Question

Twenty years ago a small town in Texas had a population of 10,000. The population has increased 8% each year since then. Predict what the population of the town will be in 10 more years.

Answer 1

Answer: After 10 more years i.e. after 30 years, the population of the town will be 100626.57 approx.

Step-by-step explanation:

Since we have given that

Initial population 20 years ago = 10,000

Rate of growth = 8%

Number of years = 10+20 = 30 years

According to question, we will use "Compound interest":

$Amount=P(1+\frac{r}{100})^n\\\\Amount=10000(1+\frac{8}{100})^{30}\\\\Amount=10000(1+0.08)^{30}\\\\Amount=10000(1.08)^{30}\\\\Amount=100626.57$

Hence, After 10 more years i.e. after 30 years, the population of the town will be 100626.57 approx.

Answer 2

Answer:

100,626 is the answer.

Step-by-step explanation:

Twenty years ago a small town in Texas had a population of 10,000.

The population has increased 8% each year since then.

This is exponential growth. Each years population is multiplied by 1.08 to obtain next years population. We use the formula :

$P=a\times (b)^{x}$

where a = initial value

b = growth rate in decimal form ( 1+r )

8% = 1 + 0.08 = 1.08

x = time (30years)

given population of 10,000 is 20 years ago and we have to find the population of the town will be in 10 more years.

It would be 30 years since then.

Just put 30 into the formula to get the population.

$P(t)=10,000\times 1.08^{30}$ = 100,626

The population of the town would be 100,626 in 10 more years or after 30 years.