SOLUTION: A candy store merchant mixes 25 kg of two different brands of jube-jubes. Brand A costs $2.25/kg and Brand B costs $3.75/kg. If he sells the mixture of jube-jubes for $3.03/kg, how

Question 1145792: A candy store merchant mixes 25 kg of two different brands of jube-jubes. Brand A costs $2.25/kg and Brand B costs $3.75/kg. If he sells the mixture of jube-jubes for $3.03/kg, how many kg of each brand of jube-jube are needed to make the mixture?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

A setup for a traditional algebraic solution....

x kg of Brand A plus (25-x) kg of Brand B makes 25 kg of the mixture

I'll let you finish the problem by that method. It's a few steps with some ugly decimals; but the process is straightforward.

Here is an easier way to solve mixture problems like this, if an algebraic solution is not required.

(1) Find where the $3.03 per kg cost of the mixture lies between the $2.25 and $3.75 per kg costs of Brand A and Brand B.
From 2.25 to 3.75 is a difference of 1.50
From 2.25 to 3.03 is a difference of 0.78
The $3.03 per kg cost of the mixture is 78/150 = 13/25 of the distance from $2.25 to $3.75.

(2) That means 13/25 of the mixture needs to be the higher priced Brand B.

ANSWER: 13/25 of 25kg, or 13kg, of Brand B; the other 12kg of Brand A.

You should of course get the same answer when you finish solving the problem algebraically, as described above.
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