Mixing pure acid

Question

How much pure acid should be mixed with 5 gallons of a 60% acid solution in order to get an 80% acid solution? 

Answer 1

Let' x gallons of acid is required

100 gallons of 60% acid solution contains 40 gallons of water and 60 gallons of pure acid

adding x gallons will increase the solution's volume to (100+x) gallons which contains 40 gallons of water and (60+x) gallons of pure acid hence the proportion of acid to water in this solution will be (60+x)/40 and acid to whole solution (60+x)/(100+x)

100 gallons of this new solution will contain 20 gallons of water and 80 gallon of acid hence the proportion of acid to water in this solution will be 80/20 and acid to whole solution 80/100

Thus we obtain two equations,

(60+x)/40=80/20 ...(1) and (60+x)/(100+x)=80/100 ...(2) solving any of them we derive the required result,

Thus, multiplying both sides by 40 in (1), 60+x=160

=> Subtracting 60 from both sides,              x=100

 

 

Alternative, multiplying both sides in (2) by 100, (6000+100x)/(100+x)=80

=> multiplying both sides  by (100+x)                            (6000+100x)=8000+80x

=> Subtracting 6000 from both sides,                                        100 x=2000+80x

=>Subtracting 80x from both sides,                                             20x=2000

=> Dividing both sides by 20,                                                          x=100