swerve wrote:In what proportion must flour at $0.8 per pound be mixed with flour at $0.9 per pound so that the mixture cost $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
Answer: E
For more information on weighted averages, you can watch this video:
https://www.gmatprepnow.com/module/gmat- ... ics?id=805
Here are some additional practice questions related to weighted averages:
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https://www.beatthegmat.com/weighted-ave ... 17237.html
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https://www.beatthegmat.com/weighted-ave ... 14506.html
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https://www.beatthegmat.com/average-weig ... 57853.html
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https://www.beatthegmat.com/averages-que ... 87118.html
Cheers,
Brent
