GRE Statistics: Conceptualizing Weighted Averages

Understanding Statistics (mean, median, and mode) questions on the GRE is becoming more sophisticated. You will not simply be asked to find the mean, but to think conceptually. For instance, let’s compare the following two problems:

Q1. What is the mean of 17, 23, 25, 36, 49?

(A) 25

(B) 26

(C) 30

(D) 34

(E) 40

Q2. Set A consists of 30 numbers, the average of which is 60. Set B consists of 29 numbers, the average of which is 70.

Column A: The average of Set A and B

Column B: 65

(A) The quantity in Column A is greater

(B) The quantity in Column B is greater

(C) The two quantities are equal

(D) The relationship cannot be determined from the information given

The first question is an example of number crunching, and is probably the type of problem you would encounter in a math textbook. The GRE, however, isn’t looking for your ability to crunch numbers, per se – especially with the inclusion of a calculator on the test – but rather the way you are able to deconstruct a problem. Sometimes, the answer to a problem can require very little math.

The second problem favors thinking over number crunching, and is thus a GRE-esque problem. To solve it, you could find the total for both sets, and then divide by the number of elements that pertain to both sets. Or…you reason that because Set A has more elements (or numbers), when you combine the two sets the average is going to skew towards the average of Set A, which is 60. If the two sets contained the same number of elements, then you could take the average of the averages of each set, which, in this case, would be 65 (the number that is in the middle of 60 and 70, the respective average of each set).

But, as I mentioned, the average leans more towards 60 than it does 70. So, the answer would have to be less than 65. We do not need to know this exact number – we simply know that it is less than Column B. Because Quantitative Comparison asks you to determine which column is larger, the answer to this question is (B).

This concept of thinking is called Weighted Average. It is tested far more often on the GRE than a straightforward average question.

Let’s try another example.

Q3. Set S and Set T both contain x elements. The average of Set S is 40. If the average of Set S and Set T combined is 50, which of the following must be true?

I. The average of Set T is 60.

II. The range of Set T is greater than that of Set S.

III. x is an even number

(A) I only

(B) I & II

(C) II only

(D) I & III

(E) All of the above

This question requires even more thinking than the previous question. However, if you understood the last question, then you shouldn’t have too much difficulty with this one. Taking the concept we just learned, you should be able to see that, regardless of the number of elements, Set T must have an average of 60. If Set T had more elements, and had an average of 60, then the combined average would be greater than 50. If Set T had fewer elements than Set A, then the combined average of the two sets would be less than 50. Therefore, (I) must be true.

As for (II), we do not know anything about the range of elements. For example, Set T can have an average of 60 if it contains 55 and 65, or if it contains 0 and 120, the latter of which has a much larger range. Therefore, (II) cannot be true.

Finally, as long as the number of elements, x, is the same for both sets, whether there is an odd number or even number is unimportant.

Therefore the answer is (A).


Know your basic statistics, but also know that crunching numbers is secondary to being able to conceptualize properly.

Answer = C