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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
If a town of 25,000 people is growing at a rate of approx. 1% per year
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28 Dec 2022, 09:43
28 Dec 2022, 09:43
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Expert Reply
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Question Stats:
100% (01:08) correct
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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
a.26000
b.27000
c.28000
d.29000
e.30000
GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
If a town of 25,000 people is growing at a rate of approx. 1% per year
[#permalink]
30 Dec 2022, 23:19
30 Dec 2022, 23:19
Expert Reply
To calculate (1.01)^5* 25 000 is indeed a bad idea)
How can we quickly estimate the closest value of this expression?
Compound interest gives 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 than 26 250+ 50=26 300, which is closer to 26 000. So A.
Answer: A
How can we quickly estimate the closest value of this expression?
Compound interest gives 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 than 26 250+ 50=26 300, which is closer to 26 000. So A.
Answer: A
gmatclubot
If a town of 25,000 people is growing at a rate of approx. 1% per year [#permalink]
30 Dec 2022, 23:19
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