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# some of the %50 intensity red paint is replaced with %25

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some of the %50 intensity red paint is replaced with %25 [#permalink]

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05 Jun 2007, 15:15
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some of the %50 intensity red paint is replaced with %25 solution of red paint such that the new paint intensity is 30%. What fraction of the original paint was replaced?
1/30
1/5
2/3
3/4
4/5
CEO
Joined: 17 May 2007
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05 Jun 2007, 15:31
This problem can get extremely nasty if one is not thinking clearly.

The way I solved it is by visualizing the resultant mixture.

This mixture has some 25% paint and some 50% paint and the resultant is 30%. So lets assume the total amount of paint is 100

Then
(25x/100) + (50(100-x)/100) = 30
Solving this gives x = 80

implying that 80/100 of the mixture is made of 25% paint and this is also the amount that must have been replaced.

80/100 = 4/5 Option E
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05 Jun 2007, 15:45
Note that when the 50% solution is combined with the 25% solution the result is a 30% solution... So you can just put (50-30)/(50-25) = 4/5

This is a weighted average question.
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07 Jun 2007, 16:40
baer wrote:
Note that when the 50% solution is combined with the 25% solution the result is a 30% solution... So you can just put (50-30)/(50-25) = 4/5

This is a weighted average question.

Could you please explain that formula? what do the numbers represent?

Thanks[/b]
07 Jun 2007, 16:40
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