Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai

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

Using alligation:

30 %.................60 %

...............50 %

10......................20

1 : 2

Hence both the mixtures are mixed in 1 : 2.

Let us say 10 grams of Suel's and 20 grams of Jane

Dried Fruit = 7

Total = 30

Percentage= 7/30 = 23.33 %

If you want to learn alligation you can watch the video by clicking on the links in the signature.

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

.3S+.6J=.5(S+J)
S/J=1/2
S/(S+J)=1/3
(1/3)(.7)=.2333=23.33%

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!

The first thing you should notice is that the combined mixture is 50% and Sue is at 30%. We can quickly see what would happen if the mixture was 1/1. It would be 0.3 + 0.6 then divided by 2 or 45%.

In order to get to 50% we must have more of Jane's mixture then Sue. With this in mind, it will make it easier to solve. We are going to figure out the proportion of S mixture to J= jane mixture. In my chart I used N1=Sue mixture amount and N2= Jane

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

Hi All,

This question can be solved with a bit of algebra:

Sue's Mix: 30% nuts and 70% dried fruit
Jane's Mix: 60% nuts and 40% chocolate

Combined mix is 50% nuts…so we can create the follow equation:

X = number of servings of Sue's Mix
Y = number of servings of Jane's Mix

(.3X + .6Y)/(X + Y) = .5

.3X + .6Y = .5X + .5Y

.1Y = .2X

Y = 2X

This means that the mixture is "1 part" Sue's Mix and "2 parts" Jane's Mix, with a Total of 3 parts.

The question asks what percent of the Total Mix is dried fruit….

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

Since .75/3 = 25%…

.7/3 has to be a little less than 25%

Final Answer:

GMAT assassins aren't born, they're made,
Rich

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

IMO the best way to solve this question is to use allegation.

ratio of Sue : Jane's trail mix must be in a 1:2 ratio.

That means total will be 100% Sue and 200% Jane, total mixture is 300% original size.

Sue's mixture has 70% dried fruit so \(\frac{70}{300} = 23.33%\)

Let Sue's mixture be x, and Jane's mixture be y.
Nuts %age => 0.3x+0.6y/(x+y) * 100 = 50
This simplifies into y=2x
Dry fruits %age => 0.7x/x+y * 100
=> 0.7x/3x * 100 = 23.33

Posted from my mobile device

See image for Step 1.
Note: we assumed a total weight of 100 units for step 1.

Step 2 - Determine the amount that Sue's mix contributes to final mix i.e. determine x

0.3x +0.6(100-x)= 50
0.3x+60-0.6x=50
-0.3x=-10
3x/10=10
3x=100
x=100/3 (this is the weight of Sue's mix)

Step 3 - determine the amount of fruit
Since we know the weight of Sue's mix, we can use the proportion given to us.
Fruit = 70%

7/10 * 100/3
=70/3
=23 1/3

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 1/3

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) = 23 1/3

So, of the 100 pounds of combined mix, there are 23 1/3 pounds of dried fruit.
In other words, 23 1/3% of the combined mixture is dried fruit?

Answer: B

Cheers,
Brent

60 - 50 / 50 - 30 = 10/20 = 1/2

Therefore the combined mix is 33% Sue's mix and 66% Jane's mix.

Out of Sue's 33% mix, 70% is dried fruit.

Therefore 23.3% contains dried fruit. Answer is B.

Lets let x = fraction of trail mix that is sues, and 1-x = fraction of trail mix that is janes, and we will convert everything to decimals in our final calculation

then 30X + 60(1-x) = 50. Everything in our equation is in terms of percents so looks good

so -30x = -10 and x = 1/3. So we then needs to take .33 of .7 which gives .2331, which is closest to B

OA is B

We have established that the ratio of Jane's trail mix (J) to Sue's trail mix (S) is 2:1.

Now, let's determine the overall composition of the combined mixture:

For nuts: Since Jane's mix has twice as many nuts as Sue's mix (60% vs. 30%), the overall nuts in the combined mixture will be weighted more toward Jane's mix. It will be more than 50%, but we can't determine the exact percentage with the given information.

For dried fruit: Since Sue's mix is 70% dried fruit (which is higher than Jane's mix), the overall dried fruit in the combined mixture will also be weighted more toward Sue's mix. It will be more than 50%, but we can't determine the exact percentage with the given information.

Given the information provided, we can conclude that the percent of the combined mixture that is dried fruit is more than 50%.

So, if the answer is indeed among the options given, then option (B) "23.33%" would be the closest approximation. However, it's important to note that we cannot calculate the exact percentage based on the information provided. If option (B) is considered the correct answer, it is an approximation.

As I like to do with ratio problems, I like to represent them as equations if possible because it makes it easier to visualize and solve. Firstly, we know the percentages of nuts in both the individual and the combined mixtures. Moreover, we also know that Sue's the only one who had any dried fruit. Thus, if we can figure out how much total amount we had from combining them, and then taking Sue's fruit amount from that, we will have the answer.

Let's say Sue's mixture amount is \(s\) and Jane's is \(j\). We can represent the nuts ratio as an equation:

\(\frac{.3s + .6j}{s+j} = 0.5\)

We can then do some variable manipulating to get the following:

\(.3s + .6j = .5s + .5j\)
\(.1j = .2s\)
\(j = 2s\)

Thus, we now know that Jane had twice as much trail mix as Sue. Therefore, we know that in terms of \(s\), the total amount of trail mix in the combined mixture was \(s + 2s = 3s\). And since we know that the dried fruit amount was \(.7s\), we can now construct our ratio of fruit to total as:

\(\frac{.7s}{3s} = \frac{.7}{3} = \frac{7}{30} = 0.233333... = 23.3333...\)%

Hence our answer is B