SOLUTION: joan has one solution that is 18% acid and second solution that is 43% acid.How many liters should she mix together in order to get an 80 liters solution that is 26.5% acid

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: joan has one solution that is 18% acid and second solution that is 43% acid.How many liters should she mix together in order to get an 80 liters solution that is 26.5% acid      Log On


   



Question 1043926: joan has one solution that is 18% acid and
second solution that is 43% acid.How many
liters should she mix together in order to get
an 80 liters solution that is 26.5% acid

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = liters of 18% solution needed
Let +b+ = liters of 43% solution needed
-----------------------------------
+.18a+ = liters of acid in 18% solution
+.43b+ = liters of acid in 43% solution
-----------------------------------
(1) +a+%2B+b+=+80+
(2) +%28+.18a+%2B+.43b+%29+%2F+80+=+.265+
-----------------------------------
(2) +.18a+%2B+.43b+=+.265%2A80+
(2) +.18a+%2B+.43b+=+21.2+
(2) +18a+%2B+43b+=+2120+
---------------------------
Multiply both sides of (1) by +18+
and subtract (1) from (2)
(2) +18a+%2B+43b+=+2120+
(1) +-18a+-+18b+=+-1440+
--------------------------
+25b+=+680+
+b+=+27.2+
and
(1) +a+%2B+27.2+=+80+
(1) +a+=+52.8+
----------------------
52.8 liters of 18% solution are needed
27.2 liters of 43% solution are needed
--------------------------------
check answer:
(2) +%28+.18a+%2B+.43b+%29+%2F+80+=+.265+
(2) +%28+.18%2A52.8+%2B+.43%2A27.2+%29+%2F+80+=+.265+
(2) +%28+9.504+%2B+11.696+%29+%2F+80+=+.265+
(2) +21.2%2F80+=+.265+
(2) +21.2+=+.265%2A80+
(2) +21.2+=+21.2+
OK