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# An investment of d dollars at k percent simple annual

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Manager
Joined: 14 Mar 2006
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An investment of d dollars at k percent simple annual [#permalink]

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29 May 2006, 06:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

Can't seem to drive the equation.

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Director
Joined: 13 Nov 2003
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Location: BULGARIA

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29 May 2006, 06:54
You do not need an equation. D/300\$=X/2400\$. Then amount 8d will yield 2400\$ in one year, respectively 8d/3 will yield 2400 in 3 years

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VP
Joined: 29 Dec 2005
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29 May 2006, 11:14
BG wrote:
You do not need an equation. D/300\$=X/2400\$. Then amount 8d will yield 2400\$ in one year, respectively 8d/3 will yield 2400 in 3 years

agree with you nice solution, BG.

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Manager
Joined: 04 May 2006
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30 May 2006, 05:40
Using the data give for first scenario, 2*d*k/100 = 600 --> (1)

Using the data given for second scenario (assuming X as amount invested), 3*X*k/100 = 2400 --> (2)

Substituting k/100 = 300/d (derived from (1) ) into 2, it gives -
3*X*300/d = 2400
==> X = 8d/3.

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SVP
Joined: 30 Mar 2006
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31 May 2006, 02:22
8d/3

Given :

600 = d*k*2/100
K = 600*50/d

Now to get interest of 2400 on say P

2400 = P*k*3/100
P = 2400*100*d/600*50 = 8D/3

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31 May 2006, 02:22
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# An investment of d dollars at k percent simple annual

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