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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
An investment of d dollars at k percent simple annual interest yields
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12 Dec 2022, 06:04
12 Dec 2022, 06:04
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62% (01:36) correct
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An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 over a 3 year period?
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3
A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3
Re: An investment of d dollars at k percent simple annual interest yields
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14 Dec 2022, 05:14
14 Dec 2022, 05:14
Expert Reply
Simple interest = principal * interest rate * time, where "principal" is the starting amount and "rate" is the interest rate at which the money grows per a given period of time (note: express the rate as a decimal in the formula). Time must be expressed in the same units used for time in the Rate.
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)
What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).
Answer: E.
Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.
Answer: E.
An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period --> \(d*\frac{k}{100}*2=600\) --> \(k=\frac{30,000}{d}\)
What (x) dollar amount invested at the same rate will yield $2,400 over a 3 year period --> \(x*\frac{k}{100}*3=2,400\) --> \(x*\frac{30,000}{100d}*3=2,400\) --> \(x=\frac{8d}{3}\).
Answer: E.
Or another way: In order interest to be 4 times more in 1.5 times longer time period then investment must be 4d/1.5=8d/3.
Answer: E.
Re: An investment of d dollars at k percent simple annual interest yields
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07 Jan 2024, 03:03
07 Jan 2024, 03:03
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