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# gmatprep word problem (in terms of)

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Director
Joined: 23 May 2008
Posts: 757
gmatprep word problem (in terms of) [#permalink]

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29 Jul 2009, 17:47
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pls solve

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Current Student
Joined: 03 Aug 2006
Posts: 112
Re: gmatprep word problem (in terms of) [#permalink]

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29 Jul 2009, 20:17
1
KUDOS
The key here is that this is simple interest.

Given:
Scenario 1
Principal Amount: $$d$$
Simple Interest:$$k%$$
Term: 2 years

Interest for Each year:

$$\Rightarrow d \times \frac{k}{100}$$

Interest for 2 years combined

$$\Rightarrow \frac{dk}{100} + \frac{dk}{100} = \frac{2dk}{100}$$

Also given Interest over 2 years: $$600$$

$$\Rightarrow \frac{2dk}{100} = 600$$

$$\Rightarrow k=\frac{600 \times 100}{2d}$$

$$\Rightarrow k=\frac{30000}{d}$$

Scenario 2
Let Principal Amount: U
Simple Interest:k%
Term: 3 years

Interest for Each year:

$$\Rightarrow U \times \frac{k}{100}$$

Interest for 3 years combined

$$\Rightarrow \frac{Uk}{100} + \frac{Uk}{100} + \frac{Uk}{100} = \frac{3Uk}{100}$$

Replace$$k=\frac{30000}{d}$$

$$\Rightarrow \frac{3Uk}{100} = \frac{3U \times \frac{30000}{d}}{100} = \frac{900U}{d}$$

Also give total interest for 3 years = 2400

$$\Rightarrow \frac{900U}{d} = 2400$$

Solving for U gives

$$U = \frac{8d}{3}$$

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re: gmatprep word problem (in terms of)   [#permalink] 29 Jul 2009, 20:17
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