12 L of 20% acid solution must be mixed with 8 L of 40% solution to obtain 20 L of 28% solution.
Solution:
Let us first set up a table. Fill in the table with known values
The table is attached below
From the attached table, we can set up two equations,
Sum of values of two acids = Value of mixture
0.2x + 0.4y=5.6
For convenience, we'll multiply the entire equation by 10,
2 x + 4 y = 56 ------> (1)
Now, Sum of amounts of each acid = Amount of mixture
x + y = 20 ---------> (2)
Multiply eqn (2) with 2 to derive the equation into one variable
eqn ( 2) × 2 => 2x + 2y = 40
Subtracting equation eqn (2) from (1),
( + ) 2 x + 4 y = 56
( − ) 2 x + 2 y = 40
− − − − − − − −
( = ) 0 + 2y = 16
Thus, 2y = 16
y = 8
Substituting y = 8 in eqn (2),
x + 8 = 20
x = 20 – 8 = 12
So, we have x = 12 and y = 8
We can conclude that 12 L of 20% acid solution must be mixed with 8 L of 40% solution to obtain 20 L of 28% solution.
