Question

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 1

Let x be the quantity of the 1st solution (at 10%) and y, the 2nd (at 4%)
At the end we will have a TOTAL = x + y = 500 ml (mind you this is a volume)
But we need 10% of x and 4% of y to have a TOTAL mixture of 6%
Or 0.1x + 0.04 y = 0.06 (of 500)
Solve :
x + y = 500
0.1 x + 0.04 y =30
You will find:

x= 500/3 = 166.66 ml
y = 1000/3 = 333.33 ml

Apply this methodology for all other items to get the right answer