SOLUTION: A chemist has one solution that is 25 percent acid and a second that is 50 percent acid. How many liters of each should be mixed to get 10 L of a solution that is 40 percent acid?

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Question 432534: A chemist has one solution that is 25 percent acid and a second that is 50 percent acid. How many liters of each should be mixed to get 10 L of a solution that is 40 percent acid?
Found 2 solutions by josmiceli, rwm:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = liters of 25% acid to be used
Let b = liters of 50% acid to be used
In words:
( acid in solution)/(total solution) = 40%
given:
.25a = acid in 25% solution
.5b = acid in 50% solution
+a+%2B+b+=+10+
----------------
(1) +%28.25a+%2B+.5b%29+%2F+10+=+.4+
(2) a+%2B+b+=+10+
From (1):
(1) +.25a+%2B+.5b+=+4+
(1) +25a+%2B+50b+=+400+
Multiply both sides of (2) by 25
and subtract (2) from (1)
(1) +25a+%2B+50b+=+400+
(2) -25a+-+25b+=+250+
+25b+=+150+
+b+=+6+
and
(2) a+%2B+b+=+10+
+a+=+10+-+6+
+a+=+4+
4 liters of 25% acid must be used
6 liters of 50% acid must be used
check:
(1) +%28.25a+%2B+.5b%29+%2F+10+=+.4+
(1) +%28.25%2A4+%2B+.5%2A6%29+%2F+10+=+.4+
(1) +%28+1+%2B+3%29%2F10+=+.4+
(1) +4+=+.4%2A10+
(1) +4+=+4+

Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
x+y=10 liters
.25x+.5y=.4*10
x=4 liters y=6 liters
naturally you need more of 50/50 solution.