Percentages
Aptitude

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Q.

$B$ as a percentage of $A$ is equal to $A$ as a percentage of $(A + B)$. Find $B$ as a percentage of $A$.

 A.

$62\%$

 B.

$73\%$

 C.

$41\%$

 D.

$57\%$

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Solution:
Option(A) is correct

From the question stem, we know

\(\dfrac{B}{A}=\dfrac{A}{A+B}\)

As $B$ is a percent of $A$, let us assume $B=Ax$

Then, equation (1) can be re-written as

\(x=\dfrac{1}{1+x}\)

\(\Rightarrow x(1+x)=1\)

After this step, trial and error works best. The solution is $62\%$, answer choice ($A$) is the correct one.

Alternatively, solving, we get

\(x^2+x+1=0\)

\(\Rightarrow x=\dfrac{-1+\sqrt{5}}{2}\)

or

\(\Rightarrow x=\dfrac{-1-\sqrt{5}}{2}\)

So, 

\(\Rightarrow x=\dfrac{-1+\sqrt{5}}{2}\)

\(=0.62=62\%\)


(1) Comment(s)


Kotresh
 ()

If the percentage are different then how to solve such type of questions?