Mixture problem

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Mixture problem

by Mo2men » Sun Jan 22, 2017 6:28 pm
A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?

1.Total quantity of the resulting solution is 350 mililiters.
2.The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.

Source: e-gmat

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by Jay@ManhattanReview » Sun Jan 22, 2017 10:07 pm
Mo2men wrote:A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?

1.Total quantity of the resulting solution is 350 mililiters.
2.The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.

Source: e-gmat
We are given Alcohol : Water = 7 : 3

Say Alcohol = 7x milliliters, water = 3x milliliters, and added water = y milliliters

Thus, after mixng additonal water, we have

Alcohol : Water = 7x : (3x+y) = 60% : 40% = 3 : 2

7x/(3x+y) = 3/2 ...(1)

We have to get the value of y.

S1: Total quantity of the resulting solution is 350 milliliters.

=> 7x+3x+y = 350
=> 10x+y = 350 ...(2)

Solving (1) and (2), we get the unique value of y = 50 milliliters. Sufficient.

S2: The original solution contains 10.5 milliliters of alcohol for every 4.5 milliliters of water.

It is a rephrased version of the information we already know.

The statement means Alcohol : Water = 10.5 : 4.5 = 105 : 45 = 7 : 3. It cannot help us. Insufficient.

Answer: A

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by GMATGuruNY » Mon Jan 23, 2017 3:47 am
Mo2men wrote:A solution consists of only water and alcohol such that the ratio of alcohol to water in the solution is 7:3. How much amount of water should be added to the solution (in mililiters) so that the resulting solution contains 60% alcohol?

1.Total quantity of the resulting solution is 350 mililiters.
2.The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.

Let W = the pure water, S = the original solution, and M = the mixture.

The following approach is called ALLIGATION -- a very good way to handle MIXTURE PROBLEMS.
Alligation can be performed only with PART TO WHOLE RELATIONSHIPS (percents or fractions).

Step 1: Convert any ratios to PERCENTS.
W:
Here, alcohol/total = 0%.
S:
Since alcohol/water = 7:3, and 7+3=10, alcohol/total = 7/10 = 70%.
M:
Here, alcohol/total = 60%.

Step 2: Plot the 3 percentages on a number line, with the percentages of W and S on the ends and the percentage for M in the middle.
W 0-----------------60-------------------70 S

Step 3: Calculate the distances between the percentages.
W 0-------60--------60---------10--------70 S

Step 4: Determine the ratio in the mixture.
The ratio of W to S in the mixture is equal to the RECIPROCAL of the distances in red.
W:S = 10:60 = 1:6.

Since W:S = 1:6, for every 7 liters of mixture, 1 liter of pure water must be combined with 6 liters of original solution.
Thus, the added pure water must constitute 1/7 of the final mixture.

Statement 1:
W = (1/7)(350) = 50mm.
SUFFICIENT.

Statement 2:
Thus, alcohol to water in the original solution = (10.5)/(4.5) = 21/9 = 7/3.
This information is given in the prompt.
INSUFFICIENT.

The correct answer is A.

For two other problems that I solved with alligation, check here:

https://www.beatthegmat.com/ratios-fract ... tml#484583
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by [email protected] » Mon Jan 23, 2017 10:57 am
Hi Mo2men,

This question can be solved without doing much math at all (as long as you recognize the concepts involved).

We're told that a solution has a ratio of alcohol to water of 7:3. This means that for every 7 'measures' of alcohol, there are 3 'measures' of water (for example, 7 ounces of alcohol and 3 ounces of water). By extension, this means that the solution is currently 70% alcohol and 30% water. We're asked how many milliliters of water must be added so the solution becomes 60% alcohol and 40% water.

To properly answer this question, we need to know ANY of the individual measures in the solution (total alcohol OR total water OR the overall total) - then we can figure out all of the other numbers and answer the question.

1) Total quantity of the resulting solution is 350 mililiters.

With this Fact, we have the TOTAL volume, so we can figure out the amount of alcohol and the amount of water. With those numbers, we COULD calculate how much water would need to be added (but we don't have to do that work here).
Fact 1 is SUFFICIENT

2) The original solution contains 10.5 mililiters of alcohol for every 4.5 mililiters of water.

This Fact is interesting - it gives us a RATIO of alcohol to water.... but we were already given that ratio. At first glance, this is almost certainly 'redundant' information, but we can prove it with a little work.

10.5: 4.5 =
21:9 =
7:3

This is information that we already had from the prompt, so simply repeating it does not provide anything new - and as such, it will not help us to answer the question.
Fact 2 is INSUFFICIENT

Final Answer: A

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