If a 7% saline solution and a 4% saline solution are mixed to make 500 milliliters of a 5% saline solution, how much of each solution, to the nearest milliliter, is needed?

Question

anonymous · Answer

Let x be the number of milliliters of the 7% saline solution and y be the number of milliliters of the 4% saline solution.

They need to add up to 500 milliliters total, so x + y = 500.

The amount of salt in the x milliliter 7% solution is 0.07x milliliters. The amount of salt in the y milliliter 4% solution is 0.04y milliliters. Together, they need to add up to 5% of the total 500 milliliters, which is 0.05(500) = 25 milliliters, so 0.07x + 0.04y = 25.

Solve the system by substitution.

$$x + y = 500 $$
$$x = 500 - y$$

$$0.07x + 0.04y = 25$$ (substitute 500 - y for x)
$$0.07(500 - y) + 0.04y = 25$$
$$35 - 0.07y + 0.04y = 25$$
$$-0.03y + 35 = 25$$
$$-0.03y = -10$$
$$y = 333.333...$$
y = about 333

$$x = 500 - y = 500 - 333 = 167$$

ANSWER: 167 milliliters of the 7% solution and 333 milliliters of the 4% solution (to the nearest milliliter)