In 1996, Town M had a population of 30,500 people, and it has grown at a rate of 2% every year. In 1996, Town N had 32,500 people and the population has increased by a constant rate of 600 each year. In what year were the populations of the two towns approximately equal?

Question

Jase Washington

Mathematics

3 May, 18:51

In 1996, Town M had a population of 30,500 people, and it has grown at a rate of 2% every year. In 1996, Town N had 32,500 people and the population has increased by a constant rate of 600 each year. In what year were the populations of the two towns approximately equal?

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Answers (1)

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Nancy Long · Answer 1

Town M initial population as at 1996 was 30500

it increases by 2%, ie (2/100 * 30500)

= 610 (increase by year)

so we find an arithmetic progressive expression for Town M

which is

Tn = 30500 + (n-1) 610 = 30500+610n-610

tn = 29890+610n

now for Town N we have

tn = 32500 + (n-1) 600 = 32500+600n-600

tn = 31900+600n

equate both M and N

29890+610n = 31900+600n

610n-600n=31900-29890

10n = 2010

n=201years

so 201years after 1996 = year 2197

oops that's quite far