$P$ in 2001 $P_{2001}= \frac{a}{b}$ where $a$ is the population of $X$ and $b$ that of $Y.$
$P$ in 2011 $P_{2011}=\frac{(1+0.0x)a}{(1+0.0y)b}$
Percentage Increase in $P = \frac{{P_{2011}} – {P_{2001}}}{P_{2001}} \times 100$
$=\frac{\frac{(1+0.0x)a}{(1+0.0y)b} – \frac{a}{b}}{\frac{a}{b}}= \left[\frac{1+0.0x}{1+0.0y}-1\right] \times 100 = \frac{x-y}{1+0.0y}= \frac{100(x-y)}{100+y}$
Correct Answer: D