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

  1. $\dfrac{x}{y} \\$
  2. $x-y \\$
  3. $\dfrac{100(x-y)}{100+x} \\$
  4. $\dfrac{100(x-y)}{100+y}$
in Quantitative Aptitude retagged by
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Answer: OPTION D

Let ratio of population X to population Y (Pold) = $\frac{Px}{Py}$ = P

New Population of X = $Px+ \frac{(x*Px)}{100}$  = $(1+\frac{x}{100})Px$

New Population of Y = $Py+ \frac{(y*Py)}{100}$  = $(1+\frac{y}{100})Py$

Ratio of new population of x and y (Pnew) =$\frac{(100+x)Px }{(100+y)Py }$  = $\frac{(100+x)P}{(100+y)}$

Percentage increase in P = (Pnew -Pold)/Pold  = $\frac{ \frac{(100+x)P }{(100+y)} -P }{P}$ $\times 100$   

                                                                             =($\frac{(100+x) }{(100+y)}$ -1) $\times 100$ 

                                                                             = $\frac{x-y}{100+y} \times 100$


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