Question

I have solved similar linear equations previously, but this one stumps me. The answer is 4L per the book, but I cannot get to that answer. Here is the question:
How many liters of a 60% acid solution must be mixed with a 75% acid solution to get 20L of a 72% solution?

Answer 1

Answer:

Step-by-step explanation:

We need to find quantity of both solution

So we will assume for now and form equations

Let x L of 60% acid solution and y L of 75% acid solution is taken

x + y = 20

Acid in 60% soln is x * 60/100 = 0.6x

Acid in 75% soln is y * 75/100 = 0.75y

Acid in mixed solution is 20*72/100 = 14.4

0.6x + 0.75y = 14.4

Now solve both equations

We will get

x =4 and y =16

4L of 60% acid solution is mixed

Hope it helps :-)