Aptitude Discussion

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. |
$57\%$ |

**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,

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

\(=0.62=62\%\)

**Kotresh**

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