Question

The present population of a town is 10000 and was 8000 two years ago. If it grows at the same rate, what will be the population 2 years hence?

Answer 1

Step-by-step explanation:

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Answer 2

The population becomes 12544 2 years hence.

Step-by-step explanation:

Since we have given that

Present population = 10000

Population two years ago = 8000

So, it becomes,

$A=P(1+\dfrac{r}{100})^n\\\\10000=8000(1+\dfrac{r}{100})^2\\\\\dfrac{10000}{8000}=(1+\dfrac{r}{100})^2\\\\1.25=(1+\dfrac{r}{100})^2\\\\\sqrt{1.25}=1+\dfrac{r}{100}\\1.12-1=\dfrac{r}{100}\\\\0.12=\dfrac{r}{100}\\\\r=12\%$

So, after two years , the population becomes,

$A=10000(1+\dfrac{12}{100})^2\\\\A=10000(1+0.12)62\\\\A=10000(1.12)^2\\\\A=12544$

Hence, the population becomes 12544 2 years hence.

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The present population of a town is 10000 and was 8000 two years ago if it grows at the same rate what will be the population two years hence

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