Discussion

How many ounces of nuts costing 80 cents a pound must be mixed with nuts costing 60 cents a pound to make a 10-ounce mixture costing 70 cents a pound?
(A)  3
(B)...
(C)...
(D)...
(E)...
(F)...
*This question is included in Nova Press: Set S - Word Problems, question #3

The solution is

Posted: 08/28/2012 19:46
Is there a mistake with the units here? The explanation sets the cost of the mixture at 70(10). This would mean the mixture is 10 pounds and the problem asks for a 10 oz mixture.
Posted: 08/28/2012 20:15
Jaime, if the explanation is confusing, follow this.

Formal algebraic way of solving. Notice that it's useful in problems like this to write out the units of measure.
x = weight of nut 1 in the mix, in ounces
y = weight of nut 2 in the mix, in ounces

Cost of x = 80 cents / 16 ounces * x ounces = 80x/16 cents
Cost of y = 60 cents / 16 ounces * x ounces = 60x/16 cents
Cost of (x+y) = 70 cents / 16 ounces * 10 ounces = 700/16 cents

Cost of x + Cost of y = Cost of (x+y)
80x/16 cents + 60x/16 cents = 700/16 cents; multiply both sides by 16 cents
80x + 60x = 700
140x = 700
x=5

However, if we simply look at the problem, we see that 70 cents / pound is the exact average of 80 cents / pound and 60 cents / pound, so logically if we mix half of each in a 10 ounce mixture, we will get the target cost. So 5 ounces of x and 5 ounces of y! Without even doing any algebraic setup.

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