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# Math Strategies: Averages Are Rarely Average in Difficulty

An average day on the blog posts, so to speak. Today, we’ll deal with a topic that many students give short shrift – averages/mean. Oftentimes, the reason students don’t give them much consideration is that averages seem straightforward. Indeed, you may cavalierly shrug your shoulders, thinking “anyone can find the mean – just add up the total and divide by the number of numbers”.

So, let’s put this to the test, starting with the first question:

1. The mean of twenty-five consecutive positive integers numbers is what percent of the total?

(A)  4%

(B)  5%

(C)  20%

(D) 25%

(E)  Cannot be determined by the information provided.

Explanation

You may leap into this question, thinking that the answer is (E). After all, we do not know any of the numbers, and wouldn’t the answer depend upon the numbers? Maybe you are still unsure, and so decide to test this assumption. Sedulously, you begin writing out twenty-five consecutive numbers, hoping to be able to find what percent the average is of the total. Around the 15th number, an incipient cramp building in your right-hand, you realize that this method is going to take too long. So, by default, you end up choosing (E).

However, if you know how to solve complicated GRE math problems, it’s not as tough as it looks. If you read carefully, the problem is asking what percent of the total is the mean/average. The average is always a fraction of the total that corresponds to the number of numbers. Meaning, to find the mean, we always divide the total by the number of numbers you are adding up.  If you have four numbers, then the mean will be ¼ of the total. If you five numbers, the mean will be 1/5 of the total.

In the problem above we have 25 numbers (it doesn’t really matter that they are consecutive). The mean will be 1/25 of the total, or (A) 4%.

The key is not to automatically choose (E)—it’s usually a trap. The second thing to remember is that if you are furiously scribbling away, there is always a better way to do the problem. Remember, the GRE is not testing whether you can add up twenty-five numbers. It’s seeing whether you can pick up on the fact that the mean is always going to be a percent of the total that is fixed by the number of numbers you are adding up.

Time for another Problem

2. Set A contains 120 terms with an average of 8.2. Set B contains 240 terms with an average of 10.6. If Set A and B are combined what is the resulting average?

(A)  8.4

(B)  8.8

(C)  9.0

(D) 9.8

(E)  10.1

Approach

Before you go scrambling for the GRE calculator, try to work out the problem in your head, logically. The key word is logically. So, do not try to do the mental math in your head, but, instead, look at the answer choices. Notice that 8.4 and 10.1 are very close to 8.2 and 10.6, respectively. Or, phrased another way—if Set A and B contained the same number of numbers the average of the two sets, were you to mix them, would be in the middle of 8.2 and 10.6 (which is 9.4).

This concept is called weighted averages—that is, if we add the same number of elements from each set, then the two sets are balanced. But, let’s say I were to add more numbers from Set A. What would happen to the average? Well, it would become less and less, approaching 8.4 the more numbers you put into Set A.

Because the problem calls for twice as many elements in Set B as in Set A, the average is skewed towards that of SET B, 10.6. Therefore, (A), (B), and (C) could not be the answers, because they are less than the average if we had an equal number of numbers in each set. And, as I noted before, 10.1 is very close to 10.6 meaning that the number of numbers in SET B would have to be many times that of Set A.

(D) 9.8 is closer to the average of the two, which, if you remember, was 9.4. So, it most likely is the answer.

Another Approach

One more way to think about this problem is to look at the difference between the averages of Set A and Set B: 10.6 – 8.2 = 2.4. That distance can be thought of as a sliding scale. The very middle would be 1.2 greater than 8.2, and 1.2 less than 10.6, both of which equal 9.4, the average.

If I had twice as many members in Set B, the ratio between Set A and Set B is 1:2. Or, the ratio between Set A and the total is 1:3, and the ratio between Set B and the total is 2:3. Next, break up 2.4 (the distance along the scale), into 1/3 and 2/3—0.8 and 1.6. 10.6, the average of Set B, minus 0.8 = 9.8 (D), the average whenever there are twice as many numbers in Set B as in Set A.

This last approach takes a little practice to build confidence, but, once you get the hang of it, you will not need to go running—or in the Revised GRE context, clicking—for the calculator.

Takeaways

-Most of the times when you encounter averages/means on the GRE, they will not be straightforward.

-Work your way logically through a problem. Do not get caught trying to do laborious calculations, either on paper or with the calculator.

-Like any skill, thinking logically about averages take a little practice.