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

1: 1

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1: 2

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1: 3

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1: 4

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Solution

The correct option is D ( $1 : 4$ )
Let us name the cube of volume $64 c m^{3}$ as cube A and the cube with volume $1 c m^{3}$ as cube B.

Length of side of cube A $= \sqrt[3]{64} = 4 c m$

Surface area of cube A $= 6 \times (4)^{2} = 6 \times 16 = 96 c m^{2}$

Length of side of cube B $= 1 c m$

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

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

Required ratio of surface areas $= 96 : 384$

$= 1 : 4$

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