Question

At a function, the ratio of the number of men to the number of women was $3:4$. When another $5$ men and $5$ women joining the function, the ratio become $7:9$. How many men were there at the function at first?

Answer 1

Let the ration of men and women $=x:y$

Acceding to given question.

$\frac{x}{y}=\frac{3}{4}$

$4x=3y$

$x=\frac{3y}{4}-----------\left(1\right)$

Now,

$5$ men and $5$ women joining the function.

So,

$\frac{x+5}{y+5}=\frac{7}{9}$

$9x+45=7y+35$

$9x-7y=35-45$

$9x-7y=10$

$9x-7y+10=0-----------\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ to

$9\times \frac{3y}{4}-7y+10=0$

$27y-28y+40=0$

$-y+40=0$

$-y=-40$

$y=40$

Women $=40$

Put the value of $y$ in equation $\left(1\right)$ and we get

$x=\frac{3y}{4}$

$x=\frac{3\times 40}{4}$

$x=3\times 10$

$x=30$

Men $=30$

Hence, $30$ men were there at the function at first.