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Difficulty: HardLinear Equations in One Variable

A chemist mixes two saline solutions. Solution A is 12%12\% salt by mass, and Solution B is 30%30\% salt by mass. The mass of Solution B used in the mixture is 20 grams more than 13\frac{1}{3} of the mass of Solution A used. If the resulting mixture is 18%18\% salt by mass, what is the mass, in grams, of Solution A used?

Answer: 120 grams

Answer

The mass of Solution A used is 120 grams.
The correct answer is 120. By setting the mass of Solution A to xx, the mass of Solution B is 13x+20\frac{1}{3}x + 20. Equating the total salt content from both individual solutions to the total salt content of the final mixture gives the linear equation 0.12x+0.30(13x+20)=0.18(x+13x+20)0.12x + 0.30\left(\frac{1}{3}x + 20\right) = 0.18\left(x + \frac{1}{3}x + 20\right). Simplifying both sides yields 0.22x+6=0.24x+3.60.22x + 6 = 0.24x + 3.6. Solving this equation gives 0.02x=2.40.02x = 2.4, which simplifies to x=120x = 120.

Step-by-Step Solution

1
Define the variable xx as the mass, in grams, of Solution A used in the mixture, and express the mass of Solution B in terms of xx.
Mass of Solution A = xx grams; Mass of Solution B = 13x+20\frac{1}{3}x + 20 grams.
To set up expressions representing the mass of each solution in the mixture.
2
Calculate the mass of salt contributed by each solution and write an expression for the total mass of salt.
Salt from Solution A = 0.12x0.12x grams; Salt from Solution B = 0.30(13x+20)=0.10x+60.30\left(\frac{1}{3}x + 20\right) = 0.10x + 6 grams; Total salt = 0.22x+60.22x + 6 grams.
To find the total amount of salt before mixing.
3
Express the total mass of the mixture and the total salt content using the final mixture's percentage.
Total mass of mixture = 43x+20\frac{4}{3}x + 20 grams; Total salt in final mixture = 0.18(43x+20)=0.24x+3.60.18\left(\frac{4}{3}x + 20\right) = 0.24x + 3.6 grams.
To write the total salt content in terms of the final mixture's concentration.
4
Equate the two expressions for the total mass of salt and solve the linear equation for xx.
0.22x+6=0.24x+3.6  ⟹  2.4=0.02x  ⟹  x=1200.22x + 6 = 0.24x + 3.6 \implies 2.4 = 0.02x \implies x = 120.
To find the mass of Solution A that satisfies the mixture conditions.

Key Concept

Setting up and solving a linear equation in one variable from a mixture word problem.
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Topics

Linear Equations in One Variable

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