 ## 3 thoughts on “3 Simple Steps for Solving Mixture Problems”

1. aajayunlimited says:

Here’s a way that you’ve probably never thought of to solve some mixture problems and these will be kind of tricky for some folks. Out of my college algebra book. Cross-multiplication.

Question 1. A tank contains 20 gallons of a mixture of alcohol which is 40% alcohol by volume. How much of the mixture should be removed and replaced by an equal volume of water so that the resulting solution will 25% alcohol by volume?

.40[alcohol]/.25= 20/20 -x[resulting expression common to mixture problems]. So, by cross multiplying, the equation becomes: .40(20 – x) = .25(20) | 8 – .40x = 5 | 40 – 2x = 25 | -2x = -15 | Answer: x = 7.5 gallons

Question 2.
What weight of water must be evaporated from 40 lb of a 20% salt solution to produce a 50% solution? All percents are by weight.

.20/.50 = 40/40 – x | Cross- multiplying again: 20 – x/2 = 8 | 40 – x = 16 | -x = -24 | Answer: x = 24 lb.
Just like the chart makes certain mixture problems far more simple, Cross-multiplying makes these types of problems easier. Note: must be lined up right to its corresponding part or you will get the wrong answer in all likelihood.

1. Suzy S., TakeLessons Blog Manager says:

Thanks for sharing this additional strategy!

2. aajayunlimited says:

| Correction: Question 2–It is 40 – x/40; work is right, though. Sorry, for the mistake.