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Q as a percentage of P is equal to P as a percentage of (P + Q). Find Q as a percentage of P.

62%
50%
75%
66%
Explanation:

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

$=100k\%=(100×.62)\%=62\%$

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