Question

The ratio of surface areas of 1 cube of volume $64c{m}^3$ and 64 cubes of volume $1c{m}^3$ is:

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 4

Answer 1

The correct option is D

1 : 4

Let us name the cube of volume $64c{m}^3$ as cube A and the cube with volume $1c{m}^3$ as cube B.

Length of side of cube A $=\sqrt[3]{364}=4cm$

Surface area of cube A $=6\times (4{)}^2=96c{m}^2$

Length of side of cube B $=1cm$

Surface area of a cube of side 1 cm $=6c{m}^2$

Therefore, surface area of 64 such cubes $=64\times 6=384c{m}^2$

Required ratio of surface areas = 96 : 384

⇒ 1 : 4