Question 741277: Hi! I need help on this word problem (It has to do with Linear systems but i ant even figure out how to set it up)
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 2 liters of the 50% solution.
10% solution =___L
20% solution =___L
(b) Use as little as possible of the 50% solution.
10% solution = ___ L
20% solution = ___L
50% solution = ___L
(c) Use as much as possible of the 50% solution.
10% solution ___L
20% solution ___L
50% solution ___L
A little help on just setting it up would even be enough. Thank you!
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?
let a = amt of 10% solution
let b = amt of 20%
let c = amt of 50%
:
(a) Use 2 liters of the 50% solution.
Then a + b = 10, therefore:
a = (10-b)
.10a + .20b + .50(2) = .25(12)
.10a + .20b + 1 = 3
.10a + .20b = 3 - 1
.10a + .20b = 2
.10(10-b) + .20b = 2
1 - .1b + .2b = 2
.1b = 2-1
.1b = 1
b = 10 liters of 20% solution
therefore
10% solution =_0_L
20% solution =_10_L
Check this:
.10(0) + .20(10) + .50(2) = .25(12)
0 + 2 + 1 = 3
;
(b) Use as little as possible of the 50% solution.
Then use 0 liters of 10% solution
.20b + .50c = .25(12)
.20b + .50c = 3
replace b with (12-c)
.2(12-c) + .50c = 3
2.4 - .2c + .5c = 3
.3c = 3 - 2.4
.3c = .6
c = .6/.3
c = 2 liters of 50% is minimum
10% solution = _0_ L
20% solution = _10_L
50% solution = _2_L
:
(c) Use as much as possible of the 50% solution.
Use 0 liters of 20% solution then
.10a + .50c = .25(12)
.10(12-c) + .50c = 3
1.2 - .1c + .5c = 3
.4c = 3 - 1.2
.4c = 1.8
c = 1.8/.4
c = 4.5 liters of 50% is max
10% solution _7.5__L
20% solution __ 0__L
50% solution _4.5__L
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