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E -> [2c + 4b + 5(2c + b)]/11

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Bunuel
If 2 pounds of almonds that cost c dollars per pound are mixed with 4 pounds of peanuts that cost b dollars per pound and 5 pounds of cashews that cost 2c + b dollars per pound, what is the average (arithmetic mean) cost, in dollars, per pound of the mixture?

A. 11/(c + b)
B. 40/(b + c)
C. 11(c + b)/9
D. 12(b + c)/11
E. (12c + 9b)/11

2 Pound of Almond Costs = $ 2c
4 Pound of Peanuts Costs = $ 4b
5 Pound of Cashews Costs = $ 5(2c + b)

Total Cost of 11 pounds of mixture is 2c + 4b + 10c + 5b

Or, Total Cost of 11 pounds of mixture is 12c + 9b

So, Average Cost of 11 pounds of mixture is \(\frac{12c + 9b}{11}\)

Thus, Answer must be (E) \(\frac{12c + 9b}{11}\)
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Total Price = 2c+4b+5(2c+b) = 2c + 4b+ 10c + 5b = 12c + 9b
Total weight of the mixture = 2 + 5 +4 = 11
Price per pound = (12c + 9b)/11
IMO E
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Bunuel
If 2 pounds of almonds that cost c dollars per pound are mixed with 4 pounds of peanuts that cost b dollars per pound and 5 pounds of cashews that cost 2c + b dollars per pound, what is the average (arithmetic mean) cost, in dollars, per pound of the mixture?

A. 11/(c + b)
B. 40/(b + c)
C. 11(c + b)/9
D. 12(b + c)/11
E. (12c + 9b)/11

We can create the following equation:

average = [2c + 4b + 5(2c + b)]/(2 + 4 + 5)

average = (2c + 4b + 10c + 5b)/11

average = (12c + 9b)/11

Answer: E
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Bunuel
If 2 pounds of almonds that cost c dollars per pound are mixed with 4 pounds of peanuts that cost b dollars per pound and 5 pounds of cashews that cost 2c + b dollars per pound, what is the average (arithmetic mean) cost, in dollars, per pound of the mixture?

A. 11/(c + b)
B. 40/(b + c)
C. 11(c + b)/9
D. 12(b + c)/11
E. (12c + 9b)/11

Average cost per pound = 2*c+4*b+5*(2c+b)/11= 2c+4b+10c+5b/11 = 12c+9b/11

Ans:E
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