SOLUTION: Hi! I need help on this word problem
A chemist needs 12 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%
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Question 863134: Hi! I need help on this word problem
A chemist needs 12 liters of a 25% acid solution. The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%. How many liters of each solution will satisfy each condition?
(a) Use 4 liters of the 50% solution.
10% solution =___L
20% solution =___L
Found 2 solutions by ewatrrr, ankor@dixie-net.com:
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
.20x + .10(8L-x) + .50*4L = .25*12L |solve for x
.10x = .25*12L - .10*8L - ,50*4L
= 4L -.8L - 2L
x = 1.2L/.10
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A chemist needs 12 liters of a 25% acid solution.
The solution is to be mixed from three solutions whose concentrations are 10%, 20%, and 50%.
How many liters of each solution will satisfy each condition?
:
(a) Use 4 liters of the 50% solution.
10% solution =___L
20% solution =___L
:
Let a = amt to 10% solution
Let b = amt of 20% solution
the equation
.50(4) + .10a + .20b = .25(12)
2 + .10a + .20b = 3
.10a + .20b = 3 - 2
.10a + .20b = 1
:
we know that a + b = 12 - 4
a + b = 8
a = (8-b)
:
Replace a with (8-b)
.10(8-b) + .20b = 1
.8 - .1b + .2b = 1
.1b = 1 - .8
.1b = .2
b = .2/.1
b = 2 liters of the 20% solution
then
8-2 = 6 liters of the 10% solution
:
:
See if that checks out
.50(4) + .10(6) + .20(2) = .25(12)
2 + .6 + .4 = 3
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