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# mixture of antifreeze and water

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Intern
Joined: 05 Feb 2009
Posts: 7
mixture of antifreeze and water [#permalink]

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25 Feb 2009, 06:41
An automobile radiator contains 60 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze??

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CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
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Re: mixture of antifreeze and water [#permalink]

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25 Feb 2009, 07:12
$$\frac{\frac{30%}{100%} * (60L - x) + x}{60L} =\frac{50%}{100%}$$

$$\frac{18L + 0.7x}{60L} =0.5$$

$$x =\frac{30L-18L}{0.7} = ~ 17.1L$$
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SVP
Joined: 07 Nov 2007
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Location: New York
Re: mixture of antifreeze and water [#permalink]

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25 Feb 2009, 07:32
refreshment wrote:
An automobile radiator contains 60 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so that there will be 50% antifreeze??

LET X is Leters of mixture after drained.

Water % in the mixture = 70
0.7*X/60 + (0) =0.5

X = 5/7*60

mixture drained= 60 - (5/7)*60 = 2/7*60 = 17.1
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Joined: 04 Jan 2008
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Re: mixture of antifreeze and water [#permalink]

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25 Feb 2009, 07:41
thanks
I got trapped
walker wrote:
$$\frac{\frac{30%}{100%} * (60L - x) + x}{60L} =\frac{50%}{100%}$$

$$\frac{18L + 0.7x}{60L} =0.5$$

$$x =\frac{30L-18L}{0.7} = ~ 17.1L$$

--== 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.

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Re: mixture of antifreeze and water   [#permalink] 25 Feb 2009, 07:41
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# mixture of antifreeze and water

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