How many liters of pure water should be mixed with a 19-L solution of 80% acid to produce a mixture that is 80% water

Question

This is a linear equation involving a mixture. I am unsure how to solve

 

Answer 1

Hi Sidney,

 

Couple ways to think about this one. Here is one of them. I would be drawing a picture too.

 

Let x = number of additional liters of water to be added, realizing that the number of liters of water in the original solution is 20% of 19 (an 80% solution means 80% acid--so 20% water), i.e. 3.8 liters of water to start.

 

If we add x liters to the original 19 liter solution there will then be a total of x+19 liters. But we want 80% of the new solution to be equal to the new amount of water.

 

So,

 

.8 (x + 19) --80% of the new solution 

 

must equal the new amount of water. Since there were .2 (19) liters of water to start (3.8) the new amount of water is x+3.8.

 

So, the equation is: 

 

.8(x+19) = x+3.8

 

.8x + 15.2 = x + 3.8 (At this point I might multiply both sides by 10 to eliminate the decimals, but...)

 

11.4 = .2x

 

x = 57

 

Check:

If I add 57 liters to 19, there will be 76 liters. 80% of 76= 60.8.

There were 3.8 liters of water to start, so new amount of water is 57 + 3.8 = 60.8. Q.E.D.