If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
[spoiler]
80%[/spoiler]
please explain
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alchol and water- mixtures
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cans
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initially 50% alcohol solution. (let 100 ml).
we replace x ml with 25% soln.
Thus total alcohol = 50(100-x) + 25x = 5000 - 25x = 3000 (resulting is 30%)
thus x = 80
we replace x ml with 25% soln.
Thus total alcohol = 50(100-x) + 25x = 5000 - 25x = 3000 (resulting is 30%)
thus x = 80
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Cans!!
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Cans!!
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shankar.ashwin
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Using allegations rule, first determine the ratio both the solutions are mixed to produce the 30% solution.
To find the ratio.
Ratio of 25% : Ratio of 50% = (50-30):(30-25) = 4:1. (In general, (Higher - Desired : Desired - Lower)
There the 30% soln has 4 parts of 25% solution and 1 part of 50% solution,
therefore % of 50 solution replaced = % of 30 solution added
So 4/5 * 100 = 80%
To find the ratio.
Ratio of 25% : Ratio of 50% = (50-30):(30-25) = 4:1. (In general, (Higher - Desired : Desired - Lower)
There the 30% soln has 4 parts of 25% solution and 1 part of 50% solution,
therefore % of 50 solution replaced = % of 30 solution added
So 4/5 * 100 = 80%
venmic wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
[spoiler]
80%[/spoiler]
please explain
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GMATGuruNY
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We can use alligation, which dictates the following:venmic wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
[spoiler]
80%[/spoiler]
please explain
The proportion of each element in the mixture is equal to the distance between the percentage attributed to the other element in the mixture and the percentage attributed to the mixture.
Proportion of 50% solution = |percentage in 25% solution - percentage in mixture| = |25-30| = 5.
Proportion of 25% solution = |percentage in 50% solution - percentage in mixture| = |50-30| = 20.
Ratio of 50% solution:25% solution = 5:20 = 1:4.
The sum of the values in the ratio = 1+4 = 5.
The proportion of 25% solution in the ratio = 4.
Thus, 4/5 = 80% of the original alcohol was replaced by the 25% solution.
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saketk
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Just read the topic on Mixtures and Allegations ... I am sure you wont face much problem after that.venmic wrote:If a portion of a half water/half alcohol mix is replaced with 25% alcohol solution, resulting in a 30% alcohol solution, what percentage of the original alcohol was replaced?
[spoiler]
80%[/spoiler]
please explain
Please see the attachment for a 2 step process using a small diagram ..
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