Model 3 Simplification based on cube & cube root Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 4 EXERCISES
The following question based on simplification topic of quantitative aptitude
(a) 0.005
(b) 0.5
(c) 0.05
(d) 0.15
The correct answers to the above question in:
Answer: (c)
$√^3{0.000125} = √^3{0.05 × 0.05 × 0.05}$
= 0.05
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Read more problems based on cubes and roots Based Quantitative Aptitude Questions and Answers
Question : 1
If cube root of 175616 is 56, then the value of $√^3{175.616} + √^3{0.175616} + √^3{0.000175616}$ is equal to :
a) 6.116
b) 0.168
c) 6.216
d) 62.16
Answer »Answer: (c)
Here, $√^3{175616}$ = 56
$√^3{175.616}$ = 5.6
$√^3{0.175616}$ = 0.56
and $√^3{0.000175616}$ = 0.056
Required sum
= 5.6 + 0.56 + 0.056 = 6.216
Question : 2
$(√^3{1000} + √^3{0.008} - √^3{0.125})$ is equal to
a) 9.9997
b) 9.7
c) 9.997
d) 9.97
Answer »Answer: (b)
$√^3{1000} + √^3{0.008} - √^3{0.125}$
= 10 + 0.2 - 0.5 = 9.7
Question : 3
The least number, that must be added to 1720 so as to obtain a perfect cube, is
a) 13
b) 7
c) 11
d) 8
Answer »Answer: (d)
12 × 12 × 12 = 1728
∴ Required number
= 1728 - 1720 = 8
Question : 4
The square of a natural number subtracted from its cube is 48. The number is :
a) 4
b) 8
c) 5
d) 6
Answer »Answer: (a)
Let number be x
According to question,
$x^3 - x^2$ = 48 ⇒ ∴ x = 4
Question : 5
The sum of the cubes of the numbers 22, –15 and –7 is equal to
a) 0
b) 6930
c) 3
d) 9630
Answer »Answer: (b)
Here, 22 - 15 - 7 = 0
We know that
$a^3 + b^3 + c^3$ = 3abc,
if a + b + c = 0
$(22)^3 + (–15)^3 + (–7)^3$
= 3 × 22 × (–15) (–7) = 6930
Question : 6
Which of the following is a perfect square as well as a cube? 343, 125, 81, or 64
a) 64
b) 81
c) 343
d) 125
Answer »Answer: (a)
343 = 7 × 7 × 7
125 = 5 × 5 × 5
81 = 3 × 3 × 3 × 3
64 = 8 × 8 = 4 × 4 × 4
We see that 343 and 125 are only perfect cubes of 7 and 5 respectively.
81 is only a perfect square of 9. 64 is a perfect square of 8 as well as a perfect cube of 4.
GET simplification PRACTICE TEST EXERCISES
Model 1 Simplification using VBODMAS
Model 2 Simplification based on square & square root
Model 3 Simplification based on cube & cube root
Model 4 Simplification with continued fraction
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