Question

the population of the town increased at rate of 5 %per year consecutively for 2 years . then due to epidemic in the third year decreased at the rate of 1.5% for next 3 years what is the current population of that town if it was 23,45,000, 5 years ago​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Suppose present population = P

1. If population increased by r % per year. Then the population after n years will be

$= \displaystyle \: P {(1 + \frac{r}{100} \: ) }^{n}$

2. If population decreased by r % per year. Then the population after n years will be

$= \displaystyle \: P {(1 - \frac{r}{100} \: ) }^{n}$

EVALUATION

The population before 5 years ago = 2345000

Now for the first two years the population of the town increased at rate of 5 % per year consecutively.

So after 2 years population will be

$= \displaystyle \: 2345000 \times {(1 + \frac{5}{100} \: ) }^{2}$

$= \displaystyle \: 2345000 \times {( \frac{105}{100} \: ) }^{2}$

$= \displaystyle \: 2345000 \times {( \frac{21}{20} \: ) }^{2}$

$= \displaystyle \: 2345000 \times \frac{441}{400}$

$= 258536.25$

Again for the first next 3 years the population of the town decreased at rate of 1.5 % per year consecutively.

5 % per year consecutively. So after 3 years population will be

$= \displaystyle \: 258536.25 \times {(1 - \frac{1.5}{100} \: ) }^{3}$

$= \displaystyle \: 258536.25 \times {( \frac{98.5}{100} \: ) }^{3}$

$= \displaystyle \: 258536.25 \times {(1 - \frac{1.5}{100} \: ) }^{3}$

$= 247075.76$

Answer 2

Answer:

2470758 this is answer .