SOLUTION: three solutions contain a certain acid. the first contains 15% acid, the second 35%, and the third 40%. A chemist wishes to use all three solutions to obtain a 75-liter mixture con

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: three solutions contain a certain acid. the first contains 15% acid, the second 35%, and the third 40%. A chemist wishes to use all three solutions to obtain a 75-liter mixture con      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 744940: three solutions contain a certain acid. the first contains 15% acid, the second 35%, and the third 40%. A chemist wishes to use all three solutions to obtain a 75-liter mixture containing 34% acid. If the chemist wants to use twice as much of the 35% solution as of the 40% solution, how many liters, to the nearest 10th, of each solution should be used?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = liters of 40% solution needed
+2a+ = liters of 35% solution needed
Let +b+ = liters of 15% solution needed
---------------
+.4a+ = liters of acid in 40% solution
+.35%2A2a+=+.7a+ = liters of acid in 35% solution
+.15b+ = liters of acid in 15% solution
----------------
(1) +a+%2B+2a+%2B+b+=+75+
(2) +%28+.4a+%2B+.7a+%2B+.15b+%29+%2F+75+=+.34+
----------------
(1) +3a+%2B+b+=+75+
and
(2) +1.1a+%2B+.15b+=+.34%2A75+
(2) +1.1a+%2B+.15b+=+25.5+
(2) +110a+%2B+15b+=+2550+
----------------
Multiply both sides of (1) by +15+
and subtract (1) from (2)
(2) +110a+%2B+15b+=+2550+
(1) +-45a+-+15b+=+-1125+
+65a+=+1425+
+a+=+21.9+
+2a+=+43.8+
and, since
(1) +3a+%2B+b+=+75+
(1) +3%2A21.9+%2B+b+=+75+
(1) +65.8+%2B+b+=+75+
(1) +b+=+9.2+
21.9 liters of 40% solution are needed
43.8 liters of 35% solution needed
9.2 liters of 15% solution are needed
--------------
check:
(2) +%28+.4a+%2B+.7a+%2B+.15b+%29+%2F+75+=+.34+
(2) +%28+1.1%2A21.9+%2B+.15%2A9.2+%29+%2F+75+=+.34+
(2) ++24.1+%2B+1.4+=+.34%2A75+
(2) +25.5+=+25.5+
OK