SOLUTION: pure acid is to be added to a 10%acid solution to obtain 90l of 22% solution . what amounts of each should be used?

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Question 973749: pure acid is to be added to a 10%acid solution to obtain 90l of 22% solution .
what amounts of each should be used?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = liters of pure acid needed
Let +b+ = liters of 10% acid solution needed
+.1b+ = liters of acid in 10% solution
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(1) +%28+a+%2B+.1b+%29+%2F+90+=+.22+
(2) +a+%2B+b+=+90+
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(1) +a+%2B+.1b+=+19.8+
---------------------
Subtract (1) from (2)
(2) +a+%2B+b+=+90+
(1) +-a+-+.1b+=+-19.8+
----------------------
+.9b+=+70.2+
+b+=+78+
and
(2) +a+%2B+78+=+90+
(2) +a+=+12+
12 liters of pure acid are needed
78 liters of 10% acid solution are needed
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check:
(1) +%28+a+%2B+.1b+%29+%2F+90+=+.22+
(1) +%28+12+%2B+.1%2A78+%29+%2F+90+=+.22+
(1) +%28+12+%2B+7.8+%29+%2F+90+=+.22+
(1) +19.8+=+.22%2A90+
(1) +19.8+=+19.8+
OK