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? How

Question 486933: 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? How do i work this out?
Found 2 solutions by John10, josmiceli:
Answer by John10(297) (Show Source): You can put this solution on YOUR website!
Hint: it is good to draw a picture.
First, you call x to be the amount of 50% alcohol
y---------------------80%-------
The total is 10.5 L
x + y = 10.5 (1)
The mixture is:
0.5x + 0.8y = 10.5(0.70) = 7.5 (2)
Solve (1) and (2) to find the amount of each solution.
John10:)
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of 50% alcohol solution needed
Let = liters of 80% alcohol solution needed
given:
(1)
= liters of alcohol in the 50% alcohol solution
= liters of alcohol in the 80% alcohol solution
(2)
------------------------
From (2):
(2)
(2)
(2)
Multiply both sides of (1) by and
subtract (1) from (2)
(2)
(1)

and, from (1),
(1)
(1)
(1)
3.5 liters of 50% alcohol solution are needed
7 liters of 80% alcohol solution are needed
check answer:
(2)
(2)
(2)
(2)
(2)
OK
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