R.S Aggarwal Class 8 Chapter 4 Cube and Cube Roots Test paper Solution
The 4th Chapter of RS Aggarwal book of Class 8 is about Cube and Cube Roots and in this post we have provided complete written solution for the test paper 4 . The solutions are prepared by our team of expert teachers.
Complete RS Aggarwal solution for class 8 is also available in our website.
Test Paper 4
A) 1) Evaluate [1(2/5)]3
Ans:
[1(2/5)]3
= [7/5]3
= 73/53
= (7×7×7) / (5×5×5)
= 343/125
2) Evaluate ∛4096
Ans: By prime factorization of 4096 we get
4096 = 2×2×2×2×2×2×2×2×2×2×2×2
= (2×2×2) × (2×2×2) × (2×2×2) × (2×2×2)
∛4096 = ∛[(2×2×2) × (2×2×2) × (2×2×2) × (2×2×2)]
= 2 × 2 × 2 × 2 = 16
3) Evaluate ∛ (216×343)
Ans: ∛ (216×343)
= ∛216 × ∛343
= ∛ (6×6×6) ×∛ (7×7×7)
= 6 × 7
= 42
4) Evaluate ∛ (-64/125)
Ans : ∛ (–64/125)
= ∛ (–64) / ∛ (125)
= – ∛ (64) / ∛ (125)
= – ∛ (4×4×4) / ∛ (5×5×5)
= – 4/ 5
B) Mark (✓) against the correct answer in each of the following:
5) [1(3/4)]3= ?
a) 1(27/64)
b) 2(27/64)
c) 5(23/64) (✓)
d) none of them
Ans: [1(3/4)]3
= (7/4)3
= 73/43
= (7×7×7) / (4×4×4)
= 343/64
= 5(23/64)
6) Which of the following numbers is a perfect cube?
a) 121
b) 169
c) 196
d) 216 (✓)
Ans:
a) By prime factorizing 121 we get
121 = 11×11
The factors do not form triplets
So we can say that 121 is not a perfect cube.
b) By prime factorizing 169 we get
169 = 13×13
The factors do not form triplets
So we can say that 169 is not a perfect cube.
c) By prime factorizing 196 we get
196 = 2×2×7×7 = (2×2) × (7×7)
The factors do not form triplets
So we can say that 196 is not a perfect cube.
d) By prime factorizing 216 we get
216 = (2×2×2) × (3×3×3)
The factors 2 and 3 form triplets
∴∛216 = ∛ [(2×2×2) × (3×3×3)] = 2×3 = 6
So 216 is a perfect cube of 6.
7) ∛ (216×64) =?
a) 64
b) 32
c) 24 (✓)
d) 36
Ans: ∛ (216×64)
= ∛216 × ∛64
= ∛(6×6×6) × ∛(4×4×4)
= 6 × 4
= 24
8) ∛ (-343/729) =?
a) 7/9
b) -7/9 (✓)
c) -9/7
d) 9/7
Ans: ∛ (-343/729)
= ∛ (-343)/ ∛729
= – ∛(343) / ∛ (729)
= – ∛(7×7×7)/ ∛(9×9×9)
= –7/9
9) By what least number should 324 be multiplied to get a perfect cube?
a) 12
b) 14
c) 16
d) 18 (✓)
Ans: By prime factorizing 324 we get
324 = 2×2×3×3×3×3 = (2×2) × (3×3×3) × (3)
Form this we can see that 324 is not a perfect cube
To make 324 a perfect cube we have to multiply with (2×3×3) = 18
∴324×18 = (2×2) × (3×3×3) × (3) × 18
= (2×2) × (3×3×3) × (3) × (2×3×3) = (2×2×2) × (3×3×3) × (3×3×3)
∛(324×18) = ∛[(2×2×2) × (3×3×3) × (3×3×3)] = 2×3×3= 18
So we can see that the factors of the newly formed number form triplets. So the minimum that has to be multiplied with 324 to get a perfect cube is 18.
10) ∛128/∛250
a) 3/5
b) 4/5 (✓)
c) 2/5
d) none of these
Ans: ∛128/∛250
= ∛ (2×2×2×2×2×2×2) / ∛ (2×5×5×5)
= ∛ [(2×2×2) × (2×2×2) × 2)] / ∛ [2×(5×5×5)]
= 4∛2/5∛2
= 4/5
11) Which of the following is a cube of an odd number?
a) 216
b) 512
c) 343 (✓)
d) 1000
Ans:
a) ∛216 = ∛(6×6×6) = 6
b) ∛512 = ∛ (8×8×8) = 8
c) ∛343 = ∛(7×7×7) = 7
d) ∛1000 = ∛(10×10×10) = 10
From the above we can observe that 343 is a cube of 7 which is an odd number.
C) 12) Fill in the blanks.
i) ∛ab = ∛a × (_∛b_)
ii) ∛(a/b) = (_∛a/∛b_)
iii) ∛(-x) = (_– ∛x_)
iv) (0.5)3 = (_125_)