SOLUTION: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?

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Question 419887: A chemist has one solution that is 80% acid and another solution that is 30% acid. How much of the first (80%) solution is needed to make a 400 L solution that is 62% acid?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = liters of 80% solution needed
Let b = liters of 30% solution needed
given:
(1) a+%2B+b+=+400 L
acid in final solution is .62%2A400+=+248 L
(2) .8a+%2B+.3b+=+248
----------------
Multiply both sides of (1) by 3
and subtract from (2)
----------------
(2) +8a+%2B++3b+=+2480
(1) -3a+-+3b+=+-1200
+5a+=+1280+
+a+=+256+
256 L of 80% solution are needed
check answer:
(2) .8a+%2B+.3b+=+248
(2) .8%2A256+%2B+.3b+=+248
+204.8+%2B+.3b+=+248+
.3b+=+43.2+
b+=+144
and
a+%2B+b+=+400
256+%2B+144+=+400
+400+=+400+
OK