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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 ________

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

+1 vote

Best answer

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

$P$ in 2011 $P_{2011}=\frac{a+\frac{x}{100}a}{b+\frac{y}{100}b}=\frac{(1+0.0x)a}{(1+0.0y)b}$

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

$P$ in 2011 $P_{2011}=\frac{a+\frac{x}{100}a}{b+\frac{y}{100}b}=\frac{(1+0.0x)a}{(1+0.0y)b}$

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

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