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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
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Bunuel wrote:
A club wants to mix 20 pounds of candy worth $8.00 per pound with candy worth $5.00 per pound to reduce the cost of the mixture to $6.00 per pound. How many pounds of the $5.00 per pound candy should be used?

A. 20
B. 30
C. 40
D. 50
E. 60


One option is to use the weighted average formula:
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 = the number of pounds of CHEAP ($5/lb) candy needed
This means the TOTAL weight of candy = 20 + C

So, the PROPORTION of expensive candy = 20/(20+C)
And the PROPORTION of cheap candy = C/(20+C)

Now apply the formula:
6 = [20/(20+C)](8) + [C/(20+C)](5)
Simplify: 6 = 160/(20+C) + 5C/(20+C)
Multiply both sides by (20+C) to get: 6(20+C) = 160 + 5C
Simplify: 120 + 6C = 160 + 5C
Solve to get: C = 40

Answer:

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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
(20*8+5*x)=6(20+x)
160+5x=120+6x
x=40
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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
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A club wants to mix 20 pounds of candy worth $8.00 per pound with candy worth $5.00 per pound to reduce the cost of the mixture to $6.00 per pound. How many pounds of the $5.00 per pound candy should be used?

A. 20
B. 30
C. 40
D. 50
E. 60
--------------------------------------------------------------------------------------------
Cost of Candy A = $8.00 per pound
Cost of Candy B = $5.00 per pound
Cost of Candy A and Candy B Mixture = $6.00 per pound

Using Alligation Method

Candy A----------------Candy B

$8.00------------------$5.00
-----------$6.00------------
$1.00------------------$2.00

Ratio of Candy A to Candy B is 1:2

In order to find the weight of Candy B in the mixture, notice that the weight of Candy A in the mixture is already given, i.e., 20 pounds and as we know the ratio of Candy B is two times the ratio of Candy A in the mixture,

therefore the weight of Candy B in the mixture is 40 pounds.

Answer is 'C'

--------------------------------------------------------------------------------------------
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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
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Expert Reply
Bunuel wrote:
A club wants to mix 20 pounds of candy worth $8.00 per pound with candy worth $5.00 per pound to reduce the cost of the mixture to $6.00 per pound. How many pounds of the $5.00 per pound candy should be used?

A. 20
B. 30
C. 40
D. 50
E. 60



We can create the equation:

[8(20) + 5x]/(20 + x) = 6

160 + 5x = 120 + 6x

40 = x

Answer: C
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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
Expert Reply
Bunuel wrote:
A club wants to mix 20 pounds of candy worth $8.00 per pound with candy worth $5.00 per pound to reduce the cost of the mixture to $6.00 per pound. How many pounds of the $5.00 per pound candy should be used?

A. 20
B. 30
C. 40
D. 50
E. 60



If I see a chance to use PITA (Plugging In The Answers), I'm going for it! I like trying B and D.
Let's try B. 20 pounds of $8 candy is $160. 30 pounds of $5 candy is $150. That's a total of $310 and 50 pounds. That's more than $6/pound. Eliminate.
Let's try D. 20 pounds of $8 candy is $160. 50 pounds of $5 candy is $250. That's a total of $410 and 70 pounds. That's less than $6/pound. Eliminate.
We need something between the two.

Answer choice C.


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Re: A club wants to mix 20 pounds of candy worth $8.00 per pound with cand [#permalink]
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