Question

The annual rate of growth in population of town is 5% . If its present population is 4000, what will be its population after 2 years?

Answer 1

Given :

population of town = 4000,

r = 5%

Time (n) = 2years.

To find :

Population after 2 years ?

Solution:

$Amount = 4000(1 + \frac{r}{100} ) {}^{n}$

$Amount = 4000(1 + \frac{5}{100} ) {}^{2}$

$Amount = 4410$

Hence, the population after two years = 4410.

Note that :

A rate, number, or amount in each hundred is known as percentage.

Answer 2

Given

• Annual rate of growth in population of a town is 5%
• Present population is 4000

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To Find

• The population of the town after 2 years.

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Solution

Population = 4000

Rate = 5%

Time Period = 2 years

To find the population after 2 years we will use this formula ⇒ $\sf Principal \times (1+\dfrac{Rate}{100})^{n}$

Let's find the population of the town after 2 years, step-by-step.

$\sf 4000(1+\dfrac{5}{100} )^{2}$

Step 1: Solve the brackets.

⇒ $\sf 4000(1+\dfrac{5}{100} )^{2}$

⇒ $\sf 4000(\dfrac{100}{100} +\dfrac{5}{100} )^{2}$

⇒ $\sf 4000(\dfrac{105}{100} )^{2}$

⇒ $\sf 4000(\dfrac{21}{20})^{2}$

Step 2: Square the numbers in bracket.

⇒ $\sf 4000(\dfrac{21}{20})^{2}$

⇒ $\sf 4000(\dfrac{441}{400})$

Step 3: Multiply.

⇒ $\sf 4000\times \dfrac{441}{400}$

⇒ $\sf 10\times 441$

⇒ $\sf 4410$

∴ The population of the town after 2 years would be 4410

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