If a town of 25,000 people is growing at a rate of approx. 1% per year

To calculate (1.01)^5* 25 000 is indeed a bad idea, especially on a gmat test day:)

How can we quickly estimate the closest value of this expression?
Compound interest gives a greater growth because it is "an interest of interests".
We can easily calculate the amount after five years for a simple interest: 25 000* (1.01*5)=26 250. The increase=26 250- 25 000=1 250.
Thus, the compound interest should increase by lower than 0.01*(1 250)*4=50 (the compound interest becomes different from simple interest beginning with the 2nd year, thus, 5-1=4).
The increase with the compound interest rate will be lower that 26 250+ 50=26 300, which is closer to 26 000. So A.

This question is quite easy, but it appears difficult if we focus too much on the formula, and the fact that it is difficult to execute 1.01^5 without a calculator. Beware of "formula tunnel vision": instead of worrying about how to calculate the answer using the interest formula, we should instead understand that a 1 percent increase is really quite easy to calculate by hand (just remove the final two digits of the original number, and then add it to the original number):

Year 1: 25,000 + 1% = 25,000 + 250 = 25,250

Year 2: 25,250 + 1% = 25,250 + 252 = 25,502

Year 3: 25,502 + 1% = 25,502 + 255 = 25,757

Year 4: 25,757 + 1% = 25,757 + 257 = 26,014

Year 5: 26,014 + 1% = 26,014 + 260 = 26,274

Thus the answer is closest to Choice A (26,000).
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1% is quite small and hence even simple interest calculation with it'll yield result quite close to comp interest. So ans A

But how far can we assume SI=CI? 1%? 2%? 3%?

Let population in the beginning = 100
at the end of 1st year population = 100+1% of 100 = 100+1 = 101
at the end of 2nd year population = 101+1% of 101 = 101+1 approx = 102 approx (on the lower side because 1% of 101 is greater than 1)
at the end of 3rd year population = 102+1% of 102 = 102+1 approx = 103 approx (on the lower side because 1% of 102 is greater than 1)
at the end of 4th year population = 103+1% of 103 = 103+1 approx = 104 approx (on the lower side because 1% of 103 is greater than 1)
at the end of 5th year population = 104+1% of 104 = 104+1 approx = 105 approx (on the lower side because 1% of 104 is greater than 1)

i.e. 5% greater than 25000 = 25000+5% of 25000 = 25000+1250 = 26250 (on the lower side)

i.e. Population must be greater than 26250 but since the closest option is 26000 so

Answer: option A
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After 1 year, the population will be:

25,000 + 250 = 25,250

After 2 years, the population will be:

25,250 + 252 = 25,502

After 3 years, the population will be:

25,502 + 255 = 25,757

So essentially, we see that the population growth is a little more than 250 people a year.

25,000 + 250 x 5 = 26,250, so about 26,000.

Alternate solution:

The population of the town in 5 years is:

25,000 x (1 + 0.01)^5 = 25,000 x 1.01^5 = 26,275, which is closest to 26,000.

Answer: A
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1% of 25,000 = 250

Since 1% compounded is really insignificant, we can use simple interest: \(250 * 5 = 1,250 \)

\(25,000 + 12,250 = 16,250\)

Closest answer is A.
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We can use Simple Interest formula - SI = PTR/100. Here , P = 25000, T=5, R=1 . therefore, PTR/100 = 1250. Therefore, the population of the town after 5 years would be closest to 26000. (25000+1250* = 26000*)

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