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yet another ratios/mixture problem...

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yet another ratios/mixture problem...

by topspin360 » Sun Jul 08, 2012 8:57 pm
but i assure you i'm getting close to mastering this topic.

please provide some insight on why my method doesn't work... again, if there is scalable methodical way to do it, that'd be great.

why cant i do the following: since original alcohol is 24%, originally water is 76%.
so, 76x/100 + 200 = 84x/100 (since 1/3 of 24 is 8, add that to water concentration). solve for x.

If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?
180 grams
220 grams
250 grams
350 grams
400 grams
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by topspin360 » Sun Jul 08, 2012 9:16 pm
i also just tried another method that doesn't work either:

24x/(76x+200) = 16/84. solve for x.
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by Anurag@Gurome » Sun Jul 08, 2012 9:57 pm
topspin360 wrote:If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?
Say, x grams of 24%-solution was used.
Hence, amount of alcohol in x grams of 24%-solution = 0.24x grams

Now, concentration of new solution = 24% - 1/3 of 24% = 24% - 8% = 16%
Hence, amount of alcohol in (x + 200) grams of 16%-solution = 0.16(x + 200) grams

So, 0.16(x + 200) = 0.24x
--> (0.24x - 0.16x) = (0.16)*(200)
--> 0.08x = 32
--> x = 32/(0.08) = 400

The correct answer is E.
topspin360 wrote:why cant i do the following: since original alcohol is 24%, originally water is 76%.
so, 76x/100 + 200 = 84x/100 (since 1/3 of 24 is 8, add that to water concentration). solve for x.
The amount of new mixture is (x + 200) grams.
Hence, amount of water in the new mixture is 84% of (x + 200) grams = 0.84(x + 200)
Hence, the correct equation will be: 76x/100 + 200 = 84(x + 200)/100
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by eagleeye » Sun Jul 08, 2012 11:53 pm
topspin360 wrote:but i assure you i'm getting close to mastering this topic.

please provide some insight on why my method doesn't work... again, if there is scalable methodical way to do it, that'd be great.

why cant i do the following: since original alcohol is 24%, originally water is 76%.
so, 76x/100 + 200 = 84x/100 (since 1/3 of 24 is 8, add that to water concentration). solve for x.

If after 200 grams of water were added to the 24%-solution of alcohol, the strength of the solution decreased by one-third, how much of the 24%-solution was used?
180 grams
220 grams
250 grams
350 grams
400 grams
If you are stuck or unsure, you can do it two ways. One is plugging in numbers, and checking.
The other one is algebra.
Let me describe the algebraic approach first:
(This approach is scalable to many a problem)
We need to find the weight of 24% solution.

Step 1:
Assume the required variable to be x. So let x be the weight of 24% solution.

Step 2:
Write down the expression for the new values.
Here we are told that 200 grams of water was added.
So new weight of solution = 200 + x
The amount of alcohol remains unchanged in the two.

Step 3:
Apply the constraint, or relationship.

We are told that the strength of the old solution decreased by 1/3 on addition of water. Since it decreased "BY" 1/3, it decreases "TO" 2/3.

So (new amount of alcohol)/(new amount of solution) = 2/3*(old amount of alcohol)/(old amount of solution). Since alcohol amount remains unchanged, it will cancel out on both sides of the equation.
Then:

1/(x+200) = 2/3*(1/x) => 2x+400 = 3x => x=400.
Couple of things are to be seen from this.

A. Not all data is relevant. The fact that alcohol was 24% has no bearing on the final answer.
B. A stepwise approach can sometimes be faster.

Anyways, these general steps may be helpful. Good luck !
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