Question

The population of a town was 1,60,000 three years ago. if it increased by 3%, 2.5% and 5% respectively in the last three years, then the present population of the town is

Answer 1

Given Population of a town = 160000.

R1 = 3%, R2 = 2.5%, R3 = 5%.

Then the present population of the town = 160000(1+3/100)(1+2.5/100)(1+5/100)

                                                                     = 177366.


Hope this helps!

Answer 2

Answer: The present population of the town is 177,366.

Step-by-step explanation:

Since we have given that

Population of a town three years ago = 1,60,000

Rate of interest 3%, 2.5% and 5%.

As we know the formula for "compound interest":

$Amount=P(1+\dfrac{r_1}{100})(1+\dfrac{r_2}{100})(1+\dfrac{r_3}{100})\\\\Amount=160000(1+\dfrac{3}{100})(1+\dfrac{2.5}{100})(1+\dfrac{5}{100})\\\\Amount=160000\times 1.03\times 1.025\times 1.05\\\\Amount=177,366$

Hence, the present population of the town is 177,366.