SOLUTION: A 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. How many liters of each solution is needed? I tried setting up a table bu
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Question 301827: A 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. How many liters of each solution is needed? I tried setting up a table but that didn't work. Can you please explain how you derive at your answer. Thanks and have a wonderful weekend.
Found 3 solutions by josmiceli, richwmiller, mananth:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!
Let = liters of 60% solution needed
Let = liters of 80% solution needed
given:
= acid in 60% solution
= acid in 80% solution
-----------------------
(1) liters
(2)
Multiply both sides of (1) by and
subtract from (2)
(2)
(1)
15 liters of 60% solution and 5 liters of 80% solution are needed
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
We want to end up with 20 liters.
x+y=20
we will have three different mixtures
.6x+.8y=.65*20
now we have our two equations in two unknowns
solve the first for x or y. It doesn't matter and then sub that into equation 2
x=20-y
.6(20-y)+.8y=.65*20
.6*20-.6y+.8y=.65*20
.2y=(.65-.6)*20
.2y=.05*20
multiply by 100
20y=5*20
divide by 20
y=5
x=15
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
A 60% acid solution is to be mixed with a 80% acid solution to produce 20 liters of a 65% acid solution. How many liters of each solution is needed? I tried setting up a table but that didn't work. Can you please explain how you derive at your answer. Thanks and have a wonderful weekend.
60% acid required be x
80% acid required will be 20-x
Total 20 liters 65%
0.6x +0.8(20-x)=20*0.65
0.6x+16-0.8x=13.00
-0.2x= -16+13
-0.2x=-3
x= -3/-0.2
x=15 liters 60% acid required
balance 5 liters will be 80% acid
Ananth
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