If the number of school going children in a town is 64000 and it is increasing annually by 10%, then what will be the number of school going children in the town at the end of 3 years?

The correct option is 85184.

Let's solve the problem step-by-step using the formula for compound growth, which is similar to the compound interest formula.

We are given the following values:
- Initial number of school-going children, $P=64000$
- Annual growth rate, $R=10\%$
- Time period, $n=3$ years

The formula to find the population after $n$ years at an annual growth rate of $R$ is:

$A=P{\left(1+\frac{R}{100}\right)}^n$

Substitute the given values into the formula:

$A=64000\times {\left(1+\frac{10}{100}\right)}^3$

Simplify the term inside the brackets:

$1+\frac{10}{100}=1+0.1=1.1$

Now, calculate the expression with the exponent:

$A=64000\times {\left(1.1\right)}^3$

Evaluate ${\left(1.1\right)}^3$:

${\left(1.1\right)}^3=1.1\times 1.1\times 1.1=1.331$

Multiply the initial population by the growth factor:

$A=64000\times 1.331$

$A=64\times 1331$

$A=85184$

Thus, the number of school-going children in the town at the end of 3 years will be 85184.