GATE Civil 2019 Set 2 | GA | Question: 7

Question

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 ________

• A. $\dfrac{x}{y} \\$
• B. $x-y \\$
• C. $\dfrac{100(x-y)}{100+x} \\$
• D. $\dfrac{100(x-y)}{100+y}$

Comments

If anyone got 100(x-y)/y you likely made a mistake when taking increase % we should take old + new rather then increase alone .

Answer 1

$P_{2001}= \frac{a}{b}$ where $a$ is the population of $X$ and $b$ that of $Y.$

$P_{2011}=\frac{a+\frac{x}{100}a}{b+\frac{y}{100}b}=\frac{(1+0.01x)a}{(1+0.01y)b}$

Percentage Increase in $P = \frac{{P_{2011}} – {P_{2001}}}{P_{2001}} \times 100$

$\qquad =\frac{\frac{(1+0.01x)a}{(1+0.01y)b} – \frac{a}{b}}{\frac{a}{b}}\times 100= \left[\frac{1+0.01x}{1+0.01y}-1\right] \times 100 = \frac{x-y}{100+y}\times 100= \frac{100(x-y)}{100+y}$

Correct Answer: D

Answer 2

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) $\times 100$ 

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

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