)
Solution 4(vii)
Solution 4(viii)
Solution 4(ix)
Solution 5(i)
Solution 5(ii)
Solution 5(iii)
Solution 5(iv)
Solution 5(v)
Solution 5(vi)
Solution 6
26224 = 2 × 2 × (3 × 3 × 3) × (3 × 3 × 3) × 3 × 3
Hence, the required smallest number = 2 × 2 × 3 × 3 = 36
Solution 7
30375 = 3 × 3 × (3 × 3 × 3) × (5 × 5 × 5)
Hence, the required smallest number to be multiplied is 3.
Solution 8(i)
Solution 8(ii)
Solution 8(iii)
Solution 8(iv)
Solution 8(v)
Solution 8(vi)
Solution 8(vii)
Cubes and Cube Roots Exercise TEST YOURSELF
Solution 1(i)
Correct option: (d) 56x3
Solution 1(ii)
Correct option: (a) 9
32 = 9
Hence, if a number is multiplied by 3, it square will be multiplied by 9.
Solution 1(iii)
Correct option: (a) 5 and 4
Let the two numbers be 5a and 4a respectively.
Then, (5a)3 – (4a)3 = 61
⇒ 125a3 – 64a3 = 61
⇒ 61a3 = 61
⇒ a3 = 1
⇒ a = 1
⇒ 5a = 5 and 4a = 4
Solution 1(iv)
Correct option: (c) 3.6
Solution 1(v)
Correct option: (b) –6
Solution 2
(i) False – Cube of an odd number is odd.
(ii) True
(iii) False – 152 = 225 and 153 = 3375
Note: Back answer is incorrect
(iv) False – The cube of the smallest two-digit number 10 is 1000, that is. 4 digits.
(v) True
Solution 3(i)
Solution 3(ii)
Note: Back answer incorrect
Solution 4
Let the edge of the cube = a cm
Surface area of a cube = 486 cm2
⇒ 6a2 = 486
⇒ a2 = 81
⇒ a = 9
Volume of a cube = a3 = 93 = 729 cm3
Solution 5(i)
Solution 5(ii)
Solution 6
Let the three numbers be 2a, 3a and a respectively.
Then, (2a)3 + (3a)3 + a3 = 288
⇒ 8a3 + 27a3 + a3 = 288
⇒ 36a3 = 288
⇒ a3 = 8
⇒ a = 2
⇒ 2a = 2(2) = 4 and 3a = 3(2) = 6
Hence, the numbers are 4, 6 and 2 respectively.
Solution 7
14,580 = 2 × 2 × (3 × 3 × 3) × (3 × 3 × 3) × 5
Hence, the required smallest number to be multiplied = 2 × 5 × 5 = 50
And,
Solution 8
8232 = (2 × 2 × 2) × 3 × (7 × 7 × 7)
Hence, the required smallest number to be divided = 3
And,
Solution 9(i)
Solution 9(ii)
Solution 10
Let the greater number = a
Then, the smaller number = a – 387
⇒ Smaller number = 512 – 387 = 125
