SOLUTION: How many ounces of a 15% alcohol solution must be mixed with 8 ounces of a 20% alcohol solution to make a 19% alcohol​ solution?

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Question 1192851: How many ounces of a 15% alcohol solution must be mixed with 8 ounces of a 20% alcohol solution to make a 19% alcohol​ solution?
Found 3 solutions by Boreal, greenestamps, Edwin McCravy:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
x oz*0.15=0.15x pure
8 oz*0.20=1.6 oz pure
8+x oz of 0.19=1.52+0.19x pure
-
set them equal
0.15x+1.6=0.19x+1.52
0.04x=0.08 oz
multiply by 25
x=2 oz. of 15%

Answer by greenestamps(13203)   (Show Source): You can put this solution on YOUR website!


The traditional formal algebraic approach is to write and solve an equation that says (very logically!) that the sum of the amounts of alcohol in the two ingredients is equal to the amount of alcohol in the mixture.

In this problem, you are starting with 8 ounces of which 20% is alcohol, adding x ounces of which 15% is alcohol, to end up with (8+x) ounces, of which 19% is alcohol. So the equation would be



This is solved using basic algebra....





ANSWER: 2 ounces of 15% alcohol

CHECK:
.20(8)+.15(2)=1.6+.3=1.9
.19(10)=1.9

If a formal algebraic solution is not required, you can try either of these other two methods, closely related to each other but very different from the formal algebraic method.

(1) The target percentage 19 is "4 times as close to 20% as it is to 15%", so the mixture must use 4 times as much of the 20% alcohol as it does 15% alcohol. Since the mixture uses 8 ounces of the 20% alcohol, the amount of 15% alcohol must be 1/4 of 8 ounces, which is 2 ounces.

OR....

(2) The target percentage 19% is 4/5 of the way from 15% to 20%, so 4/5 of the mixture must be the 20% alcohol. Since that 4/5 of the mixture that is 20% alcohol is 8 ounces, the 1/5 of the mixture that is the 15% alcohol must be 8/4 = 2 ounces.
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
This is the way mixture problems are usually taught, using charts.

We're mixing a weaker solution and a stronger solution to get a
medium strength (final) solution.

Make this chart:

                  -------------------------------------------------
                  |   ounces   | % as decimal | oz. of pure alcohol
-------------------------------------------------------------------
weaker solution   |            |              |     
-------------------------------------------------------------------
stronger solution |            |              |    
===================================================================
medium strength   |            |              |
-------------------------------------------------------------------


Fill in the given percentages expressed as decimals, and the given
amount of the stronger solution:


                  -------------------------------------------------
                  |   ounces   | % as decimal | oz. of pure alcohol
-------------------------------------------------------------------
weaker solution   |            |      0.15    |     
-------------------------------------------------------------------
stronger solution |     8      |      0.20    |     
===================================================================
medium strength   |            |      0.19    |
-------------------------------------------------------------------

We want to know the number of ounces of the weaker solution, so we
put x there.


                  -------------------------------------------------
                  |   ounces   | % as decimal | oz. of pure alcohol
-------------------------------------------------------------------
weaker solution   |     x      |      0.15    |     
-------------------------------------------------------------------
stronger solution |     8      |      0.20    |     
===================================================================
medium strength   |            |      0.19    |
-------------------------------------------------------------------

We add the number of ounces of the weaker (x) and stronger solutions
(8) and that gives x+8 ounces of medium strength (final) solution.


                  -------------------------------------------------
                  |   ounces   | % as decimal | oz. of pure alcohol
-------------------------------------------------------------------
weaker solution   |     x      |      0.15    |    
-------------------------------------------------------------------
stronger solution |     8      |      0.20    |    
===================================================================
medium strength   |    x+8     |      0.19    |
-------------------------------------------------------------------

We multiply each of the ounces by the percentage expressed as a
decimal to get the ounces of pure alcohol contained in each solution:


                  -------------------------------------------------
                  |   ounces   | % as decimal | oz. of pure alcohol
-------------------------------------------------------------------
weaker solution   |     x      |      0.15    |     0.15x
-------------------------------------------------------------------
stronger solution |     8      |      0.20    |     0.20(8)
===================================================================
medium strength   |    x+8     |      0.19    |     0.19(x+8)
-------------------------------------------------------------------

The equation comes from the right column, the ounces of pure alcohol:



Move the decimal 2 places right in each term:



Solve that for x.







2 ounces of the weaker 15% solution.

Edwin
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