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Mixtures

by singhmaharaj » Tue Apr 29, 2014 9:26 am
A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2 percent of the gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x?

1) y = 10

2) z = 16
I need some generalized method of solving mixture problems
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by Matt@VeritasPrep » Tue Apr 29, 2014 5:56 pm
singhmaharaj wrote:A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2 percent of the gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x?

1) y = 10

2) z = 16
I need some generalized method of solving mixture problems
The general method:

Suppose a% of A is x, b% of B is x, and c% of (A + B) is x. We then have the equation

(a/100)A + (b/100)B = (c/100)(A+B)

or

(aA + bB)/100 = (cA +cB)/100

or

A(a-c) = B(c-b)

or

A/B = (c-b)/(a-c)

Using that approach on this equation, we have

(10% of x) + (2% of y) = (5% of z)

The wording is sloppy, but I'll assume that the question means to say x + y = z (in other words, that the mixture is exactly x tons + y tons, with nothing added or removed). In that case, we have

.1x + .02y = .05z = .05(x+y)

or

10x + 2y = 5(x+y)

or

5x = 3y

S1 tells us y = 10, so 5x = 3*10, or x = 6. SUFFICIENT.

S2 tells us x + y = 16, so 5x = 3(16-x), or x = 6. SUFFICIENT.
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by Tushar14 » Tue Apr 29, 2014 6:56 pm
Using scale method here, since 10% and 2% give weighted average of 5%, x:y = 3:5
We also know x + y = z.

1. y = 10.
If y = 10, x = 6 since their ratios must be 3:5. Sufficient.

2. z = 16
If sum of x and y is 16, x must be 6 and y must be 10 to give a ratio of 3:5.
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by Tushar14 » Tue Apr 29, 2014 6:57 pm
Set the equation: 0.1x+0.02y=0.05(x+y), where x+y=z --> 5x=3y -->

(1) y=10 --> 5x=3y=30 --> x=6. Sufficient.

(2) z=x+y=16 --> y=16-x --> 5x=3y=3(16-x) --> x=6. Sufficient.
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by GMATGuruNY » Wed Apr 30, 2014 2:43 am
singhmaharaj wrote:A contractor combined x tons of a gravel mixture that contained 10% gravel G, by weight, with y tons of a mixture that contained 2 percent of the gravel G, by weight, to produce z tons of a mixture that was 5 percent gravel G, by weight. What is the value of x?

1) y = 10

2) z = 16
I need some generalized method of solving mixture problems
x = 10% gravel
y = 2% gravel.
z = the MIXTURE of x and y = 5% gravel.

To determine the required ratio of x to y, use ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the percentages for x and y on the ends and the percentage for mixture z in the middle.
x 10%-----------5%-----------2% y

Step 2: Calculate the distances between the percentages.
x 10%-----5-----5%----3-----2% y

Step 3: Determine the ratio in the mixture.
The required ratio of x to y is equal to the RECIPROCAL of the distances in red.
x:y = 3:5.

Since x:y = 3:5, and 3+5 = 8, every 8 tons of mixture z is composed of 3 tons of x and 5 tons of y.

Statement 1: y=10
Since x:y = 3:5 = 6:10, x=6.
SUFFICIENT.

Statement 2: z=16
Since x:y = 3:5 = 6:10, and 6+10 = 16, the 16 tons of mixture z must be composed of 6 tons of x and 10 tons of y.
SUFFICIENT.

The correct answer is D.

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