Question 897499: How many liters of an 8% solution of salt should be added to a 24% solution in order to obtain 560 liters of an 16% solution?
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Please help! These problems confuse me so much, and I won't have access to help at school until the weekend. I don't have a lot of experience in mixture problems and the examples in my notes don't follow the structure of this particular problem.
I have tried to make charts but I can't seem to make one that works with this problem. My charts have listed: the percent solution, the total liters, and I don't know what else to include. I don't think I was filling the chart out right.
Feel free to change the numbers in the problem to answer it, but PLEASE follow the same structure, that is what is confusing me most!
Thank you!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! How many liters of an 8% solution of salt should be added to a 24% solution in order to obtain 560 liters of an 16% solution?
:
Here's simple method, you don't even need a chart.
:
Let x = amt of 24% solution required
The resulting total is to be 560 liters, therefore
(560-x) = amt of 8% solution
:
A simple mixture equation using the decimal equiv of the percent values
.24x + .08(560-x) = .16(560)
.24x + 44.8 - .08x = 89.6
.24x - .08x = 89.6 - 44.8
.16x = 44.8
x = 44.8/.16
x = 280 liters of the 24% solution
but they want to know the the amt of the 8% solution
560 - 280 = 280 liters of the 8% solution
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