Percentages
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Q.

The marks scored in History by $P,Q,R$ and $S$ form a geometric progression in that order. If the marks scored by $R$ were (dfrac{275}{9}%)less than the sum of the marks scored by $P$ and $Q$, then marks scored by $S$ were what percent more than the marks scored by $Q$? (Assume that everyone scored positive marks).

 A.

$56.25 \%$

 B.

$55.55 \%$

 C.

$64 \%$

 D.

$67.75 \%$

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Solution:
Option(A) is correct

Let the marks scored by $P,Q,R$ and $S$ be $a, ak, ak^2$ and $ak^3$ respectively.

Marks scored by $R$ were \(\dfrac{275}{9}\%\)  less than the marks scored by $P$ and $Q$.

So, if marks scored by $P$ and $Q$ were 36 then those scored by $R$ is 25.

Marks scored by $P$ and $Q =a+ak=36$

Marks scored by $R =ak^2=25$

\(\Rightarrow \dfrac{(a+ak)}{a k^2}=\dfrac{36}{25}\)

\(\Rightarrow k=\dfrac{5}{4}\)

\(P=a,Q=\dfrac{5a}{4}, R=\dfrac{25a}{16}, S=\dfrac{125a}{64}\)

$\Rightarrow P=64a$, $Q=80a$, $R=100a$ and $S=125a$

$S$ scored \(\dfrac{125a-80}{80a}\times 100\%\)more than $Q$

\(\Rightarrow \dfrac{45}{80}\times 100\)

\(=56.25\%\)

$S$ scored \(56.25 \% \)  more than $Q$.


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