How many gallons of each of a 4% and an 11% acid solutions should be mixed to obtain 35 gallons of a 7% solution?
1). 20 gallons, 15 gallons
2). 22 gallons, 13 gallons
3). 13 gallons, 22 gallons
4). 15 gallons, 20 gallons
Let us denote by y gallons the amount of 4% solution.
⇒ The amount of solution of 11% acid solution = 35 - y gallons
|
Amount of solution (gallons) |
Percentage |
Amount of solvent (gallons) |
Component 1 |
y |
0.04 |
0.04y |
Component 2 |
35 - y |
0.11 |
0.11 (35 - y) |
Mixture |
35 |
0.07 |
0.07 (35) = 0.04y + 0.11 (35 - y) |
Solving 0.07 (35) = 0.04y + 0.11 (35 - y),we get
y = 20 gallons
If y = 20 gallons; then the other amount, denoted by 35 - y must be 35 - 20 = 15 gallons.