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Re: In What proportion must flour at $0.8 per pound be mixed [#permalink]

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06 Sep 2014, 02:10

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Using weighted average method: Let x be the proportion in which the $0.8 per pound flour is mixed with $0.9 per pound flour. Thus 0.8*x + 0.9*(1-x) = 0.825 0.9 – 0.1x = 0.825 0.1x = 0.075 x = 0.75

Thus ¾ of the mixture contains $0.8 per pound flour. Thus ratio of both flours is 3:1

If one has knowledge of weighted averages… the method by acegmat1 solves the questions in approx. 45 seconds…
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In What proportion must flour at $0.8 per pound be mixed with flour at $0.9 per pound so that the mixture costs $0.825 per pound?

A. 1:3 B. 1:2 C. 1:1 D. 2:1 E. 3:1

We can solve this using weighted averages

Weighted average of groups COMBINED = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let C = pounds of Cheap flour ($0.8 per pound) needed Let E = pounds of Expensive flour ($0.9 per pound) needed So, the TOTAL weight = C + E The proportion of Cheap Flour in the final mix = C/(C+E) The proportion of Expensive Flour in the final mix = E/(C+E)

Plug all values into formula to get: 0.825 = [C/(C+E)][0.8] + [E/(C+E)][0.9] Simplify: 0.825 = 0.8C/(C+E) + 0.9E/(C+E) Simplify: 0.825 = (0.8C + 0.9E)/(C+E) Multiply both sides by (C+E) to get: 0.825(C+E) = 0.8C + 0.9E Expand: 0.825C + 0.825E = 0.8C + 0.9E Subtract 0.8C from both sides: 0.025C + 0.825E = 0.9E Subtract 0.825E from both sides: 0.025C = 0.075E This is the same as: 25C = 75E And this is the same as C = 3E

This tells is that C is 3 TIMES the value of E So, the ratio C : E = 3 : 1