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# An investment of d dollars at k percent simple annual

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Senior Manager
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An investment of d dollars at k percent simple annual [#permalink]

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12 Jul 2007, 12:03
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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

I am terrible with interest rates (even worse, I want to be a investment banker ) Please help!

Thanks!
Senior Manager
Joined: 11 Jun 2006
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12 Jul 2007, 12:15
E.

2dk=600
k=300/d

now set an equation for what we're looking to get

3xk=2400
substitute for k

3x(300/d)=2400
x=8d/3
Senior Manager
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12 Jul 2007, 12:40
baer wrote:
E.

2dk=600
k=300/d

now set an equation for what we're looking to get

3xk=2400
substitute for k

3x(300/d)=2400
x=8d/3

Wow, that seemed simple.

So, the equation is:

interest = # of years x principle x interest rate

I overcomplicated the question by looking into the A = (1+r/c)^(n*c) equation.

Thank you 40oz. I'll have a OE tonight.
Senior Manager
Joined: 11 Jun 2006
Posts: 254
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Kudos [?]: 11 [0], given: 0

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12 Jul 2007, 13:01
leeye84 wrote:
baer wrote:
E.

2dk=600
k=300/d

now set an equation for what we're looking to get

3xk=2400
substitute for k

3x(300/d)=2400
x=8d/3

Wow, that seemed simple.

So, the equation is:

interest = # of years x principle x interest rate

I overcomplicated the question by looking into the A = (1+r/c)^(n*c) equation.

Thank you 40oz. I'll have a OE tonight.

You were thinking about *compounded* interest, while the question asked for *simple* interest. Enjoy the 40. I will definitely need one after I take my GMAT tomorrow! Hopefully a celebratory one!
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12 Jul 2007, 14:56
Here is how I do it (same concept as Baer)

Since it is simple interest we know that d at k% gives us 600 in 2 yrs.
So in 1 yr -->> 300
in 3 yrs --->>>900

We need some amount in terms of d (call it x) such that 900x=2400
x=8/3 times d.
12 Jul 2007, 14:56
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