Algebra Demystified 2nd Ed

Chapter 8 linear appliCaTionS 263

peanut mixture to obtain a 50% peanut mixture. You might have a two-cycle

engine requiring a particular oil and gas mixture. Or you might have a recipe

calling for 1% fat milk and all you have on hand is 2% fat milk and^12 % fat milk.

We solve these problems with the following method.

Mixture problems involve three quantities—the two concentrations being

mixed together and the final concentration. Of these three quantities, one of

them is a fixed number. We let the variable represent one of the two concentra-

tions being mixed. The other unknown quantity is written as some combination

of the variable and the fixed quantity. If one of the quantities being mixed is

known, then we let x represent the other quantity being mixed and the final

solution would be “x + known quantity.” If the final solution is known, we again

let x represent one of the quantities being mixed, the other quantity being

mixed would be of the form “final solution quantity –x.” For example, in the

following problem, the amount of one of the two concentrations being mixed

is known: “How many liters of 10% acid solution should be mixed with 75 liters

of 30% acid solution to yield a 25% acid solution?”

If we let x represent the number of liters of 10% acid solution, then x + 75

represents the number of liters of the final solution. If the problem were stated,

“How many liters of 10% acid solution and 30% solution should be mixed

together with to produce 100 liters of 25% solution?” We can let x represent

either the number of liters of 10% solution or the 30% solution. Here, we let x

represent the number of liters of 10% solution. How do we represent the num-

ber of liters of 30% solution? For the moment, let “?” represent the number of

liters of 30% solution. We know that the final solution must be 100 liters, so

the two amounts must sum to 100:

x +? = 100.
x
xx
x

+=

−−

=−

?

?

100

100

Now we see that 100 – x represents the number of liters of 30% solution.

Many mixture problems can be represented by drawing three boxes. We

write the percentages given above the boxes and the volume inside the boxes.

Below the boxes, we multiply the percentages (converted to decimal numbers)

and the volume below the boxes; this gives us the equation to solve. Inciden-

tally, the product of the percent and the volume, is the amount of pure acid/

alcohol/milk-fat, etc. in that particular concentration.