SOLUTION: A chemist has two different solutions. One has 50% alcohol and 50% water and the other has 75% alcohol and 25% water. How many liters of each should be mixed to obtain 8 liters of

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Question 323585: A chemist has two different solutions. One has 50% alcohol and 50% water and the other has 75% alcohol and 25% water. How many liters of each should be mixed to obtain 8 liters of solution comprised of 60% alcohol and 40% water?
Found 2 solutions by mananth, modi.kk91:
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
One has 50% alcohol and 50% water
other has 75% alcohol and 25% water.
Mixture = 8 liters of solution 60% alcohol and 40% water?
..
let 50% alcohol required be x liters
75% acohol required will be 8-x liters
Total mixture = 8 liters 60%
..
0.5x+0.75(8-x)=8*0.6
0.5x+6-0.75x=4.8
-0.25x=-1.2
x= -1.2/ -0.25
x=4.8 liters 50% alcohol
75% alcohol required = 3.2 liters

Answer by modi.kk91(20)   (Show Source): You can put this solution on YOUR website!
this solution is given by allegation rule

here only takes only alcohol solution
50% 75%

60%

75-6o=15 and 60-50=10
ratio=15:10=3:2===3x+2x=5x

ans=3*1.6=4.8
2*1.6=3.2
then 5x=8
x= 8/5=1.6

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