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mixture of antifreeze and water

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

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New post 25 Feb 2009, 06:41
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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Re: mixture of antifreeze and water [#permalink]

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New post 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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Re: mixture of antifreeze and water [#permalink]

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New post 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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Re: mixture of antifreeze and water [#permalink]

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

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


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

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