# 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.

### 15 Responses to Math Strategies: Averages Are Rarely Average in Difficulty

1. Tee February 15, 2016 at 1:17 pm #

This blog also by Chris explains how to answer #2 better than it was explained here:

https://magoosh.com/gre/2011/gre-math-mean-and-median-sorry-mode-you-werent-invited/

I was confused with the explanation provided here!

2. Abi June 19, 2015 at 7:16 am #

Hi Chris!

Questions of this sort leave me confused as I’m always left with 2 equations and 3 unknowns:

‘The average age of 8 people increases by 2 years when 2 women are included in place of 2 men, aged 20 and 24. Find the average age of the women.’

(When I attempt this, I write 2 equations, one is in terms of the old average and one in terms of the new average. My problem is that the former equation consists of the old average and the sum of the ages of the men (both unknown), and the latter equation consists of the additional unknown of the women’s age.)

Would be grateful if you could help!

Many thanks,

Abi

• Shruti July 16, 2015 at 12:43 pm #

suppose average age of 8 men is x
i.e (m1+m2……+m8)/8=x ==>m1+m2+……+m8=8x…………………………eq 1.
as given age of 2 of the mens is 20 and 24,putting in eq 1 give
m1+m2+……+m6+20+24=8x ===>m1+m2+m3…..+m6=8x-44…………………eq1
now adding 2 women age makes avg 2 more then x
i.e (m1+……+m6+w1+w2)/8=x+2==>m1+……+m6+w1+w2=8x+16………..eq2
now substituting the value of 6 men from eq1 to eq 2 will give us sum of age of two women
i.e 8x-44+w1+w2=8x+16 ===>w1+w2=60 dividing both side by 2 for the average of two women’s age i.e (w1+w2)/2=30,gives us the average age of the women.

• Sneha July 19, 2015 at 5:46 am #

Hi Abi,

The question you posted can also be solved as below.
Average age of 8 people increases by 2 years i.e. total age is increased by (8*2) years = 16 when two women are added.
The 2 women are included in place of 2 men, aged 20 and 24. i.e the sum of 2 men removed is 20+24 = 44 years.
now, to find the total age of 2 women, 44 years + 16 years = 60 years
average = 60/2=30.

I hope this helps.

3. Nishith December 8, 2013 at 6:30 am #

For problem 2 the second approach makes more mathematical sense.

Since difference between the averages of Set A and Set B: 10.6 – 8.2 = 2.4

The combined average needs to be somewhere close to the higher end of the scale.

Since the the two sets are in the ratio 1:2 we can also divide the difference in averages by 3

2.4/3 =0.8

and the combined average needs to 2 steps from 2.4 = 0.8+ 0.8 = 1.6 from the lower end of averages of 2 which is 8.2 + 1.6 = 9.8

I had initially calculated the whole thing manually

(8.2 * 120 + 10.6 * 240 ) / 360 = 9.8 (approx)

I wonder if this is a good idea as I avoid all confusion and just did total /N formula which is applicable for all average problems.

• Chris Lele December 9, 2013 at 11:40 am #

Hey Nishith,

You can use either approach. The weighted average approach is just much faster. So if you are short on time, use that one. Also, the weighted average can come in quite handy on mixture problems too.

4. Hundanol November 2, 2013 at 3:22 pm #

your answer for Q. no. 1 seems to be wrong in one case -12, -11,…..0……11, 12
The average and the total are both 0 in this case which amounts to the average being 100% of the sum. 🙂
May be the question should be restricted to positive consecutive numbers to exclude choice E.

• Chris Lele December 9, 2013 at 11:41 am #

Good catch! I changed it accordingly :). So now the answer only refers to positive integers.

5. kishor October 1, 2013 at 8:59 am #

I like the explanation of the first problem. Thanks Chris.

P.S. : I don’t mean that the second problem explanation is not good.

6. John August 27, 2013 at 5:09 pm #

Hi Chris,
For question #1, would the answer of 4% be true if the 25 consecutive numbers include Zero or several negative numbers?
John

• Chris Lele August 28, 2013 at 1:15 pm #

Hi John,

Yes, even with negative numbers and zero, it seems the answer still remains the same. Let’s just take a smaller subset of numbers, say -1, 0, 1, 2, 3. Total = 5 Average = 1, or 1/5 = 20% of total. The only exception is if the middle number is zero. If you choose -2, -1, 0. 1, 2 = Total = 0, Mean = 0, then, technically, the answer could still be 20%, or just about any number, since any percent of zero will be zero.

7. Bijal August 3, 2013 at 7:46 pm #

Hi Chris,

For question #2, when you are using the weight average I can understand answer choices A-C being incorrect, I thought that 10.1 would be the answer because the combined average will be leaning towards the mean of set B.

• Chris Lele August 5, 2013 at 1:26 pm #

Hi Bijal,

Using the concept of weighted averages, you can’t eliminate 10.1 as easily, because it seems as though it could be the right answer. When answers are close, a good technique is to subtract the two numbers (10.6 – 8.2 = 2.4). Then, take the ratio of the numbers, 1:2 and split 2.4 into that same ratio (1.6 and 0.8). Then add the higher number to the number that has the lower number in the ratio (in this case, 8.2 + 1.6 = 9.8).

Let me know if that makes sense!

• Bijal August 10, 2013 at 10:32 am #

Makes sense, but still need some clarification:

1. Do we always take a ratio of 1:2 in these type of questions?
2. Do you mean to say add the highest number within the ratio and the lowest number of the two given averages. In that case 1.6 + 8.2?

Am I understanding this correctly?

Thanks.

• Chris Lele August 12, 2013 at 1:06 pm #

Sure, for the first point: the reason we have a ratio of 2:1 is the terms 240 and 120 are in a ratio of 2:1. Had the question said 360 terms and 120 terms, then the ratio would be 3:1.

For the second point, you want to subtract both average to find the difference, which equals 2.4. Then you want to split that number into the corresponding ratio (in this case 2:1) so that you get 0.8 and 1.6. Add the higher number of ratio (1.6) to the lower average, or subtract the lower number of the ratio (0.8) by the higher average. In both cases, you’ll get the average of both sets (in this case 9.8).

Hope that helps!

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