Question

The present population of a town is 60,000. The population increases annually at 10%. Find the population after 3 years. ( A ) 80000 ( B ) 79860 ( C ) 77860 ( D ) 86000

Answer 1

Answer:

(B) 79860

Step-by-step explanation:

After 1 year Population would be

60000*1/10= 6000

60000+6000 = 66000

After 2 year Population would be

66000*1/10 = 6600

66000 + 6600 = 72600

After 3 year Population would be

72600*1/10 = 7260

72600 + 7260 = 79860

Thanks for patient reading.

Answer 2

Answer:

Option (B) 79,860

The population after 3 years will be 79,860.

Step-by-step explanation:

Given :

Present population = 60,000:

Increases annually at = 10%

To find :

Population after 3 years

Solution :

✯ $\textsf{\underline{\underline{Population after 3 years -}}}$

$\bigstar \: {\boxed{\tt{A= P\left( 1 + \frac{R}{100}\right)^{N}}}}$

Here :

• P = 60,000
• R = 10%
• N = 3

$\tt{\leadsto} \:A= 60000\left( 1 + \dfrac{10}{100}\right)^{3} \\ \\ \tt{\leadsto} \:A=60000\left( \dfrac{110}{100}\right)^{3} \\ \\ \tt{\leadsto} \:A=60000 \times \dfrac{110}{100} \times \dfrac{110}{100} \times \dfrac{110}{100} \\ \\ \tt{\leadsto} \:A=6 \times \dfrac{110}{1} \times \dfrac{11}{1} \times \dfrac{11}{1} \\ \\ \tt{\leadsto} \:A=6 \times 110 \times 11 \times 11 \\ \\ \tt{\leadsto} \:A=660 \times 121 \\ \\ \tt{\leadsto} \:A=79860$

Population = 79860

$\therefore$ The population after 3 years will be 79,860.