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:

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

(A) 25

(B) 26

(C) 30

(D) 34

(E) 40

**2. 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 | Column B |
---|---|

The average of Set A and 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.

**3. 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).

**Takeaways**

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

For Q2,

the answer i got is 64.91 which is nothing but 65

so the answer should be C right ?

Hi Indu,

For the GRE, you should never round the numbers unless you are specifically told that you should estimate the answer. Even if the answer were 64.9999, it is still less than 65! It’s also important to realize that you don’t need to do any calculations to answer this question, as long as you recognize the basic idea behind weight averages. The group with the average of 60 is slightly larger than the group with the average of 70, which means that the 60 group will “weigh down” the average and pull it closer to 60. If the group sizes were equal, the answer would be 65. Since the 60 group is “heavier,” the answer must be less than 65. Does that make sense?

Hello Chris,

For Q3, I do not understand why the average of Set T would be 60. How do we come to that calculation?

Thank you

Best,

Isha

When the problem says that Set S and Set T both have

xelements, it means that Set S and Set T both have the same number of things… and thus both have the same “weight” in a calculated average.To help you picture how this allows them to be averaged together for an average of 50, let’s plug in some very simple numbers. Let’s say that Set S gets is average of 40 from having 3 items, each of which has a value of 40. Thus, Set S = 40 + 40 + 40. And (40 + 40 + 40)/ 3 = 120/3 = 40 = the average of Set S. In this scenario, Set T must ALSO have exactly three items, because both sets have the same number of items, that common

xvalue. We’ll say that Set T gets its average of 60 from three items with a value of 60. (60 + 60 + 60)/3 = 50.Between Set S and Set T, you have a total of 6 items. Add all 6 items up, divide the sum by 6, and you have the combined average for the two sets: [(40+40+40)+(60+60+60)]/6 = (120+180)/6 = 300/6 = 50.

Or to put it more simply if Set S and Set T are equally weighted, then you cna simply add the two together, divide by 2, and get an average for the pair of sets. If you know that Set S is 40 and that the average is 60, them your algebraic equation to find the value of Set S would be (40 + T)/2 = 50.

Solve by: (40 + T)/2 = 50 >>> [(40+T)/2]*2 = 50*2 >>> 40+T = 100 >>> 40+ T-40 = 100 -40 >>> T=100-40 >> T = 60.

(60 + 60 + 60)/3 = 60 (not 50)

Yes you are correct–it looks like we made a type there, but meant to write 60. Thanks for catching it!

When the problem says that Set S and Set T both have

xelements, it means that Set S and Set T both have the same number of things… and thus both have the same “weight” in a calculated average.To help you picture how this allows them to be averaged together for an average of 50, let’s plug in some very simple numbers. Let’s say that Set S gets is average of 40 from having 3 items, each of which has a value of 40. Thus, Set S = 40 + 40 + 40. And (40 + 40 + 40)/ 3 = 120/3 = 40 = the average of Set S. In this scenario, Set T must ALSO have exactly three items, because both sets have the same number of items, that common

xvalue. We’ll say that Set T gets its average of 60 from three items with a value of 60. (60 + 60 + 60)/3 = 50.Between Set S and Set T, you have a total of 6 items. Add all 6 items up, divide the sum by 6, and you have the combined average for the two sets: [(40+40+40)+(60+60+60)]/6 = (120+180)/6 = 300/6 = 50.

Or to put it more simply if Set S and Set T are equally weighted, then you cna simply add the two together, divide by 2, and get an average for the pair of sets. If you know that Set S is 40 and that the average is 60, them your algebraic equation to find the value of Set S would be (40 + T)/2 = 50.

Solve by: (40 + T)/2 = 50 >>> [(40+T)/2]*2 = 50*2 >>> 40+T = 100 >>> 40+ T-40 = 100 -40 >>> T=100-40 >> T = 60.https://magoosh.com/gre/wp-admin/edit-comments.php?comment_status=moderated#comments-form

For question 2, shouldn’t the answer be (A)?

I calculated the weighted mean.

(30*60 + 29*70) / (60+70) = 64.91, which is less than 65. So Column B is greater than Column A.

hey katy…you first check you calculation there must be sth wrong with formula…either you can see calculation of Manikanda pandiayan, but denominator is(30+29)

Hi Chris,

Can you please show me how you would solve Q1 without using the process of summing the numbers and dividing by 5??

Because of the comment “Sometimes, the answer to a problem can require very little math”, I figure there is a more straight forward method….

Really old comment, but since I was looking for the same, I thought I’d post my method.

1. On scrap paper, I modified the numbers to write the following.

Scrap paper: 20 – 20 – 25 – 35 – 50

Mental math: +3 and -3 for the first two numbers; +1 and -1 for the last two numbers. This will not affect the average.

2. Mental math: Reduce 20 from all numbers to get this new modified set. This makes it easier to work with.

Scrap paper: 0 – 0 – 5 – 15 – 30.

3. Modified sum = 50. Modified avg = 50/5 = 10.

Therefore, correct avg = 10 + 20 (here, we bring back the 20 from step 2).

PS:

– In step 2, instead of reducing all numbers by 20 (and adding it back in step 3), you could instead, divide all numbers by 5 (and then in step 3, multiply the modified avg by 5). Slightly messier, but the principle applies and you’d get the same answer.

– Hope the paragraph spaces show up correctly, or this isn’t gonna make any sense.