Question

Last year , there were 30 male members and 20 female members in a Chess Club . This year , the number of male members is decreased by r% while the number of female members is increased by r% . If the number of male members is equal to that of female members this year , find the value of r .
Can anyone explain to me ?
Thks \（>∀<）/

Answer 1

The r% of a quantity x is computed by dividing x in 100 parts, and considering r of such parts. So, the r% of the male is

$30\times\cfrac{r}{100} = 0.3 r$

and similarly, the r% of female is

$20\times\cfrac{r}{100} = 0.2 r$

The number of males decreased by this quantity, so now it is

$30 - 0.3r$

and the number of female increased by this quantity, so now it is

$20+0.2r$

we know that these two new counts are the same number, so we can build and solve the equality

$30 - 0.3r = 20+0.2r$

Subtract 20 and add 0.3r from both sides:

$10 = 0.5r$

Divide both sides by 0.5 to solve for r:

$r = 20$

Let's check the answer

The 20% of 30 is $30 \times \frac{20}{100} = 6$, while the 20% of 20 is 4. So, we are stating that $30-6 = 20+4$ which is true because both expressions evaluate to 24.