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# A dealer mixes a pounds of nuts worth b cents per pound with c pounds

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Math Expert
Joined: 02 Sep 2009
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A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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05 Mar 2018, 21:35
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A dealer mixes a pounds of nuts worth b cents per pound with c pounds of nuts worth d cents per pound. At what price should he sell a pound of the mixture if he wishes to make a profit of 10 cents per pound?

(A) (ab + cd)/(a + c) + 10

(B) (ab + cd)/(a + c) + 0.1

(C) (b + d)/(a + c) + 10

(D) (b + d)/(a + c) + 0.1

(E) (b + d + 10)/(a + c)
[Reveal] Spoiler: OA

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Re: A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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05 Mar 2018, 21:36
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Bunuel wrote:
A dealer mixes a pounds of nuts worth b cents per pound with c pounds of nuts worth d cents per pound. At what price should he sell a pound of the mixture if he wishes to make a profit of 10 cents per pound?

(A) (ab + cd)/(a + c) + 10

(B) (ab + cd)/(a + c) + 0.1

(C) (b + d)/(a + c) + 10

(D) (b + d)/(a + c) + 0.1

(E) (b + d + 10)/(a + c)

18. Mixture Problems

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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06 Mar 2018, 00:21
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Bunuel wrote:
A dealer mixes a pounds of nuts worth b cents per pound with c pounds of nuts worth d cents per pound. At what price should he sell a pound of the mixture if he wishes to make a profit of 10 cents per pound?

(A) (ab + cd)/(a + c) + 10

(B) (ab + cd)/(a + c) + 0.1

(C) (b + d)/(a + c) + 10

(D) (b + d)/(a + c) + 0.1

(E) (b + d + 10)/(a + c)

ANS: cost of A pounds of nuts worth B cent/ pound = AB cents
cost of C pounds of nuts worth D cent/ pound = CD cents
total cost of mixture = AB+CD cents
total cost of mixture per pound = (AB+CD)/ A+C cents
profit = 10 cents
Total = (AB+CD)/ A+C + 10
Answer (A)
Hope this is correct and of some help!
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A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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06 Mar 2018, 00:24
Since the dealer mixes
a=1 pounds of nuts worth b=30 cents per pound
and c=2 pounds of nuts at d=15 cents per pound,

he would be selling one pound of this mixture at $$\frac{1*30 + 2*15}{1 + 2} = \frac{60}{3} = 20$$ cents

For the dealer to make a profit of 10 cents per pound, he would have to sell the mixture at $$30$$ cents.

Evaluating answer options,

(A) (ab + cd)/(a + c) + 10 = 1*30 + 2*15/(1 + 2) + 10 = 30 cents.

Therefore, Option A((ab + cd)/(a + c) + 10) is the price at which the dealer should sell a pound of the mixture.
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Re: A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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07 Mar 2018, 07:32
Type 1:
Quantity: 'a' lbs
Unit Price: 'b' cents/lbs
Total Cost: 'ab' cents

Type 2:
Quantity: 'c' lbs
Unit Price: 'd' cents/lbs
Total Cost: 'cd' cents

Type 1+2:
Net Cost Price: (ab+cd) cents
Net Quantity: (a+c)
Net Unit Price: (ab+cd)/(a+c)

Profit on unit price: 10 cents

Therefore,
Selling Price per unit:
{(ab+cd)/(a+c)} + 10

Ans A

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Re: A dealer mixes a pounds of nuts worth b cents per pound with c pounds [#permalink]

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08 Mar 2018, 11:38
Bunuel wrote:
A dealer mixes a pounds of nuts worth b cents per pound with c pounds of nuts worth d cents per pound. At what price should he sell a pound of the mixture if he wishes to make a profit of 10 cents per pound?

(A) (ab + cd)/(a + c) + 10

(B) (ab + cd)/(a + c) + 0.1

(C) (b + d)/(a + c) + 10

(D) (b + d)/(a + c) + 0.1

(E) (b + d + 10)/(a + c)

The mixed nuts are worth (ab + cd)/(a + c) cents per pound. Thus if he wants to make a profit of 10 cents per pound, he should sell them for (ab + cd)/(a + c) + 10 cents per pound.

Answer: A
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Re: A dealer mixes a pounds of nuts worth b cents per pound with c pounds   [#permalink] 08 Mar 2018, 11:38
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# A dealer mixes a pounds of nuts worth b cents per pound with c pounds

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