The population of a town increases 8% annually. If its present population is 1,42,560, what was the population one year ago?

The correct option is 1,32,000.

To find the population of the town one year ago, we can set up a simple algebraic relationship based on the annual percentage increase.
Let the population of the town one year ago be represented by $P$.
The population increases by 8% annually. This means the current population is 108% of the population one year ago.

We can write this relationship as:
$\text{Present Population}=P\times (1+\frac{8}{100})$
Given that the present population is 1,42,560, we substitute this value into our equation:
$\mathrm{1,42,560}=P\times 1.08$
To solve for $P$, we divide the present population by 1.08:
$P=\frac{\mathrm{1,42,560}}{1.08}$
Simplifying the division:
$P=\frac{\mathrm{14,256,000}}{108}$
$P=\mathrm{1,32,000}$
Therefore, the population of the town one year ago was 1,32,000.