Question

Suppose the population of a town increases by 3% each year. the population of the town today is 25,000. use compound interest table to find its population ten years from now. (Must give the answer to the nearest thousand)

Answer 1

Explanation:

Let the population after 10 years be x.

`\therefore \: x =25000 * (1 + (3)/(100) )^(10) \\ \\ \therefore \: x = 25000 * (1 + 0.03 )^(10) \\ \\ \therefore \: x =25000 * (1.03 )^(10) \\ \\ \therefore \: x = 25000 * 1.34391638 \\ \\ \therefore \: x =33,597.9095 \\ \\ \therefore \: x \approx \: 33598`

Thus, the population of the town after 10 years would be 33598.

Answer 2

Answer: The population after 10 years is 33500

Explanation:

The current population of the town is 25000. It increases at a rate of 3%

The formula for compound interest is expressed as

A = P(1+r/n)^nt

Where

A represents the final value at the end of t years.

t represents time in years.

n represents the periods of increase.

r represents growth rate

P represents initial or current value.

From the information given,

r = 3%

t = 10 years

Looking at the compound interest table, the columns represents the growth rate while the the rows represents the number of years.

Therefore, the compound interest multiplier for 3% and 10 years is 1.34

Therefore

The population after 10 years would be

1.34 × 25000 = 33500