Free Cubes and Cube Roots 02 Practice Test - 8th Grade 

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.

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 

216 is the cube of 6 as 216 can be expressed as $(6\times 6\times 6)$

729 is the cube of 9 as 729 can be expressed as $(9\times 9\times 9)$

$\sqrt[3]{36\times 6\times 6}$$\times$$\sqrt[3]{39\times 9\times 9}$

= $6\times 9$

So the cube root of the product will be 6 × 9 = 54.

The sum of first 5 natural numbers can be expressed as ${\left(\frac{n(n+1)}{2}\right)}^2$ where n = 5, which gives us
${\left(\frac{(5\times 6)}{2}\right)}^2$ =${\left(\frac{30}{2}\right)}^2$ = ${15}^2$
Thus the cubes of first 5 natural numbers can be expressed as the square of 15.

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 $\frac{162}{27}$ =  6

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

The cube of 3 = 27

The cube of 4 = 64

Since 30 lies between 27 & 64 so its cube root must lie between 3 & 4.
Hence, the given statement is false.

8= 2 x 2 x 2

Cube root of 8 = 2

So 2+  $\frac{1}{2}$ = 2.5

Cube of 0 is 0 (anything multiplied to 0 is 0)

Cube root of 8 is 2 and cube root of -8 is -2. When they are multiplied together, the result is -4.