SOLUTION: I get really confused when trying to set these problems up. Can you help me work this one out so I can use it for my other questions please? How many liters of a 40% alcohol soluti

Algebra ->  Percentage-and-ratio-word-problems -> SOLUTION: I get really confused when trying to set these problems up. Can you help me work this one out so I can use it for my other questions please? How many liters of a 40% alcohol soluti      Log On


   

Question 170410: I get really confused when trying to set these problems up. Can you help me work this one out so I can use it for my other questions please? How many liters of a 40% alcohol solution should be mixed with 40 liters of a 90% solution to get a 50% solution? Thanks for your help.
Found 2 solutions by Mathtut, Edwin McCravy:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
lets call x the number of liters for the 40% solution.
40%(x)+90%(40)=50%(40+x)
:
.4x+.9(40)=.5(40+x)
:
.4x+36=20+.5x
:
.1x=16
:
highlight%28x=160%29liters of 40% solution

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution with chart:
I get really confused when trying to set these problems up. Can you help me work this one out so I can use it for my other questions please? How many liters of a 40% alcohol solution should be mixed with 40 liters of a 90% solution to get a 50% solution? Thanks for your help.

Make this chart:


               | liters |  %  | pure alcohol
--------------------------------------------
1st solution   |        |     |           
2nd solution   |        |     |            
---------------|--------|-----|-------------
final solution |        |     |          

Let x = liters of 1st solution
and we are given that there are 40 liters
of the 2nd solution, so fill in x and 40 
liters.


               | liters |  %  | pure alcohol
--------------------------------------------
1st solution   |    x   |     |      
2nd solution   |   40   |     |    
---------------|--------|-----|-------------
final solution |        |     |

Add the two to get x+40 liters of final solution.


               | liters |  %  | pure alcohol
--------------------------------------------
1st solution   |    x   |     |      
2nd solution   |   40   |     |   
---------------|--------|-----|-------------
final solution | x+40   |     | 

Put in the three percentages as decimals:


               | liters |  %  | pure alcohol
--------------------------------------------
1st solution   |    x   | .40 |      
2nd solution   |   40   | .90 |     
---------------|--------|-----|-------------
final solution | x+40   | .50 | 

Now get the amount of pure alcohol contained
in each by multiplying by the percentage as
a decimal.


               | liters |  %  | pure alcohol
--------------------------------------------
1st solution   |    x   | .40 |      .40x
2nd solution   |   40   | .90 |   .90(40)  
---------------|--------|-----|-------------
final solution | x+40   | .50 | .50(x+40)

The equation comes from the last column:

.40x + .90(40) = .50(x+40)

Can you solve that for x? If not post again.

answer:  x = 160 liters

Edwin