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by Xbond » Tue Jun 09, 2009 1:33 pm
Hi there,

I found one part of the answer but I didn't find the ratio rate.
Could demonstrate the simplest way to resolve it ?

An investment of "d" dollars at "k" percent simple annual interest yield $600 interest over a 2-year period. In terms od d, what dollar amount invested at the same rate will yield $2400 interest over a 3-year period ?

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



the answer is E
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by ssmiles08 » Tue Jun 09, 2009 1:43 pm
Let d1 = dollar amount for the 600 interest
Let d2 = dollar amount for the 2400 interest.

simple interest formula = P*k*t
600 = d1*k*2

d1*k = 300

2400 = d2*k*3

d2*k = 800

so k = 800/d2

plug it back into the first equation: d1*k = 300
and we get d1*(800/d2) = 300

d2 = (8/3)*d1 (E)
Last edited by ssmiles08 on Tue Jun 09, 2009 1:43 pm, edited 1 time in total.
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by yogami » Tue Jun 09, 2009 1:43 pm
So first statement gives

600 = 2*d*k/100 i.e. dk = 30000
second one gives

2400 = 3*x*k/100 or xk = 80000

Now what multiplied to dk will give 80000? Obviously when dk is multiplied by 8/3

because (8/3)*30000 = 80000
therefore (8/3)dk = 80000
(8/3)dk = xk
or x = 8d/3
200 or 800. It don't matter no more.
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Re: Interest yield

by Stuart@KaplanGMAT » Tue Jun 09, 2009 2:45 pm
Xbond wrote:Hi there,

I found one part of the answer but I didn't find the ratio rate.
Could demonstrate the simplest way to resolve it ?

An investment of "d" dollars at "k" percent simple annual interest yield $600 interest over a 2-year period. In terms od d, what dollar amount invested at the same rate will yield $2400 interest over a 3-year period ?

A) 2d/3 B) 3d/4 C) 4d/3 D) 3d/2 E) 8d/3
Some good algebraic solutions. This question is also ripe for good ol' critical thinking (aka doing math without the algebra).

$600/2 years means we're earning $300 interest/year.

We want to earn $2400/3 years or $800 interest/year.

If we want to earn 800/300 or 8/3 as much interest, we need to invest 8/3 as much money: choose (E).
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