Solving a percentage mixture problem using a system of linear equations

Question

A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 20% salt and Solution B is 70% salt. She wants to obtain 150 ounces of a mixture that is 60% salt. How many ounces of each solution should she use?

Answer 1

Let a be the volume of solution A and b the volume of solution B.

the total volume of the mixture will be a + b = 150 ounces

the amount of salt in the mixture will be

0.2a + 0.7b = 0.6*150

The system of two equations can then be solved by any method, for example by multiplying the first equation by 2, the second equation by 10,

2a + 2b = 300

2a + 7b = 900

and then subtracting the first equation from the second

5b = 600

solving for b

b = 600/5 = 120 ounces of Solution B

and finally substituting the value of b in the first equation

a + 120 = 150

and solving for a

a = 150 - 120 = 30 ounces of Solution A

On another note, I wanted to point out that in real life this method would not work for salt solutions for a number of reasons (I am a scientist). It would however work for solutions of isopropanol (rubbing alcohol) and other similar substances. Moreover, scientists do not measure volumes of solutions in ounces, they measure them in milliliters (sometimes in microliters, and rarely in liters).

Answer 2

Or, can use substitution method: a = 150 - b

0.2(150 - b) + 0.7b = 90

30 - 0.2b + 0.7b = 90

Combining like terms and moving the 30 to the other side: 0.5b = 60 -> b = 120 oz. for Sol'n B.

Then "a" must be 150 - 120 = 30 oz. for Sol'n A.