SOLUTION: 14 oz.of mixed nuts containing 15% peanuts were mixed with 16 oz.of another kind of mixed nuts that contain 60% peanuts.Peanuts are what percent of the new mixture?

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Question 1203426: 14 oz.of mixed nuts containing 15% peanuts were mixed with 16 oz.of another kind of mixed nuts that contain 60% peanuts.Peanuts are what percent of the new mixture?

Found 3 solutions by Theo, greenestamps, josgarithmetic:
Answer by Theo(13342)   (Show Source):
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14 ounces of mixed nuts type 1 + 16 ounces of mixed nuts type 2 = 30 ounces of mixed nuts type 3, where type 3 = type 1 + type 2 combined.

.15 * 14 + .6 * 16 = 2.1 + 9.6 = 11.7 ounces of peanuts in 30 ounces of type 3 mixture.

11.7 / 30 = .39 * 100 = 39% of peanuts in 30 ounces of mixed nuts type 3.

since type 3 is the new mixture, your solution is 39% peanuts in the new mixture.

Answer by greenestamps(13200)   (Show Source):
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Here is an alternative to the standard formal algebraic solution for solving 2-part mixture problems like the one shown by the other tutor.

For this problem, it is not an efficient method; but for many similar problems it is faster and easier than formal algebra.

The fraction of the mixture that is 60% peanuts is 16/(14+16) = 16/30 = 8/15.

That means the percentage of peanuts in the mixture is 8/15 of the way from 15% to 60%.

From 15% to 60% is a difference of 45%; 8/15 of 45% is 24%. The percentage of peanuts in the mixture is 15%+24% = 39%.

ANSWER: 39%

Answer by josgarithmetic(39617)   (Show Source):
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Account for the peanuts for the two different mixtures.

Compute the expression; that is the percentage peanuts for the new mixture.