Question 816338: how many liters of a 30% alcohol solution must be mixed with 30 liters of an 80% solution to get a 60% solution? Found 3 solutions by Alan3354, richwmiller, ewatrrr:Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Do it like this one. Only the numbers are different.
-----------
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
You can put this solution on YOUR website! It is not like the other tutor suggested because you don't know the total number of liters yet.
the total number is x+30
.3x+.8*30=.6*(x+30)
x=20 for a total of 50 liters
Hi,
how many liters 'x' of a 30% alcohol solution
must be mixed with 30 liters of an 80% solution to get a 60% solution? | After multiplying thru by 100
Solve for x
