SOLUTION: A chemist needs a 26% acid mixture, but his lab only has 20% and 35% solutions. If he needs a total of 80 liters of the 26% solution, how much of each percent acid does he need?

Question 395269: A chemist needs a 26% acid mixture, but his lab only has 20% and 35% solutions. If he needs a total of 80 liters of the 26% solution, how much of each percent acid does he need?
I was thinking:
.20(x) + .35(y) = .26(80)
but I'm stuck and it just doesnt' seem like the right way to do it.
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A chemist needs a 26% acid mixture, but his lab only has 20% and 35% solutions. If he needs a total of 80 liters of the 26% solution, how much of each percent acid does he need?
I was thinking:
.20(x) + .35(y) = .26(80)
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You can use x & y
Your equation is right, and it's the way to do it.
The 2nd eqn is
x + y = 80
You can finish it, I think.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A chemist needs a 26% acid mixture, but his lab only has 20% and 35% solutions. If he needs a total of 80 liters of the 26% solution, how much of each percent acid does he need?
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If you want to use 2 variables you made a good start.
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Equations:
Quantity Equation: x + y = 80 liters
Acid Equation::::0.20x+0.35y = 0.26*80
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Multiply thru 1st by 20
Multiply thru 2nd by 100
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20x + 20y = 20*80
20x + 35y = 26*80
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Subtract 1st from 2nd and solve for "y":
15y = 6*80
y = 32 liters (amt. of 35% solution needed)
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Solve for "x":
x+y = 80
x + 32 = 80
x = 48 liters (amt. of 20% solution needed)
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Cheers,
Stan H.
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