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. The value of 'k' is

• A. $0.64$
• B. $0.512$
• C. $51.2$
• D. $64$

Answer 1

The correct option is C $51.2$

Total surface area of a cube of edge a $=6a^2$
Let side of cube A be A and side of cube B be B.
Given $6A^2=\frac{64}{100}\times 6B^2$
$=>A=\frac{8}{10}B$
Volume of a cube $={side}^3$
Hence, Volume of cube A $ = {A}^{3} ={(\dfrac{8}{10} B)}^{3} = \dfrac{512}{1000}{B}^{3} $
Volume of cube A $=B^3$
Given, $\frac{512}{1000}{B}^3=$ k % $\times B^3$
=> k % $=\frac{512}{1000}$
$=>k=\frac{512}{1000}\times 100=51.2$