Problem solving

Question

How much pure acid should be mixed with 6 gallons of 50% acid in order to get an 80% acid solution?

Answer 1

100% - 80% = 20%

80% - 50% = 30%

 

The ratio is 3/2

 

Therefore the gallons of pure acid (100%) needed is (3/2) x 6 = 9 gallons

Answer 2

Kailynn:

 

Let the amount of pure acid to be mixed (ADDED) x gallons (usually it helps to take as x what is asked).  Strength of pure acid is 100%. The total amount of solution is (x+6) gallons.  The strength of the total mixture is 80%

 

You can work with the percentages directly rather than decimals, if you are consistent.  Thus,

 

100 x + 50 (6) = 80(x+6)

 

100 x + 300 = 80 x +240

 

20 x = -60  (The minus sign is a RED FLAG!!!)

 

x = - 3,

 

This means the problem has a mistake!!!  PLEASE CHECK and post a comment.

 

Dr. G.