Breaking Down the Scale Method for Weighted Averages

Before you dive into this post, make sure you are are familiar with the Scale Method for weighted averages, which we have discussed in previous posts.

We know that the scale formula of weighted averages is the following:

w1/w2 = (A2 – Aavg)/(Aavg – A1)

One point of confusion for many test takers regarding this formula is figuring out what A1, A2, w1 and w2 actually are.

Here is the simple answer: they can be anything. You can choose to set up the solution as you want. The only thing is that it must be consistent across. A1 and w1 could be the parameters of either solution; A2 and w2 will be the parameters of the other solution. We could also work with the concentration of either ingredient of the solution. We will illustrate this point with an example GMAT question:

A container holds 4 quarts of alcohol and 4 quarts of water. How many quarts of water must be added to the container to create a mixture that is 3 parts alcohol to 5 parts water by volume?

(A) 4/3
(B) 5/3
(C) 7/3
(D) 8/3
(E) 10/3

Now, we have been given two solutions that we have to mix:

1. A container holding 4 quarts of alcohol and 4 quarts of water
2. Water (which means it has no alcohol in it)

When these solutions are mixed together, they give us a mixture that is 3 parts alcohol to 5 parts water by volume.

So, what are A1, w1, A2, w2 and Aavg? We can work with the concentration of either alcohol or water. Let’s first see how we can work with the concentration of water:

Method 1:
A1 is the concentration of water in the solution of 4 quarts of alcohol and 4 quarts of water. So A1 = 4/8.
w1 is the volume of this solution.
A2 is the concentration of water in the solution of water only. So A2 = 8/8 (we want to write this in the same format that we write A1 in.)
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (8/8 – 5/8)/(5/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water.

Method 2:
A1 is the concentration of water in pure water. So A1 is 8/8
w1 is the volume of this solution.
A2 is the concentration of water in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8
w2 is the volume of this solution.
Aavg is the concentration of water in the final solution i.e. 5/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 5/8)/(5/8 – 8/8)
w1/w2 = 1/3

So 1 part of water should be mixed with 3 parts of the solution with alcohol and water (same result as above).

Now we will see how to work with the concentration of alcohol. Of course the result will be the same.

Method 3:
A1 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A1 is 4/8.
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of water only. So A2 is 0/8 (to write in the same way as above)
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (0/8 – 3/8)/(3/8 – 4/8)
w1/w2 = 3/1

So 3 parts of the solution with alcohol and water should be mixed with 1 part of pure water (same as above).

Method 4:
A1 is the concentration of alcohol in pure water. So A1 is 0/8
w1 is the volume of this solution.
A2 is the concentration of alcohol in the solution of 4 quarts alcohol and 4 quarts water. So A2 is 4/8.
w2 is the volume of this solution.
Aavg is the concentration of alcohol in the final solution i.e. 3/8

w1/w2 = (A2 – Aavg)/(Aavg – A1)
w1/w2 = (4/8 – 3/8)/(3/8 – 0/8)
w1/w2 = 1/3

So 1 part of pure water should be mixed with 3 parts of the solution with alcohol and water (same result as above).