SOLUTION: Mixing a 10% solution with a 18% solution to produce 5 pounds of a 12% solution. How many pounds of the 10% solution are required?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Mixing a 10% solution with a 18% solution to produce 5 pounds of a 12% solution. How many pounds of the 10% solution are required?      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 703666: Mixing a 10% solution with a 18% solution to produce 5 pounds of a 12% solution. How many pounds of the 10% solution are required?
Found 2 solutions by nshah11, josmiceli:
Answer by nshah11(47) About Me  (Show Source):
You can put this solution on YOUR website!
Assume x lbs of 10% solution is mixed with (5 - x) lbs of 18% solution.
(0.1x + 0.18(5 -x))/5 = 0.12
-0.08x + 0.9 = 0.6
x = 3.75 lbs

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = pounds of 10% solution needed
Let +b+ = pounds of 18% solution needed
--------------
given:
(1) +a+%2B+b+=+5+
(2) +%28+.1a+%2B+.18b+%29+%2F+5+=+.12+
---------------------------
(2) +.1a+%2B+.18b+=+.12%2A5+
(2) +10a+%2B+18b+=+60+
(2) +5a+%2B+9b+=+30+
Multiply both sides of (1) by +5+ and
subtract (1) from (2)
(2) +5a+%2B+9b+=+30+
(1) +-5a+-+5b+=+-25+
+4b+=+5+
+b+=+1.25+
and
(1) +a+%2B+1.25+=+5+
(1) +a+=+3.75+
3.75 pounds of 10% solution are needed
1.25 pounds of 18% solution are needed
check:
(2) +%28+.1a+%2B+.18b+%29+%2F+5+=+.12+
(2) +%28+.1%2A3.75+%2B+.18%2A1.25+%29+%2F+5+=+.12+
(2) +%28+.375+%2B+.225+%29+%2F+5+=+.12+
(2) +.6+%2F+5+=+.12+
(2) +.6+=+.6+
OK