Question

two years ago the population of a town was 10000 during first year it incresed at the rate of 5% per nannum and during the second year it increased at the rate of 6% per annum what is its present population

Answer 1

Given :

Population of town (2 years ago) : 10000

Rate of incresion in first year : 5%

Rate of incresion in second year : 6%

To find :

The present population of the town.

Solution :

First, we'll find the population of the town one year ago.

Population of the town (1 year ago) :

$\sf \dashrightarrow 10000 + (5\% \: of \: 10000)$

$\sf \dashrightarrow 10000 + \bigg( \dfrac{5}{100} \times 10000 \bigg)$

$\sf \dashrightarrow 10000 + \bigg( \dfrac{1}{20} \times 10000 \bigg)$

$\sf \dashrightarrow 10000 + \bigg( \dfrac{1 \times 10000}{20} \bigg)$

$\sf \dashrightarrow 10000 + \bigg( \dfrac{10000}{20} \bigg)$

$\sf \dashrightarrow 10000 + 500$

$\sf \dashrightarrow 10500$

Now, we can find the present population of the town.

Present population of the town :

$\sf \dashrightarrow 10500 + (6\% \: of \: 10500)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{6}{100} \times 10500 \bigg)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{3}{20} \times 10500 \bigg)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{3 \times 10500}{20} \bigg)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{31500}{20} \bigg)$

$\sf \dashrightarrow 10500 + 1575$

$\sf \dashrightarrow 12075$

Hence, the present population of the town is 12075.

Answer 2

Step-by-step explanation:

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