Question

The ratio of surface area of a cube of volume 64 cm${}^3$ to that of 64 cubes of volume 1 cm${}^3$ each = .

• A. 1:1
• B. 1:2
• C. 1:4

Answer 1

The correct option is C 1:4

Let us call the cube of volume 64 cm${}^3$ as cube A and the cube with volume 1 cm${}^3$ as cube B.
Length of side of cube A
$=\sqrt[3]{364}$ = 4 cm
Surface area of cube A
$=6(side{)}^2=6\times 4^2$
= $6\times 16$
Length of side of cube B = 1 cm.
Surface area of a cube of side 1 cm
= $6(side{)}^2=6(1{)}^2=6{\text{cm}}^2$
Therefore, surface area of 64 such cubes
= $64\times 6$
Ratio of required surface areas
= $\frac{6\times 16}{64\times 6}$ .
So, ratio of required surface areas
= 1 : 4