SOLUTION: how many gallons of a 50% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture of 40% antifreeze

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Question 553942: how many gallons of a 50% antifreeze solution must be mixed with 90 gallons of 25% antifreeze to get a mixture of 40% antifreeze
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +x+ = gallons of 50% antifreeze needed
+.5x+ = gallons of antifreeze in 50% solution
given:
+.25%2A90+=+22.5+ = gallons of antifreeze in 90% solution
(1) +%28+.5x+%2B+22.5+%29+%2F+%28+x+%2B+90+%29+=+.4+
--------------------------------------------
+.5x+%2B+22.5+=+.4%2A%28+x+%2B+90+%29+
+.5x+%2B+22.5+=+.4x+%2B+36+
+.1x+=+36+-+22.5+
+.1x+=+13.5+
+x+=+135+
135 gallons of 50% antifreeze are needed
check answer:
(1) +%28+.5%2A135+%2B+22.5+%29+%2F+%28+135+%2B+90+%29+=+.4+
(1) +%28+67.5+%2B+22.5+%29+%2F+225+=+.4+
(1) +90+=+.4%2A225+
(1) +90+=+90+
OK