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## Mixture problem : with allegation method

This topic has 3 expert replies and 3 member replies
gmatquant25 Senior | Next Rank: 100 Posts
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#### Mixture problem : with allegation method

Wed Mar 13, 2013 10:33 am
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%

I am trying to solve it using Allegation method , but not quit sure how to get the end result , Please help . Also, elaborate the steps so that I can understand the approach .
many thanks !

I started with 10% and x% in the first row

16% in the second row

and x-16% and 6% in the last one

this gives me x-16/6= ? "stuck here "

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Anju@Gurome GMAT Instructor
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Wed Mar 13, 2013 10:39 am
gmatquant25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.
If we view this as an weighted average problem then weighted average of 10 and x is 16, where weight of 10 is 3/4 and weight of is 1/4.

Hence, (3/4)*10 + (1/4)*x = 16
-----> 30 + x = 4*16 = 64
-----> x = (64 - 30) = 34

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gmatquant25 Senior | Next Rank: 100 Posts
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Wed Mar 13, 2013 11:06 am
thanks - I could solve this problem in every other way. I want to do it using allegation rule , It helps me solve such problems in no time .

Last edited by gmatquant25 on Wed Mar 13, 2013 11:51 am; edited 1 time in total

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GMATGuruNY GMAT Instructor
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Wed Mar 13, 2013 1:09 pm
gmatquant25 wrote:
One-fourth of a solution that was 10% by weight was replaced by a second solution resulting in a solution that was 16% sugar by weight. The second solution was what percent sugar by weight.

34%
24%
22%
18%
8.5%
Here's how to solve with alligation.
Let O = the original solution, R = the replacement solution, and M = the mixture.

Step 1: Plot the percentages on a number line, with the two ingredients on the ends and the mixture in the middle.
O 10%----------M=16%----------R

Step 2: Plot the distances between the percentages.
(distance between O and M) : (distance between M and R) is equal to the RECIPROCAL of the ratio of O to R in the mixture.
Since R = 1/4 of the mixture, O:R = 3:1.
Plotting the reciprocal of this ratio on the number line, we get:
O 10%----x-----M=16%----3x----R

Since x is the distance between O and M:
x = 16-10 = 6.
Since 3x is the distance between M and R:
R = 16 + 3x = 16 + 3*6 = 34.

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vishugogo Master | Next Rank: 500 Posts
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Thu Mar 14, 2013 7:05 am
[quote="GMATGuruNY"]how is R = 1/4 of the mixture

vishugogo Master | Next Rank: 500 Posts
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Thu Mar 14, 2013 7:08 am
[quote="Anju@Gurome"][

how is weight of 10 is 3/4 and weight of x is 1/4

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Thu Mar 14, 2013 7:48 am
vishugogo wrote:
how is R = 1/4 of the mixture
From the problem: ONE-FOURTH of a solution...WAS REPLACED by a second solution.
Thus, the replacement solution -- the value of R -- is equal to 1/4 of the resulting mixture.

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