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 surface area of cube A is 'k'. If the length of side of cube B, and 'k', is an integer, what could
be the minimum value of 'k'?​

Answer 1

Answer:

it will easy because the minimum value of k is b and the minimum value of a

Answer 2

The minimum value of k is 36

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

Given :

Volume of cube B = side of cube A

⇒  $b^3$ = a  or  a = $b^3$

Volume of cube A = $a^3$

=  $(b^3)^3$

= $b^9$ cubic units.

Surface area of cube A =  $6a^2$

=  $6(b^3)^2$

= $6 b^6$ sq. units

we know that,

$\frac{Volume of cube A}{Volume of cube B}$ = k

⇒  k = $\frac{b^9}{6b^6}$

⇒  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.

Substituting 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.

$k=\frac{6^3}{6}$

$k=36$

Therefore value of k is 36

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