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by alex.gellatly » Mon Apr 16, 2012 11:06 pm
An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?

(A) (2d) / 3
(B) (3d) / 4
(C) (4d) / 3
(D) (3d) / 2
(E) (8d) / 3

Thanks
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by Anurag@Gurome » Mon Apr 16, 2012 11:07 pm
alex.gellatly wrote:An investment of d dollars at k percent simple annual interest yields $600 interest over a 2-year period. In terms of d, what dollar amount invested at the same rate will yield $2,400 interest over a 3-year period?

(A) (2d) / 3
(B) (3d) / 4
(C) (4d) / 3
(D) (3d) / 2
(E) (8d) / 3

Thanks

Simple interest = principal * interest rate * time, where
principal = amount invested
rate = interest rate

An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period implies d * k/100 * 2 = 600
k = 30,000/d

Let us assume that the amount invested at the same rate which will yield $2,400 over a 3 year period = $x

Then, x * k/100 * 3 = 2400
x * (30000/100d) * 3 = 2400
x = [spoiler]8d/3 [/spoiler]

The correct answer is E.
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by aneesh.kg » Mon Apr 16, 2012 11:27 pm
SI = PRT/100

Here, we have two cases. Let the Simple Interest in the two cases be I1 and I2.

I1 = 600 = (d.k.2)/100
I2 = 2400 = (x.k.3)/100,

where x = principal for the second case.

Dividing the second equation by the first,

4 = 3x/2d,

or

x = 8d/3
(E) is the answer

Please Note: In such questions, dividing the two equations is a better method than finding the rate of interest from the first case and substituting it in the second case.
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