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Population of state $\text{X}$ increased by $x\%$  and the population of state $\text{Y}$ increased by $y\%$ from $2001$ to $2011.$ Assume that $\text{X}$ is greater than $\text{Y}$. Let $\text{P}$ be the ratio of the population of state $\text{X}$ to state $\text{Y}$ in a given year. The percentage increase in $\text{P}$ from $2001$ to $2011$ is ________

  1. $\frac{x}{y}$
  2. $x-y$
  3. $\frac{100(x-y)}{100+x}$
  4. $\frac{100(x-y)}{100+y}$
in Quantitative Aptitude edited by
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Let’s assume initial population of state X is X’

Initial population of state Y is Y’

Initial ratio P1 = X’/Y’

New population of state X is : X’ (1 + x/100)

New population of state X=Y is : Y’ (1 + y/100)

New ratio P2 = [X’(1+x/100)] / [Y’(1+y/100)]

 

Change in ratio = [ P2 – P1 ] / P1  =  [ (1 + x/100) / (1 + y/100) ] – 1

Solving this will give , (x-y) / (100+y)

Percentage increased , 100 * (x-y)  / (100+y)
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