EDGE Math 8

Example 2

Glenn, a chemist, would like to make 15 liters (L) of a 20% alcohol solution by mixing the following:

Solution 1: 60% alcohol solution
Solution 2: 10% alcohol solution

How many liters of each solution should he use?

Solution

Let x be the number of liters of the 60% alcohol solution and y be the number of liters of the 10% alcohol solution.

It will be helpful to organize the given information in a table, as shown below. The illustration after the table will also help you visualize the given problem.

Number of Liters of Alcohol in a Solution

The two equations are obtained as follows.

First equation: amount of solution 1 + amount of solution 2 = 15 L, or

x + y = 15.

Second equation: amount of alcohol in solution 1 + amount of alcohol in solution 2 = 3 L, or

0.60 x + 0.10 y = 3.

You now have a system of two linear equations in two variables.

{\begin{array}{l} x + y = 15 \\ 0.60 x + 0.10 y = 3 \end{array}

To eliminate x, multiply both sides of the first equation by $- 0.60,$ and then add the result to the second equation.

\begin{array}{l} - 0.60 x - 0.60 y = - 9 \\ \underline{+ 0.60 x + 0.10 y = 3} \\ - 0.50 y = - 6 \end{array}

Solve for y in the resulting equation.

$- 0.50 y$	$= - 6$
$y$	$= \frac{- 6}{- 0.50}$
$y$	$= 12$

The solution of the system is $(3, 12) .$

Checking

With $x = 3$ and $y = 12,$ the total number of liters of the mixture is

x + y = 3 + 12 = 15 L .

The total amount of alcohol in the mixture is

0.60 (3) + 0.10 (12) = 1.8 + 1.2 = 3 L .

The percentage of alcohol in the mixture is

\frac{3}{15} \times 100 = 20 % .

The conditions of the problem are satisfied. Thus, Glenn should use 3 L of solution 1 (which is a 60% alcohol solution) and 12 L of solution 2 (which is a 10% alcohol solution) to obtain the desired mixture (which is a 20% alcohol solution).

Investment and Mixture Problems