Question 10

A laboratory employee is mixing a 10% saline solution with 4% saline solution. How much of each solution is needed to make 500 milliliters of a 6% solution.

170 milliliters of the 10% solution and 340 milliliters of 4% solution

167 milliliters of the 10% solution and 333 milliliters of 4% solution

110 milliliters of the 10% solution and 210 milliliters of 4% solution

155 milliliters of the 10% solution and 300 milliliters of 4% solution

Answer : 167 milliliters of the 10% solution and 333 milliliters of 4% solution

A laboratory employee is mixing a 10% saline solution with 4% saline solution. How much of each solution is needed to make 500 milliliters of a 6% solution.

Let x represent the milliliters of 10% saline solution

Total ml of solution is 500 ml

so 500-x represents the milliliters of 4% saline solution

The equation becomes

ml of 10% solution + ml of 4% solution = 500ml of 6%

10x + 4 (500-x) = 500 * 6

10x + 2000 - 4x = 3000

6x + 2000 = 3000

Subtract 2000 on both sides

6x = 1000

so x = 166.666 ... Hence x = 167 ml

So 167 ml is mixed with 10% of saline solution

500-x represents the milliliters of 4% saline solution

Replace x with 167

500 - 167 = 333 milliliters of 4% saline solution

So answer is

167 milliliters of the 10% solution and 333 milliliters of 4% solution