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Mixture Problem

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by GMATGuruNY » Thu Oct 28, 2010 5:51 pm
N:Dure wrote:Need help understanding how to solve it
A great way to handle ratio problems is to use a ratio box. Here's what the box looks like:

__________A____W____Total

Ratio:_____3____5_____8

Multiplier:

Actual:____4

In the top row, we put the ratio of A:W = 3:5.
The last column is the total: 3+5 = 8.
In the bottom row, we put any known values. The amount of alcohol is not changing, so we know that there are 4 quarts of alcohol in the resulting mixture.



To determine the middle row -- the multiplier -- we divide the known actual value (alcohol = 4) by its corresponding value in the top row (3): 4/3. The multiplier is placed in every column along the middle row:

__________A____W____Total

Ratio:_____3____5______8

Multiplier:__4/3__4/3____4/3

Actual:____4



Now we can complete the box by multiplying every value in the top row by the multiplier:

__________A_____W______Total

Ratio:_____3_____5_______8

Multiplier:__4/3__4/3_____4/3

Actual:____4____20/3____32/3


So the actual amount of water in the resulting mixture = 20/3 quarts.
Since we started with 4 quarts, 20/3 - 4 = 8/3 quarts must be added.

The correct answer is D.
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by N:Dure » Thu Oct 28, 2010 7:46 pm
Thank you for explaining Mitch, appreciate your help!
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by Rahul@gurome » Thu Oct 28, 2010 10:59 pm
An easy way to comprehend this kind of ratio-proportion problem is to analyze the situations before and after making the changes. Let's see how can we do that!

Before mixing water:

Amount of water = 4 quarts
Amount of alcohol = 4 quarts
Amount of mixture = (4 + 4) quarts = 8 quarts

Say, x quarts of water is mixed. Now,

Amount of water = (4 + x) quarts
Amount of alcohol = 4 quarts
Amount of mixture = (4 + 4 + x) quarts = (8 + x) quarts

According to the question,
.... (4 + x)/(8 + x) = 5/(3 + 5)
=> 8*(4 + x) = 5*(8 + x)
=> 32 + 8x = 40 + 5x
=> 3x = 8
=> x = 8/3

The correct answer is D.
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