SOLUTION: One solution contains 50% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to produce 10.5 liters of a 70% alcohol solution?

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Question 291550: One solution contains 50% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to produce 10.5 liters of a 70% alcohol solution?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = liters of 50% solution needed
Let b = liters of 80% solution needed
In words:
(liters of alcohol in final solution)/(total liters of final solution)
= 70%
given:
alcohol in 50% solution:
.5a
alcohol is 80% solution:
.8b
and
(1)%28.5a+%2B+.8b%29%2F10.5+=+.7
(2)a+%2B+b+=+10.5
----------------------
Multiply both sides of (1) by 10.5
(1) .5a+%2B+.8b+=+7.35
Multiply both sides of (2) by .5
and subtract (2) from (1)
(1).5a+%2B+.8b+=+7.35
(2)-.5a+-+.5b+=+-5.25
.3b+=+2.1
b+=+7
And from (2):
a+%2B+b+=+10.5
a+%2B+7+=+10.5
a+=+3.5
3.5 liters of 50% solution and 7 liters of 80% solution are needed
check answer:
(1)%28.5a+%2B+.8b%29%2F10.5+=+.7
(1)%28.5%2A3.5+%2B+.8%2A7%29%2F10.5+=+.7
%281.75+%2B+5.6%29%2F10.5+=+.7
7.35+=+10.5%2A.7
7.35+=+7.35
OK