SOLUTION: Suppose that one solution contains 30% alcohol and another solution contains 80% alcohol. How many liters of each solution should be mixed to make 12.5 liters of a 60% alcohol so

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Question 1184101: Suppose that one solution contains 30% alcohol and another solution contains
80% alcohol. How many liters of each solution should be mixed to make 12.5
liters of a 60% alcohol solution?

Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Using the standard formal algebraic solution method....

x = liters of 30% alcohol
12.5-x = liters of 80% alcohol

The total amount of alcohol in the two ingredients is 60% of the total 12.5 liters:

.30%28x%29%2B.80%2812.5-x%29=.60%2812.5%29

I'll let you finish the solution by that method....

Here is a quick and easy way to solve any 2-part "mixture" problem like this if a formal algebraic solution is not required....

Consider the three percentages 30, 60, and 80 on a number line and observe/calculate that 60 is 3/5 of the way from 30 to 80. (30 to 80 is a difference of 50, 30 to 60 is a difference of 30; 30/50 = 3/5.)
That means 3/5 of the mixture is the 80% alcohol.

ANSWER: 3/5 of 12.5 liters, or 7.5 liters, of 80% alcohol; the other 5 liters of 30% alcohol.

CHECK:
.80(7.5)+.30(5)=6+1.5=7.5
.60(12.5)=7.5


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Suppose that one solution contains 30% alcohol and another solution contains
80% alcohol. How many liters of each solution should be mixed to make 12.5
liters of a 60% alcohol solution?
This is mixing a weaker solution of alcohol with a stronger solution of alcohol and ending up with a medium strength solution of alcohol.

It can be done with one unknown, as the above shows, but very often, word
problems that have two answers are much easier to think through and set up
if you use two unknowns.


                          | percent as |       liters of
                   liters |  decimal   | pure alcohol contained
-------------------------------------------------------------------
weaker solution  |   x    |     0.3    |      0.3x
stronger solution|   y    |     0.8    |      0.8y
-------------------------------------------------------------------
medium-strength  |  12.5  |     0.6    | 12.5(0.6) which is 7.5
       solution  |        |            |
So the system comes from the 1st and 3rd columns:

system%28x%2By=12.5%2C0.3x%2B0.8y=7.5%29

Solve that and get x = 5 liters of 30% and y = 7.5 liters of 80%.

As a very rough check, make sure the answer makes sense.  That is, the
medium solution, 60%, is closer to the stronger, 80%, than it is to the
weaker 30%. So common sense tells you that it will take more of the stronger
than the weaker.  So make sure to observe that 7.5 liters is more than 5 liters.

Edwin