Rudy414 wrote:You currently have 10 oz of a solution that is 14% alcohol and the rest is water. How much water would you have to add to the solution to make the alcohol content equal to 10%?

Alcohol percentage in the original solution: 14%.
Alcohol percentage in the added water: 0%.
Alcohol percentage in the mixture: 10%.

Let S = the original solution and W = the added water.
The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 percentages on a number line, with the two starting percentages (14% and 0%) on the ends and the goal percentage (10%) in the middle.
S 14%----------10%-----------0% W

Step 2: Calculate the distances between the percentages.
S 14%----4-----10%----10-----0% W

Step 3: Determine the ratio in the mixture.
The required ratio of original solution to added water is the RECIPROCAL of the distances in red.
S:W = 10:4.

Since S:W = 10:4, and the volume of the original solution is 10 ounces, the final mixture must be composed of 10 ounces of original solution and 4 ounces of added water.

For two similar problems, check here:

https://www.beatthegmat.com/ratios-fract ... 15365.html

An alternate approach:

In the original solution, the amount of alcohol = .14(10) = 1.4 ounces.
After the water is added, these 1.4 ounces of alcohol must constitute 10% of the final mixture:
1.4 = .1x
x = 14.
Since the volume of the final mixture is 14 ounces, and the volume of the original solution is 10 ounces, the volume of the added water = 14-10 = 4 ounces.

Ola!

An average of 10% splits your two percentages in ratio 4:10, or 2:5 for simplicity's sake. Reverse this ratio and you've got your oz relationship, namely 5:2 of mix to water.

10/X=5/2

Or X=4

Kind Regards,

Misterholmes

Since S:W = 10:4, the final mixture must be composed of 10 ounces of original solution and 4 ounces of added water.

This is what I do not understand. 10:4 is the ratio, not the amounts correct? We could say that the new solution is 10 parts original solution and 4 parts water, but how do we go from 10 parts and 4 parts to 10 oz and 4 oz, other than just throwing the units ounces because we know that is the unit?

Yes you are right. It's just happens that the ratio and the quantities are equal. It's not that S:W is 10:4 and THEREFORE the quantities must be 10 oz to 4 oz. There is an intermediate step, which is algebra and involves setting up an equation (Units of S)/(Units of water)=10/4

Yes you are right. It's just happens that the ratio and the quantities are equal. It's not that S:W is 10:4 and THEREFORE the quantities must be 10 oz to 4 oz. There is an intermediate step, which is algebra and involves setting up an equation (Units of S)/(Units of water)=10/4

THANK YOU. That is exactly the piece I was missing and you explained it perfectly clear.

I like to use a RATIO BOX to calculate the actual amounts in the mixture.
A slightly more complex version of the problem posted above:

You currently have 28oz of a solution that is 14% alcohol and the rest is water. How much of the original solution must be REPLACED with pure water so that the resulting alcohol content is equal to 10%?

Once we have used alligation to determine that S:W = 10:4 = 5:2, set up the following box:

S--W--T
5--2-------RATIO
-----------MULTIPLIER
------28---ACTUAL VOLUMES

The rightmost column represent the TOTAL.
In the top row, we place the ratio (S:W = 5:2).
In the bottom row, we place the one known actual volume (total volume = 28 ounces).
Since in the top row 5+2 = 7, the total for the top row is 7.

S--W--T
5--2--7----RATIO
-----------MULTIPLIER
------28---ACTUAL VOLUMES

The middle row of the box is the MULTIPLIER.
Since under the total we get 28/7 = 4, the multiplier for the box is 4:

S--W--T
5--2--7----RATIO
4--4--4----MULTIPLIER
------28---ACTUAL VOLUMES

To complete the box, we multiply the values in the top row by the multiplier:

S--W--T
5--2--7----RATIO
4--4--4----MULTIPLIER
20-8--28---ACTUAL VOLUME

As indicated by the bottom row:
Since 5*4 = 20, there are 20 ounces of the original solution in the final mixture.
Since 2*4 = 8, there are 8 ounces of pure water in the final mixture.

Thus, of the original total volume of 28 ounces, 8 ounces must be replaced with pure water.