SOLUTION: A CHEMIST NEEDS TO MIX AN 18% ACID SOLUTION WITH A 45% ACID SOLUTION TO OBTAIN 12 GALLONS OF A 36% ACID SOLUTION. HOW MANY GALLONS OF EACH OF THE ACID SOLUTIONS MUST HE USE?

Algebra ->  Inequalities -> SOLUTION: A CHEMIST NEEDS TO MIX AN 18% ACID SOLUTION WITH A 45% ACID SOLUTION TO OBTAIN 12 GALLONS OF A 36% ACID SOLUTION. HOW MANY GALLONS OF EACH OF THE ACID SOLUTIONS MUST HE USE?       Log On


   



Question 658021: A CHEMIST NEEDS TO MIX AN 18% ACID SOLUTION WITH A 45% ACID SOLUTION TO OBTAIN 12 GALLONS OF A 36% ACID SOLUTION. HOW MANY GALLONS OF EACH OF THE ACID SOLUTIONS MUST HE USE?

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = gallons of 18% solution needed
Let +b+ = gallons of 45% solution needed
given:
(1) +a+%2B+b+=+12+
(2) +%28+.18a+%2B+.45b+%29+%2F+12+=+.36+
-----------------------------
(2) +.18a+%2B+.45b+=+.36%2A12+
(2) +.18a+%2B+.45b+=+4.32+
(2) +18a+%2B+45b+=+432+
Multiply both sides of (1) by +18+
and subtract (1) from (2)
(2) +18a+%2B+45b+=+432+
(1) +-18a+-+18b+=+-216+
+27b+=+216+
+b+=+8+
and, since
(1) +a+%2B+b+=+12+
(1) +a+=+12+-+8+
(1) +a+=+4+
-------------
4 gallons of 18% solution are needed
8 gallons of 45% solution are needed
check:
(2) +%28+.18%2A4+%2B+.45%2A8+%29+%2F+12+=+.36+
(2) +%28+.72+%2B+3.6+%29+%2F+12+=+.36+
(2) +4.32+=+.36%2A12+
(2) +4.32+=+4.32+
OK