# Question 4099b

Oct 6, 2016

$132 , 000$

#### Explanation:

Let us assume that the population in 1999 is $x$

So x + 8% "of " x =142,560

$x + \frac{8}{100} x = 142 , 560$

$\frac{108}{100} x = 142 , 560$

$x = 14 , 2560 \times \frac{100}{108}$

$x = 132 , 000$

Oct 6, 2016

$132 , 000$ in 1999

#### Explanation:

This is an example of a reverse percentage where we know the value AFTER an increase has happened, but we do not know what the original amount was.

Do NOT be tempted to just find 8% of the population in 2000 and subtract!

Here is a different approach from how Annie did it.

Percentages are always in direct proportion.
Let's compare what we know.

Let the population in 1999 be $x$
We can consider that as 100%

In 2000, the population had increased by 8%
The population of 142,560 represents 108%#

$\frac{x}{100} = \frac{142 , 560}{108}$

$x = \frac{142 , 560 \times 100}{108}$

$x = 132 , 000$