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by srcc25anu » Tue Apr 30, 2013 4:52 am
Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts and 40% chocolate chips. If the combined mixture of Sue and Jane's trails mix contains 50% nuts, what percent of the combined mixture is dried fruit?

A. 16.67%
B. 23.33%
C. 25%
D. 33.33%
E. 36.67%

[spoiler]OA: B[/spoiler]
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by Brent@GMATPrepNow » Tue Apr 30, 2013 6:28 am
srcc25anu wrote:Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts and 40% chocolate chips. If the combined mixture of Sue and Jane's trails mix contains 50% nuts, what percent of the combined mixture is dried fruit?

A. 16.67%
B. 23.33%
C. 25%
D. 33.33%
E. 36.67%

[spoiler]OA: B[/spoiler]
Here's one approach.
Let's say that the combined mixture weighs 100 pounds.
If the combined mix is 50% nuts, then there are 50 pounds of nuts in the combined mixture.

Let's let x = the # of pounds of Sue's trail mix
This means that 100-x = the # of pounds of Jane's trail mix

My word equation: (# of pounds of nuts in Sue's portion) + (# of pounds of nuts in Jane's portion) = 50

In other words: (30% of x) + (60% of 100-x) = 50
Rewrite as: 0.3x + 0.6(100-x) = 50
Expand: 0.3x + 60 - 0.6x = 50
Simplify: -0.3x = -10
Solve: x = 10/0.3 = 100/3 = 33.3333..

In other words, the combined (100-pound) mixture contains 33 1/3 pounds of Sue's trail mix.
If Sue's trail mix is 70% dried fruit, then the total weight of dried fruit = (0.7)(33 1/3) = [spoiler]23 1/3[/spoiler]
So, of the 100 pounds of combined mix, there are [spoiler]23 1/3 [/spoiler]pounds of dried fruit.
In other words, [spoiler]23 1/3%[/spoiler] of the combined mixture is dried fruit?

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Tue Apr 30, 2013 7:59 am
srcc25anu wrote:Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts and 40% chocolate chips. If the combined mixture of Sue and Jane's trails mix contains 50% nuts, what percent of the combined mixture is dried fruit?

A. 16.67%
B. 23.33%
C. 25%
D. 33.33%
E. 36.67%

[spoiler]OA: B[/spoiler]
Nut percentage in Sue's mix: 30%.
Nut percentage in Jane's mix: 60%.
Nut percentage in the mixture: 50%.

Let S = Sue's mix and J = Jane's mix.
To determine the ratio of S to J, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for S and J on the ends and the percentage for the mixture in the middle.
S 30%-----------50%-----------60% J

Step 2: Calculate the distances between the percentages.
S 30%----20----50%----10-----60% J

Step 3: Determine the ratio in the mixture.
The required rate of S to J is equal to the RECIPROCAL of the distances in red.
S:J = 10:20 = 1:2.

Since S:J = 1:2, let S = 100 units and J = 200 units, for a total of 300 units.
Since S is 70% dried fruit, the amount of dried fruit = .7S = .7(100) = 70 units.
Thus:
Percent of dried fruit = (dried fruit)/(total) * 100 = 70/300 *100 = 70/3 %.

The correct answer is B.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html
Last edited by GMATGuruNY on Tue Jun 11, 2013 8:08 am, edited 1 time in total.
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by zazoz » Tue Apr 30, 2013 11:55 am
I have a question about this kind of questions.

I personally solved this question by picking a smart number for Sue mix, and solve for other mix and reached to the answer. But here is my question, why cant we pick two smart numbers simultaneously? say, pick 100 for each of them? I know that lead to nowhere, but can you explain what is the logic behind this wrong approach?

thanks in advance
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