SOLUTION: You need 715 mL of a 50% alcohol solution. On hand, you have a 40% alcohol mixture and a 95% alcohol mixture. How much of each mixture will you need to obtain the desired solution?

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: You need 715 mL of a 50% alcohol solution. On hand, you have a 40% alcohol mixture and a 95% alcohol mixture. How much of each mixture will you need to obtain the desired solution?      Log On

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Question 867784: You need 715 mL of a 50% alcohol solution. On hand, you have a 40% alcohol mixture and a 95% alcohol mixture. How much of each mixture will you need to obtain the desired solution?
You will need
mL of the 40% solution
and
mL of the 95% solution.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Do it like this one. Only the numbers are different.
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Amanda wants to make 8 gal. of a 20% saline solution by mixing together a 56% saline solution and a 8% saline solution. How much of each solution must she use?
================
e = amount of 8%
f = amount of 56%
---
e+ f = 8 (total solution)
8e + 56f = 20*8 (total saline)
---
e+ f = 8
e + 7f = 20
------------ Subtract
-6f = -12
f = 2
e = 6