Practice Problems based on cubes and roots - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The sum of the cubes of the numbers 22, –15 and –7 is equal to
(a)
(b)
(c)
(d)
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
Q-2) The square of a natural number subtracted from its cube is 48. The number is :
(a)
(b)
(c)
(d)
Let number be x
According to question,
$x^3 - x^2$ = 48 ⇒ ∴ x = 4
Q-3) Sum of digits of the smallest number by which 1440 be multiplied so that it becomes a perfect cube, is
(a)
(b)
(c)
(d)
1440 = $2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 = 2^3 × 2^2 × 3^2 × 5$
To make 1440 a perfect cube, it must be mulitplied by 2 × 3 × 5 × 5 = 150.
∴The required sum = 1+5 + 0 = 6
Q-4) The smallest positive integer n, for which 864n is a perfect cube, is :
(a)
(b)
(c)
(d)
| 2 | 864 |
| 2 | 432 |
| 2 | 216 |
| 2 | 108 |
| 2 | 54 |
| 3 | 27 |
| 3 | 9 |
| 3 |
For 864n to be a perfect cube, n = 2
Q-5) If $√^3{3^n}$ = 27, then the value of n is :
(a)
(b)
(c)
(d)
$√^3{3^n}$ = 27
$(3)^{n/3} = 3^3$
$n/3$ = 3 ⇒ n = 3 × 3 = 9
Q-6) The smallest natural number, by which 3000 must be divided to make the quotient a perfect cube, is :
(a)
(b)
(c)
(d)
3000 = 3 × 1000 = 3 × $10^3$
Clearly, when we divide 3000 by natural number 3, the quotient is 1000 which is a perfect cube.
Q-7) By which smallest number 1323 must be multiplied, so that it becomes a perfect cube?
(a)
(b)
(c)
(d)
1323 = 3 × 3 × 3 × 7 × 7
It must be multiplied by 7.
Q-8) The sum of the digits of the smallest number which, when multiplied by 1800, gives a perfect cube, is :
(a)
(b)
(c)
(d)
1800 = 2 × 2 × 2 × 3 × 3 × 5 × 5 = $2^3 × 3^2 × 5^2$
To make 1800 a perfect cube, it must be multiplied by 15 (least number).
>∴ Required sum = 1 + 5 = 6
Q-9) $√^3{(333)^3 + (333)^3 + (334)^3 - 3 × 333 × 333 × 334}$ is equal to
(a)
(b)
(c)
(d)
We know that
$a^3 + b^3 + c^3$ - 3abc
= (a+b+c)$(a^2+b^2+c^2$ - ab - bc - ca)
= $1/2 (a + b + c) [(a - b)^2 + (b - c)^2 + (c - a)^2 ]$
∴ $√^3{(333)^3 + (333)^3 + (334)^3 - 3 × 333 × 333 × 334}$
= $√^3{1/2 (333 + 333 + 334)[(333 - 333)^2 + (333 - 334)^2 + (334 - 333)^2]}$
= $√^3{1/2 × 1000 × 2} = √^3{1000}$
= $√^3{10 × 10 × 10}$ = 10
Q-10) The least number, by which 1944 must be multiplied so as to make the result a perfect cube, is
(a)
(b)
(c)
(d)
| 2 | 1944 |
| 2 | 972 |
| 2 | 486 |
| 3 | 243 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 |
1944 = 2×2×2×3×3×3×3×3 = $2^3 × 3^3 × 3^2$
Clearly, 1944 should be multiplied by 3 to make the result a perfect cube.