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

Question

Mathematics

Jocelyn

20 September, 09:13

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

+2

Answers (1)

Know the Answer?

Easton Robbins · Answer 1

Let t be 10% solution and f be 4% solution ...

(0.1t+0.04f) / 500=0.06 and t+f=500 so f=500-t making our equation:

(0.1t+0.04 (500-t)) = 30

0.1t+20-.04t=30

0.06t=10

t=166 2/3 ml and since f=500-t, f=333 1/3 ml

So they obviously rounded to nearest ml ...

167ml of 10% and 333ml of 4%