SOLUTION: (1 pt) What quantity of 65 per cent acid solution must be mixed with a 30 percent solution to produce 630 mL of a 50 per cent solution?

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Question 492982: (1 pt) What quantity of 65 per cent acid solution must be mixed with a 30 percent solution to produce 630 mL of a 50 per cent solution?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a = ml of 65% solution needed
Let b = ml of 30% solution needed
given:
+.65a+ = acid in 65% solution
+.3b+ = acid in 30% solution
(1) +a+%2B+b+=+630+ ml
-------------------------
(2) +%28+.65a+%2B+.3b+%29+%2F+630+=+.5+
(2) +.65a+%2B+.3b+=+.5%2A630+
(2) +.65a+%2B+.3b+=+315+
(2) +65a+%2B+30b+=+31500+
Multiply both sides of (1) by +30+ and subtract (1) from (2)
(2) +65a+%2B+30b+=+31500+
(1) +-30a+-+30b+=+18900+
+35a+=+12600+
+a+=+360+
and, since
(1) +a+%2B+b+=+630+
(1) +360+%2B+b+=+630+
(1) ++b+=+270+
360 ml of 65% solution are needed
270 ml of 30% solution are needed
check answer:
(2) +%28+.65a+%2B+.3b+%29+%2F+630+=+.5+
(2) +%28+.65%2A360+%2B+.3%2A270+%29+%2F+630+=+.5+
(2) +%28+234+%2B+81+%29+%2F+630+=+.5+
(2) +315+%2F+630+=+.5+
(2) +315+=+315+
OK