SOLUTION: What quantity x of a 65% acid solution must be mixed with a 20% solution to produce 300 mL of a 45% solution. Amount x=_____ I set up a system of two equations: x+y=300 .65x+.

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Question 398200: What quantity x of a 65% acid solution must be mixed with a 20% solution to produce 300 mL of a 45% solution.
Amount x=_____
I set up a system of two equations:
x+y=300
.65x+.20y=.45
I solved for x and got 133.33. What am I doing wrong?
Thank you!!
Found 3 solutions by stanbon, scott8148, ewatrrr:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
What quantity x of a 65% acid solution must be mixed with a 20% solution to produce 300 mL of a 45% solution.
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Equations:
Quantity Eq: x+y=300 mL
Acid Eq::::: 0.65x+.20y=.45*300
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Multiply thru the 1st by 65
Multiply thru the 2nd by 100
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65x + 65y = 65*300
65x + 20y = 45*300
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Subtract and solve for "y":
45y = 20*300
y = (4/9)300
y = 133 1/3 mL (amt of 20% solution needed in the mixture)
-------------------
Solve for "x":
x+y = 300
x = 300 - 133 1/3
x = 166 3/4 mL (amt of 65% solution needed in the mixture)
================================
Cheers,
Stan H.
Answer by scott8148(6628)   (Show Source): You can put this solution on YOUR website!
you left the 300 ml out of the 2nd equation ___ .65x + .20y = .45(300)
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi

Let x and (300-x)represent the amounts of 65% and 20% solutions respectively

Question states***

Mixing together to form 300ml of 45% solution

  .65x + .20(300-x) = .45*300ml

solving for x

          .45x = .25*300ml

             x = .25*300ml/.45

             x = 166.6667ml of 65% solution 

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