Question

A student mixes 20% acid solution with a 40% solution produce 20 liters of 28% solution. How many liters of 20% solution acid solution was needed? Please help

Answer 1

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.