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 $=6\times (4{)}^2=6\times 16=96c{m}^2$

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 $=96:384$

$=1:4$