Question

7. Over the next ten years, Town A is expected to gain 1,000 people each year. During the same time period, the population of Town B is expected to increase by 5% each year. Both Town A and Town B currently have populations of 10,000 people. The table below shows the expected population of each town for the next three years. Expected Population Current Year 1 Year 2 10.000 11.000 12.000 10,000 10.500 11.025 Year 3 13.000 Town A Town B 11.576 Which number of years is the BEST approximation of the time until Town A and Town B once again have the same population? 31 years 41 years 16 years 27 years pocytes.coming outons.org/12727646activity Assessment_38_1990_1 717 YOS Jorld ODOC

Answer 1

Population of town A increases by 1000 each year

So population of town A in n yrs = 10000 + 1000*n

Population of town B increases by 5% each year

So population of town B in n yrs = 10000(1+5/100)ⁿ = 10000(1.05)ⁿ

We need to find n, for

10000 + 1000*n = 10000(1.05)ⁿ

1+0.1n = 1.05ⁿ

Substitute for each of the options in the equation above

For n = 31

1+0.1n = 1 + 0.1*31 = 4.1

1.05ⁿ = 1.05^31 = 4.5

Since exponential fnctionhas exceedd the linear function we need a lesser value of n

So put n = 27

1+0.1n = 1 + 0.1*27 = 3.7

1.05ⁿ = 1.05^27 = 3.7

27 years is the answer