Over the last six months, I’ve gone through quite a few Revised GRE prep books. One thing I’ve noticed missing is difficult questions relating to mean, median, and mode. Sure, most books describe how to find the average, and what the difference between the mean and the median. Many already know the above, but mean, median and mode questions on the actual Revised GRE are going to require much more practice than you are going to get from simply applying the basic formulas.

Below are five questions that should hopefully give you a good mental workout. If you whip through them, answering all correctly, then you are very well prepared for the concepts you’ll find on the GRE. If you stumble, don’t despair. Remember, we have plenty more questions like these—some that are every more difficult—waiting for you at gre.magoosh.com.

1. Set S is comprised of six distinct positive integers less than 10. Which of the following must be true?

I: The median is an integer

II: The median is less than the average

III: The range of digits in Set S is less than 8

(A) I only

(B) I & II

(C) II & III

(D) III only

(E) None of the above.

2. A list is comprised of five positive integers: 4, 4, x, 7, y. What is the range of the possible values of the medians?

(A) 2

(B) 3

(C) 6

(D) 7

(E) Cannot be determined by information provided.

3. The average of five positive integers is less than 20. What is the smallest possible median of this set?

(A) 19

(B) 10

(C) 4

(D) 3

(E) 1

4. Set S is comprised of 37 integers

Column A | Column B |
---|---|

The median of Set S | The mean of the lowest and the highest term |

- The quantity in Column A is greater
- The quantity in Column B is greater
- The two quantities are equal
- The relationship cannot be determined from the information given

1. A good idea is to choose numbers. For instance, just using 1, 2, 3, 4, 5, 6 we can see that the first condition does not need to be true. The median of this range of numbers is 3.5. For the second condition, we can also use numbers and determine that the average and the median are equal. Therefore the second condition also does not have to hold true. Finally, for the third condition, the range of digits is 9 – 1 = 8. The third condition says less than 8. Therefore the answer is (E) none of the above.

2. Here we want to find the lowest possible median and the great possible median. Then we want to subtract the least possible median from the greatest possible median to find the range of the medians. If x and y are less than or equal to 4, then the median is 4. If x and y are greater than or equal to 7, the median is 7. Therefore the range of median is 3, Answer (B).

3. Choosing numbers will help us on this one. If we choose 1, 1, 1, 1, 16, the sum is equal to 20. The median is 1, Answer (E).

4. This is a conceptual problem. Nonetheless we can still pick numbers. Imagine all 37 numbers are the number ‘1’. The mean is ‘1’ and the average of the least and greatest is 1. n this case both columns are equal (C).Now, just change the last number to a 3. That is you have the number ‘1’ 36 times and the number three. The median is still 1. But now the average of the least and greatest is greater than ‘1.’ In this case column (B) is larger. If the answer switches depending on the numbers we use, the answer is (D).

Hi. I had a doubt in finding mode of a sequence. What is the mode of the sequence 1,2,3,4,5,6 .

Sequences actually can’t have a mode. A set of numbers has a mode only if there is a number that occurs more frequently than any other number. And in a sequence, each number occurs just once.

For Q1, the median has to be an integer.

3.5 is the mean.

Actually, the median only needs to be an integer if the quantity of integers in the set is

odd. This is because in an odd quantity of integers, such as a string of five integers, there is one exact middle integer. So for example, in the number set2, 3, 4, 6, 9, a single number– 4– is in the exact middle of the set, with two numbers that are less than it, and two numbers that are more than it.On the other hand, if you have an even quantity of integers, the median will not necessarily be an integer. This is because int hat case, there will be two middle numbers, and the median is derived from adding those two middle numbers and dividing by 2, to get a quantity that’s exactly between the two middle numbers. In other words, the median of an even quantity of numbers is a

mean, but it’s a mean of just the two middle numbers and not the whole set. Let’s take the example immediately above and remove the 3. We now have four numbers, as follows:2, 4, 6, 9. Here, the median is 5, because the two middle numbers, when averaged, produce 5: (4+6)/2 = 5. Conversely, if we remove the number 6 form the original five-number set, we get:2, 3, 4, 9. In that case, the median is the average of the two middle numbers 3 and 4. (3+4)/2 = 3.5. So in this case, we have a set of integers, but a median that isn’t an integer. We have a non integer median in Q1 for similar reasons.Why not simple choose all 1’s for the set of 5 digits then the average is 1 and median is 1?

Hi there,

The example shown here is just one of many combinations that we could use! You can also use a set of five 1’s and get the same answer 🙂

I thought that zero was considered a positive integer? Why isn’t it in these solutions?

There are some problems here with Q2. It asks for “the range of possible medians” which can either be taken for (i) how many possible medians are there or (ii) the greatest median minus the smallest median.

In both cases, the answer would be E – it can’t be determined based on the information given. In case (i) the medians could 4, 7, x or even y (i.e. 4,4,x,y,7 or x,4,4,7,y or 4,4,7,x,y and 4,4,y,x,7). In case (ii) based on the info above in case (i) the information cannot be determined.

Hi Saul,

Thanks for pointing out! That ambiguity is very subtle. I’m clarifying it now 🙂

I think question 3 requires clarification.

Question says that it is “set” of numbers, accordingly, all the elements of set are distinct, so {1, 1, 1, 1, 16} is equivalent to {1, 16}.

We have to consider distinct numbers, accordingly median would be 3.

I agree that they should be distinct, since it is a set. But… if they are distinct, the least median can only be 19, since at least they can be equidistant, 1,2,3,4… the mean in this case is equal to median. However, if any of the difference is more, the mean might become more than the median… the answer is still D.

Cheers

I think the second question solution is not clear to me it says ” average of the 5 positive integers is less than 20 i.e the sum should be <100… why are we taking the sum of 16+1+1+1+1??? any one to clarify?

I agree with you. Can anybody help with this question?

Hi Malek,

Sure! 🙂 The set of 1, 1, 1, 1, 16 isn’t the only set we could have used–it is just an example that fits our criteria (average less than 20, all positive integers). This set therefore proves that the lowest median we can possibly have given those criteria is 1, so the answer is (E).

To be fair, I think question two needs to be re-worded. “A list is comprised of five positive integers: 4, 4, x, 7, y. What is the range of possible medians?” leads one to believe that these numbers have not yet been arranged in increasing order. I selected ‘E’ because those numbers could be anything.

Also, for question five, if you randomly pick the numbers 1, 3, 5 and 6, they don’t give you the same answer. Median = 4; Mean = 3.75 (which fulfills the original requirement of median being greater than mean) yet the answer to this set would be C.

I have come to realize that my greatest enemy in the GRE is my attention to detail when reading, so perhaps I’m incorrectly reading the questions and am making my own mistakes!

Hi Livia,

Thanks for the message :).

For the first question, if x and y are less than 4, then the median has to be 4, no matter how negative x and y are. If x and y are both larger than 7, even if they are really large, the median has to be 7. And we don’t care about those instances in between (where x and/or y are between 4 and 7) because we are looking for the range–which means we are looking for the extreme numbers.

For the last question, there were some other issues with it, so I took it down. Sorry for any confusion!

Hopefully, that clear up any confusion 🙂

Hi Chris,

another set of good questions…. 🙂

i solved 1 to 4 correctly but didn’t get the meaning of the question 5 , neither i get the solution, specially the digit thing. means actually what we have to do ??

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