SOLUTION: In my algrebra class, we are currently solving word problems using graphing, substitution, or elimination. I have one question that I am absolutely stumped on. I cant figure out ho
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Question 77641: In my algrebra class, we are currently solving word problems using graphing, substitution, or elimination. I have one question that I am absolutely stumped on. I cant figure out how to even begin to set the problem up. Here is the problem: A pastry chef created a 50ounce sugar solution that was 34% sugar from a 20% sugar solution and a 40% sugar solution. How much of the 20% sugar solution and how much of the 40% sugar solution were used? Can you please help me? ~Brandi~ Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A pastry chef created a 50ounce sugar solution that was 34% sugar from a 20% sugar solution and a 40% sugar solution. How much of the 20% sugar solution and how much of the 40% sugar solution were used?
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This is typical mixture problem. Learn this method and you can do all of them:
Let x = amt of 20% solution
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The resulting amt is given as 50 oz, therefore
(50 - x) = amt of 40% solution
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A simple equation using decimal equiv of %
.20(x) + .40(50-x) = .34(50)
.2x + 20 - .4x = 17
.2x - .4x = 17 - 20
-.2x = -3
x = -3/-.2
x = +15 oz of 20% solution
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We said that the 40% solution is 50 - x, therefore:
50 - 15 = 35 oz of the 40% solution
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Check our solution in the original equation:
.2(15) + .4(35) = .34(50)
3 + 14 = 17
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How about this? Does it make sense to you now? Any questions?
