Sue's trail mix is 30% nuts and 70% dried fruit. Jane's trail mix is 60% nuts

[BTGModeratorVI]

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%

Answer: B
Source: Veritas Prep

[Brent@GMATPrepNow]

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%

Answer: B
Source: Veritas Prep

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

[swerve]

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%

Answer: B
Source: Veritas Prep

Let Sue's mix \(= a\)

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

Dry Fruit \(=\dfrac{70a}{100}\)

Let Jane mix \(= b\)

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

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

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

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

\(3a + 6b = 5a + 5b\)

\(2a = b\)

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

If Sue's mix \(= 10 g\), then Jane's mix \(= 20 g\)

Total \(= 10+20 = 30 g\)

Quantity of Dry fruits \(= 70\% = 7 g\)

Percentage \(=\dfrac{7}{30} \cdot 100 = 23.33 \%\)

Therefore, B

[Scott@TargetTestPrep]

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%

Answer: B

Solution:

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