The Mixture Problem

MIXTURE PROBLEM

or

“Mixed Nuts” using the “MIX BOX”

There is a problem called the, “Mixture Problem.” It is an algebraic system of equations with two interrelating quantities and prices. It goes something like this:

PROBLEM:

Guido’s Health Store sells a mixture of raisins and nuts. Guido buys the raisins for $2.50/lb and the nuts for $3.50/lb. How many pounds of each should be mixed to make 20 lbs of this snack worth $3.00/lb?

SOLUTION:

I like to call this one “Mixed Nuts” because no matter what the SAT is mixing, it is just like the nuts problems & so since I teach problem solving techniques using goofy names (so that you will remember), go with me on this one.

The solution is easy-peasy-lemony-squeezy; all you have to do is use the MIX BOX & the answer falls out every time.

Here is the mix box:

Things to be mixed	Weight X	Per/Weight Unit =	COST
Raisins
Nuts
Total Mixture

So what goes in the box? It is a system of equations & the box simply organizes the information. The algebra solution looks like this:

[Weight of nuts x Price of raisins] + [Weight of nuts x Price of nuts] = Total Cost

What do we know, the cost of raisins per pound and the cost of nuts per pound and the total weight and cost for which we are aiming. Lets fill all that in on the chart.

Things to be mixed	Weight X	Per/Weight Unit =	COST
Raisins		$2.50
Nuts		$3.50
Total Mixture	20	$3.00

The total weight is 20, so the weight of raisins plus the weight of nuts equals 20.

Wr + Wn = 20; solving for Wn:

Wn = 20 – Wr; lets put that in the box

Things to be mixed	Weight X	Per/Weight Unit =	COST
Raisins	Wr	$2.50
Nuts	20 - Wr	$3.50
Total Mixture	20	$3.00

Now figure the cost by multiplying across the box (and this forms the system of equations).

Things to be mixed	Weight X	Per/Weight Unit =	COST
Raisins	Wr	$2.50	2.5Wr
Nuts	20 - Wr	$3.50	3.5(20-Wr)
Total Mixture	20	$3.00	60

Add ‘em up:

Raisins & Nuts = Total Mixture

2.5Wr + 3.5(20-Wr) = 60; simplify

2.5Wr + 70 – 3.5Wr = 60; combine like terms

-Wr = - 10

Wr = 10; and since Wn = 20 – Wr, Wn = 10

Variations on the mixture problem.

There are only a limited number of options on the mixture problem, you can be asked to calculate the weight or the cost of either of the two components or the total; basically giving us about six variations on this problem. Master the variations and the possible wording on this puzzle and you will have increase your confidence and ultimately your score.

Try these ones on for size: