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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.

$67.5\%$

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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:

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

Radius ratio:

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

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 $75\%$


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