Use scale method. First thing to do is to find out the ratio in which the two mixes were mixed together.

Sue's mix had 30% nuts and Jane's had 60%. Overall, it had 50% nuts

WSue/WJane = (60 - 50)/(50 - 30) = 1/2

So if Sue's mix was 100 gms (30 gm nuts, 70 gms dried fruit), it was mixed with 200 gms of Jane's mix.
So out of 300 gms of total mix, there was 70 gms of dried fruit i.e. (70/300) * 100 = 23.33%

Answer (B)
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Sol: Given that new mixture is 50 % nuts which is (50-30 =20) away from Sue's trail and (60-50=10) away from Jane's trail
Therefore, the ratio of Sue and Jane's mixture will be 1:2.

So if Sue's mixture is 100 gms then Jane's mixture will be 200 gms

From the table we can see that DF is 70 gm of Sue's mixture which is nothing 70/300*100 = 23.33%

Ans is A
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“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Another Approach:

Whenever one part of the mixture is changed by mixing track that part. This is indirect approach which leads to same solution as through alligation.

Take X from 1 mixture
Take Y from 2 mixture

{0.3X + 0.6Y}/(X+Y) = 1/2

X:Y = 1:2

30+70+120+80=300

70/300

I am using the following approach which I saw in similar problems but it seems that in no way it helps! Could anyone please tell what is wrong in the approach? Maybe I am committing a silly mistake.

N - F - Ch
30 70 0
60 0 40

Combining:

N F Ch
1800 4200 0
1800 0 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 you please tell? Thanks!

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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Bunuel Can you please link to similar questions? Thank you so much!