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In What proportion must flour at $0.8 per pound be mixed

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

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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

Source: Optimus Prep
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Intern
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Concentration: Marketing, Entrepreneurship
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Re: In What proportion must flour at $0.8 per pound be mixed [#permalink]

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New post 16 Aug 2014, 21:15
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bytatia wrote:
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

Source: Optimus Prep


:barbarian :barbarian :barbarian

IMO E
Ratio :- 0.075/0.025 = 3:1
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Re: In What proportion must flour at $0.8 per pound be mixed [#permalink]

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New post 06 Sep 2014, 03:10
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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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Re: In What proportion must flour at $0.8 per pound be mixed [#permalink]

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New post 23 Oct 2014, 22:17
Differential approach is my favourite in this type:

0.8----0.825-----------0.9

No need any counting, 0.825 is three times closer 0.8 than 0.9. So, the ratio is 3:1

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

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New post 14 Jan 2018, 09:11
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bytatia wrote:
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

Answer: E

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

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New post 14 Jan 2018, 09:34
bytatia wrote:
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

Source: Optimus Prep


8 ---- 8.25 ---- 9 = 0.75 -----------0.25

So, the required ratio is 3:1 , answer will be (E)

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

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New post 29 Jan 2018, 11:11
bytatia wrote:
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 let x = the number of pounds of the $0.8 per pound flour and y = the number of pounds of the $0.9 per pound flour and create the equation:

0.8x + 0.9y = 0.825(x + y)

800x + 900y = 825x + 825y

75y = 25x

3y = x

3/1 = x/y

Answer: E
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Re: In What proportion must flour at $0.8 per pound be mixed   [#permalink] 29 Jan 2018, 11:11
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