Range

IMO C

median of 13 nums = 30; this means that the 7th value = 30. range of 30 and be denoted by x13-x1 = 30 (where x13 is the 13th term and x1 is first term).

to get the largest value of x13 we have to try and get the largest value of x1 and still preserve the condition of the difference being 30 and all the nums being different.

counting back from the median of 30, x1 can be 24 (remember we are looking for the highest value of x1).

x13 -x1 = 30 -> x13-24 = 30 -> x13 = 54.


to get the greatest num, we have to try and make the remaining 12 nums as small as possible bearing in mind that eachnumhas to be different.


hope that helps

can't this be a 60 ?? E ??


twelve 30s in a row...and then 30+30 which gives a range of 30.


correct me if i am wrong!!

Median =30

High-Low=30

Low<=30

High=30+Low

-->High<=60

Highest possible integer is 60

Pick E

selango wrote:Median =30

High-Low=30

Low<=30

High=30+Low

-->High<=60

Highest possible integer is 60

Pick E

In such a case there should be changes in papgust's 300 question set...

Yes it's C only.

See the question stem states that 13 different integers.

This means Low value can be 24 which is highest.

-->high=54.

These tiny mistakes make it between a 50 and a 51 on the test day!!

thank god its not test day!!

Arora, Selango, etc....

This problem/discussion should be flagged for all prospective gmat takers!

So much of this test is about reading carefully and finding what the trick is in the problem. Every time I go through an official GMAT Prep Test with one of my students I am struck by how many of the quant problems have less to do with high level quant skills and more to do with reading skills and general problem solving.

I have just started participating on this forum (though have been teaching the GMAT for 7+ years) and noticed immediately what incredible resources you both are to this site. Also, both of you clearly have HIGH level quant skills so it means something if both of you make a mistake on the same problem! We are all going to make these types of mistakes (God knows I make plenty) but the key is to learn from them every single time.

This problem in particular is great because it is using what I often call the double trap! In most tricky median problems, people forget that if the median is say 30 in a set of 13 numbers that the first 7 could all be 30 or the last 7 could all be 30. Having seen that type of trick before it is easy to quickly apply the same thinking here without noticing that all the integers must be different. One thing is for sure with these test makers: as soon as they establish a pattern with a certain trap or trick they will then turn it around and trap people who think it is the same problem!

The point that I try to make with my students is that you must actively look for tricks and subtleties in reading EVERY time you solve a quant problem. Before hitting next, make sure that you have not misinterpreted the wording and if the problem came together too nicely you are probably making a mistake. I have a collection of what I call "read carefully" problems that are so important and will link to a few of my favorites at the end.

Brian, my colleague at Veritas, has emphasized in many superb posts that this test is trying to determine who is best suited for business school not who has the highest level quant skills. To accomplish this goal, the test makers specifically make questions that test the following:

1. Problem Solving
2. Reading Skills
3. Logic
4. Ability to Perform under Pressure
5. Knowledge and Understanding of Basic Quant and Grammar Skills through mid high school.

Remember to engage all aspects of your brain on every quant problem, not just your knowledge of mathematics, and it will make that difference between a 48/49 and 50/51!

My two favorite examples of these types of questions are the following:

1. This question has been discussed many times on this forum and I honestly have only seen a few people get it correct the first time that they saw it. Several students of mine who went on to score 99th percentile on the quant did not get this right. https://www.beatthegmat.com/gmat-prep-t62358.html#283687 Note that in this link the answer choices are not in boxes as they are supposed to be.
2. This seemingly simple problem is all about reading comp and most people misinterpret how the fines are supposed to be calculated. https://www.beatthegmat.com/gmat-prep-si ... tml#189779

Any way - long post - but I think some important food for thought!

Great post Chris !!

I thought this would be a one off question... But I guess I was wrong....

GMAT surely is moving in this direction.... to try and check a future MBA's Ability to Perform under Pressure

Infact https://www.beatthegmat.com/gmat-prep-t62358.html#283687 is from a GMATPrep is an indicator of this!

life is a test wrote:IMO C

median of 13 nums = 30; this means that the 7th value = 30. range of 30 and be denoted by x13-x1 = 30 (where x13 is the 13th term and x1 is first term).

to get the largest value of x13 we have to try and get the largest value of x1 and still preserve the condition of the difference being 30 and all the nums being different.

counting back from the median of 30, x1 can be 24 (remember we are looking for the highest value of x1).

x13 -x1 = 30 -> x13-24 = 30 -> x13 = 54.


to get the greatest num, we have to try and make the remaining 12 nums as small as possible bearing in mind that eachnumhas to be different.


hope that helps

hi, thank you for your post but the thing i dont understand is why do we have to fnd the highest number of x1? why its not the smallest?

I myself was oversaw the 'different integers' criteria. I recall that more than 60% of my notes on the error log begin with 'Read closely'.

Chris, thank you for this awesome post!

diebeatsthegmat wrote:

life is a test wrote:IMO C

median of 13 nums = 30; this means that the 7th value = 30. range of 30 and be denoted by x13-x1 = 30 (where x13 is the 13th term and x1 is first term).

to get the largest value of x13 we have to try and get the largest value of x1 and still preserve the condition of the difference being 30 and all the nums being different.

counting back from the median of 30, x1 can be 24 (remember we are looking for the highest value of x1).

x13 -x1 = 30 -> x13-24 = 30 -> x13 = 54.


to get the greatest num, we have to try and make the remaining 12 nums as small as possible bearing in mind that eachnumhas to be different.


hope that helps

hi, thank you for your post but the thing i dont understand is why do we have to fnd the highest number of x1? why its not the smallest?

This question asks for the greatest possible integer.
So using the above equation : x13- x1=30 we can deduce:

x13=30 + x1; in order to maximize x13 x1 has to be maximized. Max vale of x1 can be 24.

Hope this helps

I disagree that the answer is C or E.

IMO this is how to break down the question.

You have a set of 13 different integers. And 30 is the median, which is defined as the middle number in the set, which means that it must be the 7th number in the set. It also says that the range of the set is 30, which means the difference between the highest and lowest number in the set is 30. Therefore, IMO, the middle number must be 15 less than the highest number and 15 greater than the lowest number. This means that the highest number is 45.

Notice that the question doesnt ask what is the highest number in the set, it asks what is the highest number that could be in the set. Therefore I think that the you should pick B.

An expert reply would be greatly appreciated

This my solution:
Call 13 different integers in order is a1<a2<a3<...a13 so we have:

a13 - a1= 30 and a1+a2+...a13=30*13=390

Note that all 13 integers in the set are different so a2>=a1+1; a3>=a2+1>=a1+2;...;a12>=a1+11
That's why we can induce a1+...+a12+a13>=a1+(a1+1)+(a1+2)+...+(a1+11)+(a1+30) = 13a1+96
SO a1=<(390-96)/13= 23
at the end a13=<53 ----> answer B