# Difficult Percentages Solved QuestionAptitude Discussion

 Q. The volume of the sphere $Q$ is (dfrac{37}{64}%)less than the volume of sphere $P$ and the volume of sphere $R$ is (dfrac{19}{27}%) less than that of sphere $Q$. By what $%$ is the surface area of sphere $R$ less than the surface area of sphere $P$?
 ✖ A. $77.77 \%$ ✖ B. $87.5 \%$ ✔ C. $75\%$ ✖ D. $67.5\%$

Solution:
Option(C) is correct

Let the volume of sphere $P$ be 64 parts.

Therefore volume of sphere $Q$

$=64-\dfrac{37}{64}\%$ of $64$

$= 64−37= 27$ parts.

The volume of $R$

$=27-\dfrac{19}{27}\times 27$

$= 27−19=8$ parts.

Volume ratio:

$=P:Q:R = 64:27:8$

$=P:Q:R=4:3:2$

The surface area will be $16:9:5$

Surface area of $R$ is less than the surface area of sphere $P$

$16k−4k=12k$

Now,

$=\dfrac{12k}{16k}\times 100$

$=75\%$

Thus surface area of sphere $R$ is less than the surface area of sphere P by $75\%$