Another Way To Solve Median & Mean Questions

Joe Lucero —  July 15, 2013 — 2 Comments

gmat median & meanThis is the second of a series of posts that offer alternate ways to solve certain GMAT problems (check out the first here: DS Value Problems). Just like last time, if you like the method, steal it! And if you don’t, I promise not to lose any sleep. There’s a lot of ways to solve most questions on the GMAT and the best way will always be the way that works best for you. So without further ado, let’s check out a GMATPrep question and see how fast you can solve:

Last month 15 homes were sold in Town X. The average (arithmetic mean) sale price of the homes was $150,000 and the median sale price was $130,000. Which of the following statements must be true?

I. At least one of the homes was sold for more than $165,000.

II. At least one of the homes was sold for more than $130,000 and less than $150,000.

III. At least one of the homes was sold for less than $130,000.

 

(A) I only

(B) II only

(C) III only

(D) I and II

(E) I and III

First things first, if you answered this question using algebra, you’re in great company. Another one of our instructors, Stacey Koprince, has a great write up on the algebra in this question, and it’s definitely worth a read-through right here. But a lot of questions on the GMAT, including this one, can be solved by thinking of extremely simple scenarios, rather than the algebra that determines all of them.

The first thing I noticed on this question is that this is one of those awful questions where there’s a whole lot of wiggle room with the information that they give you. What was the cheapest house? What was the cost of the third most expensive house? Were any of the houses all the same price? If the second cheapest house is half as expensive as the most expensive, how does that affect the cost of the other houses? It’s easy to get lost when you start to think about how little you know in this scenario.

But before I jump around and start picking values out of thin air, the most important part of this problem are the (few) things that MUST be true. In this case, there are two: the 15 house prices have a mean of $150,000 and a median of $130,000. And on my paper, I would write out a few slots to represent the house prices like this: (note- I wouldn’t write out all 15 slots. Just the first few, the last few, and, since this is a median problem, one in the middle.)

 

____

____

____

____

____

1

2

7

14

15

 

Again, there are two things that they tell me here, but I want to start with the most restrictive element in this problem. There are lots of different ways to get a mean of $150,000, but in order to get a median of $130,000, I would need at least one house to cost EXACTLY $130,000. So I add that to my chart (ignoring the $ sign and extra zeroes):

____

____

_130_

____

____

1

2

7

14

15

 

Now here’s where my method is quite different than standard algebra. Algebraically, I could figure out the total cost of the other 14 houses. But that leaves us with an unlimited number of potential scenarios that the houses could cost. My approach is to start by finding ONE scenario that could work. And the easiest group of numbers whose median is 130? A whole bunch of 130s:

 

_130_

_130_

_130_

_130_

_130_

1

2

7

14

15

 

Now hopefully something looks amiss to you. If every house cost $130,000, the median price would be $130,000 but so would the mean. And here lies the trick to this question. If you can see that at least one of these houses would have to cost more than $130,000, you can solve this particular question without any further math:

 

BIG #!

_130_

_130_

_130_

_130_

_130_

1

2

7

14

15

 

If I actually wanted to calculate what that much larger number would be, I could (total price = mean x 15 = $2.25 million; price of the most expensive house = total price − the cost of the other houses = $2.25 million − ($130,000 x 14) = $430,000). But more importantly, I’ve figured out one scenario that COULD be true. As we look for other scenarios that could also be true, we can think about how to make other houses more or less expensive. To find the maximum price of house 15, I’d minimize the price of houses 1-6 ($0) and 8-14 ($130). To find the minimum price of house 15, I’d try to maximize the prices of those other houses. But for me, once I see one scenario that works, its much easier to visualize how to play around with those other values. And notice how useful our very simple scenario is for helping us evaluate the three statements:

 I.  At least one of the homes was sold for more than $165,000.

That does happen in our scenario, so let’s skip it for a minute.

 II.  At least one of the homes was sold for more than $130,000 and less than $150,000.

Not in our scenario. We have a whole bunch of houses that cost $130,000 and one that costs way more than $150,000. I could even minimize houses 1-6 and make houses 8-14 each cost more than $150,000. This does NOT have to be true.

 

III. At least one of the homes was sold for less than $130,000.

Not in our scenario. We had a lot of $130,000 houses and one very expensive house. This does NOT have to be true.

That’s it. We’re done. Since there’s no option that says None of the Above, we know that the first roman numeral must be true and the correct answer is (A). Like so many other questions on the GMAT, this approach isn’t about proving our answer, but instead is about finding four answers that can’t be right. It turns out that (I) has to be true, because if we maximize the prices of the other 14 houses, we would get this:

 

_130_

_130_

_130_

_165_

_185_

1

2

7

14

15

 

Houses 1-7 could cost as much as $130,000 apiece, houses 8-14 could be priced at the upper limit of $165,000, but house 15 would have to cost $185,000 in order for the mean to reach $150,000. You might also recognize that if houses 1-6 were all $130,000 and houses 8-15 were all $165,000, the median of those prices would be $145,000, less than the actual median, so one house must cost more than $165,000.

Notice that the key to this method of solving is something that is also applicable to lots of other quant questions. When the GMAT gives you several pieces of information that you need to use to find an answer, it’s hard to find a scenario that accommodates all of that information, much less to think about all the scenarios that could also potentially be true. The key is to find the easiest possible scenario that could work and then rearrange elements (in this case, shifting around the extra money from the most expensive house, or taking some money from some of the less expensive houses) to see what else could work too. What’s the easiest group of 15 numbers that has a median of 130? 15 numbers that equal 130. What’s the easiest group of 15 numbers that has a median of 130 and ALSO has a mean of 150? 14 numbers that equal 130 and one number that is much larger. When it comes to complex questions, remember to KISS: keep it simple, students.

 

GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

Joe Lucero

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Joe Lucero has both a Biology degree and a Master of Education from the University of Notre Dame. He also has a 780 on his GMAT. In the fall, you will find Joe in a much better mood during weeks after the Fighting Irish win their football game. During the rest of the year, you will find him looking for new places to travel, reading almost anything non-fiction, crossfitting, and trying to solve every challenge problem in the Manhattan GMAT Student Center.

2 responses to Another Way To Solve Median & Mean Questions

  1. I like this method a lot! It’s also a great example of how when the GMAT asks you whether something “must” be true, often the easiest approach is to find one scenario that disproves it.

  2. - Median position is no 8 and not 7 right?
    - the median of those prices would be $145,000, “higher” than actual median?

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