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

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

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New post 05 Feb 2016, 03:30
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70% (01:46) correct 30% (01:45) wrong based on 249 sessions

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If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
a.26000
b.27000
c.28000
d.29000
e.30000

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

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New post 05 Feb 2016, 08:47
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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.
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If a town of 25,000 people is growing at a rate of approx. 1% per year  [#permalink]

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New post 05 Feb 2016, 03:32
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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
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Re: If a town of 25,000 people is growing at a rate of approx. 1% per year  [#permalink]

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New post 05 Feb 2016, 03:35
NoHalfMeasures wrote:
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%?
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Re: If a town of 25,000 people is growing at a rate of approx. 1% per year  [#permalink]

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New post 05 Feb 2016, 06:27
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NoHalfMeasures wrote:
If a town of 25,000 people is growing at a rate of approx. 1% per year, the population of the town in 5 years will be closest to?
a.26000
b.27000
c.28000
d.29000
e.30000


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

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New post 17 May 2017, 18:23
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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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Re: If a town of 25,000 people is growing at a rate of approx. 1% per year  [#permalink]

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