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# Mixture A is 15% alcohol and Mixture B is 50% alcohol. If

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Mixture A is 15% alcohol and Mixture B is 50% alcohol. If [#permalink]  23 Jul 2004, 17:52
Mixture A is 15% alcohol and Mixture B is 50% alcohol. If two are poured together to create a 4gallon mixture having 30% alcohol, approximately how many gallons of Mixture A are in the mixture.

Please provide xplanations. ( Ans is 3 gallon)
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Closest i came to is:

Let A be gallons of solution A poured in
then (4-A) would be the gallons of solution B poured in

0.15A + 0.5(4-A)=0.3A

A = 2.28

Not yet 3, but close. I can't think of any other way to equate the problem.
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I got the same with the same procedure detailed by ywilfred. Infact, A = 0.8/0.35 = 2.3 (approx)

Please check with the answers again. Do let us know if we are wrong.

ywilfred wrote:
Closest i came to is:

Let A be gallons of solution A poured in
then (4-A) would be the gallons of solution B poured in

0.15A + 0.5(4-A)=0.3A

A = 2.28

Not yet 3, but close. I can't think of any other way to equate the problem.

_________________

Awaiting response,

Thnx & Rgds,
Chandra

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I got 2.28 as well.

Since the question says "approximately ", the other answer choices must be very far from 3...
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One other approach would be to solve as detailed below..

If 15% Alcohol and a 50% alcohol mixture need to give a resultant mixture that has 30% Alcohol, then the ratio of Mixture A to Mixture B would be 4:3. Let me explain how 4:3..

30-15 = 15
50-30 = 20

or 20:15 = 4:3.

So the amount of Mixture A would be 4/7 * 4(gallons) = 16/7=2.285 or approx 3 gallons.
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I am also getting 2.28 ( approx 2.3). 2.3 is one of the choices but the correct answer is ( as per the author) is 3.

The explanation provided is

Since 15% is closer to 30% than 50% is to 30% hence A should have a larger content in the mixture.

Assume 2.5( one of the choices) as A. B would then be 1.5
15% of 2.5= .375(alcohol in A)
50% of 1.5= .75(alcohol in B)
Total is .375+.75= 1.125.

Alcohol content in the mixture is 30% of 4=1.2

Hence, the A should be more than 2.5. And to add to my woes, 3 is the only higher choice.

I need to know whether this is a correct explanation or not. I am confused.

Shyam
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pawars wrote:
I am also getting 2.28 ( approx 2.3). 2.3 is one of the choices but the correct answer is ( as per the author) is 3.

The explanation provided is

Since 15% is closer to 30% than 50% is to 30% hence A should have a larger content in the mixture.

Assume 2.5( one of the choices) as A. B would then be 1.5
15% of 2.5= .375(alcohol in A)
50% of 1.5= .75(alcohol in B)
Total is .375+.75= 1.125.

Alcohol content in the mixture is 30% of 4=1.2

Hence, the A should be more than 2.5. And to add to my woes, 3 is the only higher choice.

I need to know whether this is a correct explanation or not. I am confused.

Shyam

Hi pawars,
What the answer explains directly relates to the equations we have been using:

0.15A + 0.5(4-A) = 0.3(4)

Your example of setting A to 2.5 and then B at 1.5 is the same as setting value of A using an algeraic term and then B as 4 minus that amount.

Now, getting to the choice of 3 gallons.

If A is 3 gallons, then B would be 1 gallon.

3 gallons of A provide 0.15*3=0.45 gallons of alcohol
1 gallon of B provides 0.5*1=0.5 gallons of alcohol

Adding them up gives 0.95 gallons, off the mark from the expected 0.3*4=1.2 gallons of alcohol. (Difference is 0.25)

Given a choice based on how far they deviate from the the desired value, I would go for 2.3 since this value gives 1.195 gallons of alcohol (that's off from the 30% by only 0.005 gallons) compared to 2.5 which gives 1.125 (off by 0.075) and the worse is probably 3 gallons.

I feel the way to solve such mixture problems is still the equate the terms. When you say left side = right side, you can't go wrong. So I reckon the answer is not correct this time round.

I'm still preparing for my first GMAT, so maybe someone around here might have experienced such problems on an actual GMAT. Have anyone seen such a question ?
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This question is from the PR1. I think the answer is not correct. I too feel that we should go by equations for solving such questions.

If anyone has an explanation for 3, please view it.

Thanks.
Shyam
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