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

step-by-step explanation:

107 units

the answer is 510

step-by-step explanation:

15 • 13 = 195

15 • 12 = 180

15 • 5 = 75

12 • 5 = 60

195 + 180 = 375

375 + 75 = 450

450 + 60 = 510