Let x be the amount of 30% acid and y be the amount of 10% acid.
The total volume of the solution is x + y = 20 liters.
The amount of pure acid in the 30% solution is 0.3x liters.
The amount of pure acid in the 10% solution is 0.1y liters.
The total amount of pure acid in the solution is 0.3x + 0.1y liters.
Since the total volume of the solution is 20 liters, we can write the equation:
0.3x + 0.1y = 0.11 * 20
0.3x + 0.1y = 2.2
Since x + y = 20, we can write y = 20 - x.
Substituting this into the equation above, we get:
0.3x + 0.1(20 - x) = 2.2
0.3x + 2 - 0.1x = 2.2
0.2x = 0.2
x = 1
Substituting this into the equation y = 20 - x, we get:
y = 20 - 1
y = 19
So, 1 liter of 30% acid and 19 liters of 10% acid should be mixed to obtain 20 liters of 11% acid.
How many liters each of 30% and 10% acid should be mixed to obtain 20 liters of 11% acid?
Kind of Solution
Liters of Solution
Amount of Pure Acid
0.3
x
0.1
y
0.11
20
1 answer