Mixture problem

Question

how many liters of a 60% acid solution must be mixed with the  10% acid solution to get 500 out of the 50% acid solution

Answer 1

If we have x liters of 60% acid the amount of acid is .6x

 

Then we have 500-x liters of 10% acid with .1(500-x) acid

 

.6x + .1(500-x) = .5(500)

 

.6x + 50 - .1x = 250

 

.5x = 200

 

x = 400 liters of 60% acid

500-x = 100 liters of 10% acid

Answer 2

x liter 60%and y liter 10% 

 

0.6x+.1y=.5(x+y)

.5(x+y)=500> x+y = 1000 > y=1000-x

 

.6x+.1(1000-x)=250 >.5x=250-100=150 > x=300 y= 1000-300=700