Question

If the surface area of cube A is 64% of the surface area of cube B, then the volume of cube A is ‘k’ percent of the volume of cube B. Then value of ‘k’ is
(a) 0.64
(b) 0.512
(c) 51.2
(d) 64

Answer 1

Hint:In this question, the ratio of the surface areas of the two cubes is given in terms of percentage and we are asked to find out the ratio of their volumes in terms of percentage. Therefore, we should use the definitions to calculate the surface area and volumes of a cube and use the given information to find the answer to this question.

Complete step-by-step answer:
We know that the formula for the surface area of a cube of side length a is given by
$\text{Surface Area=6}{{a}^{2}}...................(1.1)$
Let the side lengths of the cube A be $a$ and the surface are of the cube B be $b$ units. It is given that
$\dfrac{\text{Surface Area of A}}{\text{Surface Area of B}}=64%=\dfrac{64}{100}..........(1.2)$
Therefore, using equations (1.1) and (1.2), we can write
$\begin{align}
  & \dfrac{6{{a}^{2}}}{6{{b}^{2}}}=\dfrac{64}{100}\Rightarrow 100{{a}^{2}}=64{{b}^{2}} \\
 & \Rightarrow {{\left( 10a \right)}^{2}}={{\left( 8b \right)}^{2}} \\
 & \Rightarrow 10a=8b\text{ (negative value is discarded as side length cannot be negative)} \\
 & \Rightarrow \dfrac{a}{b}\text{=}\dfrac{8}{10}................\text{(1}\text{.3)} \\
\end{align}$
Now, the formula for the volume of a cube of side length a is given by
$\text{Volume=}{{a}^{3}}...................(1.4)$
Therefore, from equations (1.3) and (1.4), we obtain
$\dfrac{\text{Volume of A}}{\text{Volume of B}}=\dfrac{{{a}^{3}}}{{{b}^{3}}}={{\left( \dfrac{a}{b} \right)}^{3}}={{\left( \dfrac{8}{10} \right)}^{3}}=0.512=\dfrac{51.2}{100}=51.2%$
Which matches the value in option (c) . Hence, option (c) is the correct answer to this question.

Note: We should note that in equation (1.3), the equation ${{\left( 10a \right)}^{2}}={{\left( 8b \right)}^{2}}$ can be satisfied for both $10a=8b$ and $10a=-8b$. However, as the length of a cube cannot have negative value we should only consider the solution for which the side lengths are positive and hence the second solution should be discarded.Students should remember the formulas of surface area and volume of the cube for solving these types of questions.