Question

The numerical value of the volume of cube B is equal to the length of the side of cube A. The ratio of the volume to the surface area of cube A is ‘k’. If the length of side of cube B, and ‘k’, are integers, what could be the minimum value of ‘k’?

• A. 16
• B. 6
• C. 36
• D. 9

Answer 1

The correct option is C 36

Let 'a' and 'b' the sides of the cubes A and B respectively.

Given: Volume of cube B = side of cube A
$\Rightarrow b^3=a$ or $a=b^3$

Volume of cube A = $a^3$
= $(b^3{)}^3$
= $b^9$ cubic units.

Surface area of cube A = $6a^2$
= $6(b^3{)}^2$
= $6b^6$ sq. units

Given: $\frac{VolumeofcubeA}{SurfaceareaofcubeA}=k$
$\Rightarrow k=\frac{b^9}{6b^6}$
$\Rightarrow k=\frac{b^3}{6}$

Now we need to find the values of k and b such that they are integers.
'b' cannot be negative or zero because it is a side of a cube.
Subsituting b = 1, 2, 3, 4, 5, 6 in the expression $k=\frac{b^3}{6}$, it can be observed that the least integral value of k is obtained when b = 6.
$k=\frac{6^3}{6}$
$k=36$