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

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

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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%
[Reveal] Spoiler: OA

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Originally posted by HarveyS on 08 Apr 2014, 16:08.
Last edited by ENGRTOMBA2018 on 11 Aug 2015, 11:32, edited 1 time in total.
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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Let Sue's mix = a

Nuts \(= \frac{30a}{100}\)

Dry fruit \(= \frac{70a}{100}\)

Let Janes mix = b

Nuts \(= \frac{60b}{100}\)

Chocolate \(= \frac{40b}{100}\)

Combined mixture of Sue and Jane's trails mix contains 50% nuts

\(\frac{30a}{100} + \frac{60b}{100} = \frac{50}{100} (a+b)\)

3a + 6b = 5a + 5b

2a = b

It means quantity of Jane is twice that of Susan. Lets take numbers to solve this further

If Sue's mix = 10 gms, then Jane's mix = 20 Gms

Total = 10+20 = 30 Gms

Quantity of Dry fruits = 70% = 7 Gms

Percentage \(= \frac{7}{30} * 100 = 23.33%\)

Answer = B
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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Mountain14 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?

16.67%
23.33%
25%
33.33%
36.67%


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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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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Mountain14 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?

16.67%
23.33%
25%
33.33%
36.67%


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.
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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Mountain14 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?

16.67%
23.33%
25%
33.33%
36.67%


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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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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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
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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.3S+.6J=.5(S+J)
S/J=1/2
S/(S+J)=1/3
(1/3)(.7)=.2333=23.33%
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Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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New post 02 Nov 2016, 08:59
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!
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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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
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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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New post 07 Dec 2017, 07:36
Expert's post
HarveyS 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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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai [#permalink]

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New post 14 Mar 2018, 19:57
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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:
[Reveal] Spoiler:
B


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Re: Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trai   [#permalink] 14 Mar 2018, 19:57
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