Question

two years ago the population of atown was 10000.during first years,it increased at the rate of 5% per annum and deuring second years,it increased at the rate of 6%annum whar is ots present population?​

Answer 1

Answer:

11,100

Step-by-step explanation:

BECAUSE

TWO YEARS AGO'S POPULATION=10,000.

INCREASED IN ONE YEAR=5%.

INCREASED IN SECOND YEAR=6%.

SO,

5+6=11.

=10,000+11%=11,100

HOPE IT HELPS MARK IT AS BRAINLIEST

Answer 2

Correct Question

Two years ago, the population of a town was 10000. During first year, it increased at the rate of 5% and during second year, it increased at the rate of 6%. What is its present population?

Answer

Given :

Population of town (2 years ago) : 10000

Incresion in first year : 5%

Incresion in second year : 6%

To find :

The present population of the town.

Solution :

First, we should find the population of the town, 1 year ago.

Population of the town (1 year ago) :

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

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

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

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

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

$\sf \dashrightarrow 10000 + 500$

$\sf \dashrightarrow 10500$

Present population of the town :

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

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

$\sf \dashrightarrow 10500 + \bigg( \dfrac{3}{50} \times 10500 \bigg)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{3 \times 10500}{50} \bigg)$

$\sf \dashrightarrow 10500 + \bigg( \dfrac{31500}{50} \bigg)$

$\sf \dashrightarrow 10500 + 630$

$\sf \dashrightarrow 11130$

Hence, the present population of the town is 11130.