8th Grade > Mathematics
CUBES AND CUBE ROOTS MCQs
:
B
Number of chocolates Amrit has after going to shop A = 3
Number of chocolates Amrit has after going to shop B = 3 x 3 + 3 = 12
Number of chocolates Amrit has after shop B and eating 3 = 12 - 3 = 9
Number of chocolates Amrit has after going to shop C = 3 x 9 + 9 = 36
Hence, the number of chocolates Amrit has at the end is 36.
:
D
The sum of first 5 natural numbers can be expressed as (n(n+1)2)2 where n = 5, which gives us
((5×6)2)2 =(302)2 = 152
Thus the cubes of first 5 natural numbers can be expressed as the square of 15.
:
B
8= 2 x 2 x 2
Cube root of 8 = 2
So 2+ 12= 2.5
:
C
Cube of 1 = 1
Cube of 2 = 8
Cube of 3 = 27
Cube of 4 = 64
So total 4 numbers have their cubes that lie between 0 and 100
:
A
216 is the cube of 6 as 216 can be expressed as (6×6×6)
729 is the cube of 9 as 729 can be expressedas (9×9×9)
3√6×6×6×3√9×9×9
= 6×9
So the cube root of the product will be 6 × 9 = 54.
:
B
Volume of cuboid = 3×6×9 =162 cubic units
Volume of cube = 3×3×3= 27 cubic units
Number of cubes that can be fit in the cuboid is 16227= 6
Amrit's teacher told him "Go to shop A and buy 3 chocolates. Then go to shop B and buy thrice the number of chocolates you have and eat 3 chocolates. Then go to shop C and again buy thrice the number of chocolates you have." What will be the number of chocolates left with Amrit at the end?
:
B
Number of chocolates Amrit has after going to shop A = 3
Number of chocolates Amrit has after going to shop B = 3 x 3 + 3 = 12
Number of chocolates Amrit has after shop B and eating 3 = 12 - 3 = 9
Number of chocolates Amrit has after going to shop C = 3 x 9 + 9 = 36
Hence, the number of chocolates Amrit has at the end is 36.
:
A
The digit in the units place in 298 is 8.
cube of 8 is 512, which ends in 2.
Hence, the digit in the units place in the cube of 298 is 2
:
A
216 is the cube of 6 as 216 can be expressed as (6×6×6)
729 is the cube of 9 as 729 can be expressed as (9×9×9)
3√6×6×6×3√9×9×9
= 6×9
So the cube root of the product will be 6 × 9 = 54.