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Mixture problem using Alligation

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Mixture problem using Alligation

by Rudy414 » Wed Apr 10, 2013 6:03 pm
You currently have 10 oz of a solution that is 14% alcohol and the rest is water. How much water would you have to add to the solution to make the alcohol content equal to 10%?

I know how to solve this through general logic and statistics (i.e., 14% of 10 = 1.4; 1.4 = 10% of 14; 14-10 = 4; 4 oz of water is needed.)

My question is how to solve this using the alligation technique, with the tic-tac-toe looking board.

The way I attempted to solve this, I thought about the water rather than the alcohol and came up with this grid (it looks better on scratch paper):

100 I I 4 = 4/14
I 90 I
86 I I 10 = 10/14

Here is where I am stuck. I don't know what to think of the 4/14 and 10/14. It sounds like it is saying that the new 14 oz solution will have 10 oz of the 86% water and 4 oz of the 100% solution, which would be the right answer, but I am not sure if I am drawing an incorrect logical process because I already know the answer and am just forcing the information I have to work, or if this process is in fact correct.

Please let me know if I am approaching this right, and if not, where I am going wrong. A similar example would help too.

Thanks!
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by GMATGuruNY » Wed Apr 10, 2013 6:59 pm
Rudy414 wrote:You currently have 10 oz of a solution that is 14% alcohol and the rest is water. How much water would you have to add to the solution to make the alcohol content equal to 10%?
Alcohol percentage in the original solution: 14%.
Alcohol percentage in the added water: 0%.
Alcohol percentage in the mixture: 10%.

Let S = the original solution and W = the added water.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (14% and 0%) on the ends and the goal percentage (10%) in the middle.
S 14%----------10%-----------0% W

Step 2: Calculate the distances between the percentages.
S 14%----4-----10%----10-----0% W

Step 3: Determine the ratio in the mixture.
The required ratio of original solution to added water is the RECIPROCAL of the distances in red.
S:W = 10:4.

Since S:W = 10:4, and the volume of the original solution is 10 ounces, the final mixture must be composed of 10 ounces of original solution and 4 ounces of added water.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

An alternate approach:

In the original solution, the amount of alcohol = .14(10) = 1.4 ounces.
After the water is added, these 1.4 ounces of alcohol must constitute 10% of the final mixture:
1.4 = .1x
x = 14.
Since the volume of the final mixture is 14 ounces, and the volume of the original solution is 10 ounces, the volume of the added water = 14-10 = 4 ounces.
Last edited by GMATGuruNY on Tue Sep 15, 2015 3:23 am, edited 2 times in total.
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by misterholmes » Wed Apr 10, 2013 7:24 pm
Ola!

An average of 10% splits your two percentages in ratio 4:10, or 2:5 for simplicity's sake. Reverse this ratio and you've got your oz relationship, namely 5:2 of mix to water.

10/X=5/2

Or X=4

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by Rudy414 » Thu Apr 11, 2013 4:38 am
Since S:W = 10:4, the final mixture must be composed of 10 ounces of original solution and 4 ounces of added water.
This is what I do not understand. 10:4 is the ratio, not the amounts correct? We could say that the new solution is 10 parts original solution and 4 parts water, but how do we go from 10 parts and 4 parts to 10 oz and 4 oz, other than just throwing the units ounces because we know that is the unit?
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by misterholmes » Thu Apr 11, 2013 4:44 am
Yes you are right. It's just happens that the ratio and the quantities are equal. It's not that S:W is 10:4 and THEREFORE the quantities must be 10 oz to 4 oz. There is an intermediate step, which is algebra and involves setting up an equation (Units of S)/(Units of water)=10/4
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by Rudy414 » Thu Apr 11, 2013 5:20 am
Yes you are right. It's just happens that the ratio and the quantities are equal. It's not that S:W is 10:4 and THEREFORE the quantities must be 10 oz to 4 oz. There is an intermediate step, which is algebra and involves setting up an equation (Units of S)/(Units of water)=10/4
THANK YOU. That is exactly the piece I was missing and you explained it perfectly clear.
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by GMATGuruNY » Thu Apr 11, 2013 6:07 am
I like to use a RATIO BOX to calculate the actual amounts in the mixture.
A slightly more complex version of the problem posted above:
You currently have 28oz of a solution that is 14% alcohol and the rest is water. How much of the original solution must be REPLACED with pure water so that the resulting alcohol content is equal to 10%?
Once we have used alligation to determine that S:W = 10:4 = 5:2, set up the following box:

S--W--T
5--2-------RATIO
-----------MULTIPLIER
------28---ACTUAL VOLUMES

The rightmost column represent the TOTAL.
In the top row, we place the ratio (S:W = 5:2).
In the bottom row, we place the one known actual volume (total volume = 28 ounces).
Since in the top row 5+2 = 7, the total for the top row is 7.

S--W--T
5--2--7----RATIO
-----------MULTIPLIER
------28---ACTUAL VOLUMES

The middle row of the box is the MULTIPLIER.
Since under the total we get 28/7 = 4, the multiplier for the box is 4:

S--W--T
5--2--7----RATIO
4--4--4----MULTIPLIER
------28---ACTUAL VOLUMES

To complete the box, we multiply the values in the top row by the multiplier:

S--W--T
5--2--7----RATIO
4--4--4----MULTIPLIER
20-8--28---ACTUAL VOLUME

As indicated by the bottom row:
Since 5*4 = 20, there are 20 ounces of the original solution in the final mixture.
Since 2*4 = 8, there are 8 ounces of pure water in the final mixture.

Thus, of the original total volume of 28 ounces, 8 ounces must be replaced with pure water.
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