Stuck with Mixture problem-Help please

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Stuck with Mixture problem-Help please

by aaggar7 » Tue Jun 11, 2013 7:13 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%

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by [email protected] » Tue Jun 11, 2013 7:39 am
aaggar7 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%
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

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by GMATGuruNY » Tue Jun 11, 2013 8:09 am
aaggar7 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%
Sue's nut percentage: 30%.
Jane's nut percentage: 60%.
Nut percentage in the mixture: 50%.

The following approach is called 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% W

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 ratio 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 pounds and J = 200 pounds, for a total of 300 pounds.
Since 70% of Sue's mix is dried fruit, the amount of dried fruit = .7(100) = 70 pounds.
Thus:
Dried fruit percentage = (dried fruit)/(total) * 100 = 70/300 * 100 = 70/3 = 23.33%.

The correct answer is B.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html
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by ngalinh » Thu Jun 13, 2013 1:29 pm
GMATGuruNY wrote: The following approach is called 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% W

Step 2: Calculate the distances between the percentages.
S 30%----20-----50%----10-----60% J
Thanks Mitch, this method is awesome!

Step 3 I would do like this:
d=0.7s; j=2s --> d/(j+s) = 0.7s/3s =23.33 (%)

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by RBS » Wed Nov 02, 2016 7:32 am
Dear GMATGuruNY,

I am using the following approach which I saw in similar problems but it seems that in no way can reach to the correct solution. Could you please help me by saying what is wrong in the approach?

N F Ch
30 70
60 40

Combining:

N F Ch
1800 4200
1800 1200
-----------------
1800 4200 1200

Simplifying:

N F Ch
= 3 7 2

Asuming 100 g of total mixture:
12*x= 100 --> x= 100/12

Total g of F --> (100/12) * 7

The question ask for (F/total mixture). Hence:

(100/12)*7 / 100 = 7/12= 0.583 --> in percentage = 58.3%


Could yooy please tell? Thanks!

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by [email protected] » Tue Nov 08, 2016 4:28 pm
aaggar7 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%
To start we can define a few variables:

x = the total amount of Sue's trail mix, and y = the total amount of Jane's trail mix

We are given that Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts and 40% chocolate chips. We can represent this below:

Nuts in Sue's trail mix = 0.3x

Dried fruit in Sue's trail mix = 0.7x

Nuts in Jane's trail mix = 0.6y

Chocolate chips in Jane's trail mix = 0.4y

We are also given that when the two trail mixes are combined, the resulting trail mix will contain 50% nuts. Since nuts in Sue's trail mix = 0.3x, nuts in Jane's trail mix = 0.6y, and the total weight of the two trail mixes is x + y, we can create the following equation:

(0.3x + 0.6y)/(x + y) = 50% = 0.5

0.3x + 0.6y = 0.5x + 0.5y

0.1y = 0.2x

y = 2x

Now we can determine what percent of the combined mixture is dried fruit. Since only Sue's mixture contains dried fruit, we know that the only dried fruit in the mixture is 0.7x. We also know that the total weight of the mixture is x + y, so we can create the following ratio:

0.7x/(x + y)

0.7x/(x + 2x)

0.7x / 3x

0.7/3 = 0.2333...

So, the combined trail mix contains approximately 23.33% dried fruit.

Answer: B

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by [email protected] » Fri Nov 11, 2016 3:02 pm
RBS wrote:Dear GMATGuruNY,

I am using the following approach which I saw in similar problems but it seems that in no way can reach to the correct solution. Could you please help me by saying what is wrong in the approach?

N F Ch
30 70
60 40

Combining:

N F Ch
1800 4200
1800 1200
-----------------
1800 4200 1200
I like the idea, but I see two slight mistakes:

1) The combined nuts should be 1800 + 1800, or 3600.

2) The combined mixture should be 50% nuts. You've got Combined Nuts / Combined Total = 3600 / (3600 + 4200 + 1200), which isn't .5.