The fastest way to solve weighted average problems on the GMAT - MBA Prep Tutoring

The fastest way to solve weighted average problems on the GMAT

The concept of average weight is similar to that of the average of two or more numbers, but due to the weight of each number, the average will be closer to the number with the highest weight. The formula involves multiplying the values by their weights, adding them up and then dividing the result by the sum of the weights. But is there a faster way? In this post I'm going to show you how.


First, let me start by separating the weighted average problems into two categories:

  1. If you are given information about the values and weights, and you want to find the average weight.
  2. When you are given the values and average weight, and you want to find information about the weights of each values.

For the first type, use the regular formula. For the second type, this method works really well. I am also going to add that it must be a weighted average of two values (which is the most common type).

This method involves using the distances between the values and the actual average weight. The distance is going to be inversely proportional to the weight, because the more weight a value has, the closer it is to the average (less distance), and the less weight a value has, the farther it is from the average (more distance).

Here is a graph (assuming weight 1 > weight 2):


--({ d }_{ 1 })--Average----({ d }_{ 2 })----

\(\frac { weight\quad 1 }{ weight\quad 2 } =\frac { { d }_{ 2 } }{ { d }_{ 1 } }\)

Now let's look at an example:

On a table there are 24 cards, some with the number 8 written on them, and the rest with the number 14 written on them. If the average value of the cards on the table is 10, how many cards have the number 8 written on them?

A) 8

B) 10

C) 12

D) 16

E) 18


There are quite a few ways to answer this question: you can use the formula with two variables and solve the equations, you can plug in the answer choices and that would be quicker, but using the distance method, you can even do it mentally!

First, we start with a line. X represents the number of cards with an 8 and Y represents the number of cards with a 14.



(notice how the average, 10, is closer to 8 than to 14)

\(\frac { X }{ Y } =\frac { 14-10 }{ 10-8 } =\frac { 4 }{ 2 } =\frac { 2k }{ 1k }\)

From here we can see that X is to Y as 2 is to 1. Since \(X+Y=24\):

\(X + Y = 2k + k = 3k = 24\)

\(k = 8\)

\(X = 2k = 2(8) = 16\)

The answer is 16 (D).

As you can see, this method is very powerful. It really is possible to solve it mentally this way, and it applies for every problem that has this structure, even mixture problems (since they can be understood as a weighted average of concentrations).

Did you find this method helpful or do you have a question? Leave me a comment below, and share this post with your friends so they too can solve these problems faster.