Question

19. The population of a town increases by 5% every six months. If the present population of the town is 10,00,000,
find its population after one year.​

Answer 1

Answer:

1000000×10/100=1100000

Answer 2

Given:

A town with

• Present Population = 10,00,000
• Rate = 5 % (every 6 months) = 5 × 2 = 10 % (every 1 year)
• Time - 1 year

What To Find:

We have to find

• The population of town after 1 year.

Formula Needed:

$\bf Population \: after \: 1 \: year = Present \: Population \: \bigg(1 + \dfrac{Rate}{100} \bigg) ^{Time}$

Solution:

Using the formula,

$\sf \implies Population \: after \: 1 \: year = Present \: Population \: \bigg(1 + \dfrac{Rate}{100} \bigg) ^{Time}$

Substitute the values,

$\sf \implies Population \: after \: 1 \: year = 10,00,000 \: \bigg(1 + \dfrac{10}{100} \bigg) ^{1}$

Solve the brackets,

$\sf \implies Population \: after \: 1 \: year = 10,00,000 \: \bigg( \dfrac{110}{100} \bigg) ^{1}$

Remove the brackets,

$\sf \implies Population \: after \: 1 \: year = 10,00,000 \: \times \dfrac{110}{100}$

Cancel the zeros,

$\sf \implies Population \: after \: 1 \: year = 10,000 \: \times 110$

Multiply the values,

$\sf \implies Population \: after \: 1 \: year = 11,00,000$

Final Answer:

∴ Thus, the population after 1 year is 11,00,00.