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An investment of d dollars

by mgmat Thu Apr 24, 2008 11:55 am

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 $24,000 over a 3 year period?

A. (2d)/3
B. (3d)/4
C. (4d)/3
D. (3d)/2
E. (8d)/3 - Correct Answer

Maybe Vics may work but I couldn't approach this correctly. How do you do this?
rfernandez
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by rfernandez Thu Apr 24, 2008 5:29 pm

I suspect you may have a typo... can you double check on the amount you want to yield in the question... is it $24,000 or $2,400?
mgmat
 
 

yes

by mgmat Sat Apr 26, 2008 1:04 pm

rey, you are correct. it's 2,400. sorry.
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by rfernandez Fri May 02, 2008 3:45 pm

A VIC solution might look like this:

Choose d = 1000 and k = 30. I chose these numbers because $600 of simple interest over two years means $300 each year, and 30% of $1000 is $300.

So now, we need to find the target value, namely the size of the investment needed to yield $2400 in interest over three years, at the same rate k.

Interest = (Principal)(rate)(time)
2400 = P(.3)(3)
P=8000/3

Now, plugging in 1000 in for d in the five answer choices only gives one match, choice E.

Rey
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by kevincan Wed May 14, 2008 5:35 pm

Think that you need to generate 4 times as much interest , so multiply d by 4= 4d
However, you are given 3/2 as much time, so multiply 4d by 2/3 =8d/3

Kevin Armstrong
Madrid, Spain
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Re: An investment of d dollars

by charlottetaylor.07 Tue Oct 05, 2010 9:05 am

Can someone post the algebraic way of getting this answer please? Here's what I did and I can't seem to get the right answer...

Prt= (d)(k/100)(2)= 600
k= 30,000/d

24000=P(30,000/d)(3)
P=2d/75
Viswanathan.harsha
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Re: An investment of d dollars

by Viswanathan.harsha Wed Oct 06, 2010 12:45 am

The way I see is:

600 = d1 (k/100) (2)
2400 = d2 (k/100) (3)

Hence:
300=d1 (k/100)
800=d2 (k/100)

k/100 is constant in both, so disregard, and you can see the 8/3 relationship of d in the two equations.
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Re: An investment of d dollars

by tim Fri Oct 08, 2010 4:00 am

Looks good..
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Re: An investment of d dollars

by khanhha208 Tue Jul 24, 2012 7:51 am

Dear Tim

The problem states that:
"An investment of d dollars at k percent simple annual interest yields $600 over a 2 year period"

As I understand, the equation should be:
d*((1+(k/100))^2-1) = 600

Why is it (d)(k/100)(2)= 600? Have I misunderstood something?
tim
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Re: An investment of d dollars

by tim Tue Jul 24, 2012 10:53 am

you're calculating compound interest; the problem talks about simple interest..
Tim Sanders
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