Q as a percentage of P is equal to P as a percentage of (P + Q). Find Q as a percentage of P.
Given that $\dfrac{Q}{P}=\dfrac{P}{P+Q}\cdots(1)$
Since Q can be written as a certain percentage of P, we can assume that $Q=kP$
Hence $(1)$ becomes
$\dfrac{kP}{P}=\dfrac{P}{P + kP}$
$\Rightarrow k=\dfrac{1}{1+k}\\\Rightarrow k(k+1)=1~~\cdots(2)$
Q as a percentage of P
$=\dfrac{Q}{P}×100\\=\dfrac{kP}{P}×100=100k\%~~\cdots(3)$
From here we have two approaches.
Approach 1 - trial and error method
Here we use the values given in the choices to find out the answer.
Take 50% from the given choice.
If 50% is the answer,
$100k=50 \Rightarrow k=\dfrac{50}{100}=\dfrac{1}{2}$
But if we substitute $k=\dfrac{1}{2}$ in $(2),$
$k(k+1)=\dfrac{1}{2}\left(\dfrac{1}{2}+ 1\right)$ $=\dfrac{1}{2}×\dfrac{3}{2}=\dfrac{3}{4} \neq 1$
Now Take another choice, say 62%
If 62% is the answer,
$100k=62 \Rightarrow k=\dfrac{62}{100}$
If we substitute $k=\dfrac{62}{100}$ in $(2),$
$k(k+1)=\dfrac{62}{100}\left(\dfrac{62}{100}+ 1\right)$ $= \dfrac{62}{100}×\dfrac{162}{100}=\dfrac{10044}{10000}\approx 1$
Hence 62% is the answer.
Approach 2
lets solve the quadratic equation $(2)$
$k(k+1)=1\\k^2+k-1=0\\k=\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}=\dfrac{-1 \pm \sqrt{1^2-\left[4×1×(-1)\right]}}{2×1}=\dfrac{-1 \pm \sqrt{5}}{2}\\= \dfrac{-1 \pm 2.24}{2}\\ = \dfrac{1.24}{2}\text{ or }\dfrac{-3.24}{2}\\= 0.62$
avoiding -ve value
From $(3)$, Q as a percentage of P
$=100k\%=(100×.62)\%=62\%$