Dark Mode

26 votes

Best answer

In 2008 $\dfrac{M}{F}$ ratio is $2.5$

Assume $250$ Males, $100$ Females.

In 2009 $\dfrac{M}{F}$ ratio is $3.$ Also total no of females doubled

Females $=100\times 2 = 200.$

So, $\dfrac{M}{F} = 3 \implies \dfrac{M}{200}= 3$

$\implies M = 200\times 3 = 600.$

Increase in Male Students $= 600 - 250 = 350$

Increase $=\left(\dfrac{350}{250}\right)\times 100 \%= 140 \%$

Assume $250$ Males, $100$ Females.

In 2009 $\dfrac{M}{F}$ ratio is $3.$ Also total no of females doubled

Females $=100\times 2 = 200.$

So, $\dfrac{M}{F} = 3 \implies \dfrac{M}{200}= 3$

$\implies M = 200\times 3 = 600.$

Increase in Male Students $= 600 - 250 = 350$

Increase $=\left(\dfrac{350}{250}\right)\times 100 \%= 140 \%$

0

$Year$ $2008$ :- Let Male$=$$M_{1}$ and Female$=$$F_{1}$

Given $\frac{M_{1}}{F_{1}}=2.5$ -------->>>Equation 1

$Year$ $2009$ :- Let Male$=$$M_{2}$ and Female$=$$F_{2}$

Given $\frac{M_{2}}{F_{2}}=3$ ---------->>>Equation 2

Given $F_{2}=2*F_{1}$ ---------->>>Equation 3

From Equation 1 : $M_{1}=2.5F_{1}$

From Equation 2 and 3 : $M_{2}=3F_{2}=6F_{1}$

So % increase in number of males=$(\frac{New value-Old value}{old value})*100$

$=(\frac{6F_{1}-2.5F_{1}}{2.5F_{1}})*100=140$%

11