alchol and water- mixtures

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Mon Sep 05, 2011 4:47 am

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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Source: Beat The GMAT — Problem Solving | Mon Sep 05, 2011 4:47 am

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Mon Sep 05, 2011 4:53 am

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

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Mon Sep 05, 2011 4:58 am

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%

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

Like this post

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Sep 05, 2011 5:57 am

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

We can use alligation, which dictates the following:

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: Legendary Member; Posts: 608; Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members; Thanked: 37 times

by saketk » Mon Sep 05, 2011 8:47 am

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

Just read the topic on Mixtures and Allegations ... I am sure you wont face much problem after that.

Please see the attachment for a 2 step process using a small diagram ..

Attachments

M&A.doc: (26 KiB) Downloaded 143 times

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venmic: Master | Next Rank: 500 Posts; Posts: 222; Joined: Mon Oct 13, 2008 4:04 pmFollowed by: 2 members; Thanked: 3 times

by venmic » Mon Sep 05, 2011 12:24 pm

Thanks everyone

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