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?
a. 167 milliliters of 7% solution and 333 milliliters of 4% solution
b. 110 milliliters of the 10% solution and 210 milliliters of 4% solution
c. 155 milliliters of the 10% solution and 300 milliliters of 4% solution
d. 170 milliliters of the 10% solution and 340 milliliters of 4% solution

The correct answer is A. 167 milliliters of 7% solution and 333 milliliters of 4% solution and here is how: 
If x is the number of milliliters of the 7% saline solution and y is the number of milliliters of the 4% saline solution then add up to 500 milliliters total, so x + y = 500. 
and if we do
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 
Then you know why the answer is A. 

False because saturated fatty acids contain only single bond