Question 256007: how much water must be added to 50 gallons of a 40% alcohol solution to reduce the concentration to 16% alcohol? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of water that needs to be added
Now we know that the amount of pure alcohol before the water is added(0.40*50) has to equal the amount of pure alcohol after the water is added(0.16(50+x)). So our equation to solve is:
0.40*50=0.16(50+x) get rid of parens and simplify
20=8+0.16x subtract 8 from each side
20-8=8-8+0.16x or
12=0.16x divide each side by 0.16
x=75 gal ---amount of water that needs to be added
CK
0.40*50=0.16*(50+75)
20=0.16*125
20=20
Hope this helps----ptaylor
