Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Mar 2009
Posts: 257

Lists S and T consist of the same number of positive
[#permalink]
Show Tags
Updated on: 12 Jul 2013, 00:58
Question Stats:
65% (02:06) correct 35% (02:04) wrong based on 512 sessions
HideShow timer Statistics
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T.
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by vaivish1723 on 17 Jan 2010, 10:45.
Last edited by Bunuel on 12 Jul 2013, 00:58, edited 1 time in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 58437

Re: mean and median
[#permalink]
Show Tags
17 Jan 2010, 17:01
vaivish1723 wrote: Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T. OA is Pl explain Q: is \(Median\{S\}>Mean\{T\}\)? Given: {# of terms in S}={# of terms in T}, let's say N. (1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is \(Mean\{S\}>Mean\{T\}\)? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}. (2) \(Sum\{S\}>Sum\{T\}\). Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1). (1)+(2) From (1) question became is \(Mean\{S\}>Mean\{T\}\)? > As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is \(\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}\) true? > Is \(Sum\{S\}>Sum\{T\}\)? This is exactly what is said in statement (2) that \(Sum\{S\}>Sum\{T\}\). Hence sufficient. Answer: C.
_________________




Manager
Joined: 22 Jan 2010
Posts: 103
Schools: UCLA, INSEAD

Re: mean and median
[#permalink]
Show Tags
24 Jan 2010, 14:24
Good question and good answer.



Intern
Joined: 10 Feb 2011
Posts: 49

Re: mean and median
[#permalink]
Show Tags
02 Jul 2011, 03:59
I feel there is an ambiguity in the wording of the question itself. The question says that set S and T consist of the same number of positive integers. This statement does not exclude the possibility that Set S and/or T might contain some negative numbers or zero.
Had the question said: " Set S and T consist of only the same number of positive integers then I would have agreed with the solution. What do you guys think? Also, can the original poster mention the source of this question pls?!



Director
Joined: 25 Apr 2012
Posts: 660
Location: India
GPA: 3.21
WE: Business Development (Other)

Re: mean and median
[#permalink]
Show Tags
03 Dec 2013, 22:13
Bunuel wrote: vaivish1723 wrote: Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T. OA is Pl explain Q: is \(Median\{S\}>Mean\{T\}\)? Given: {# of terms in S}={# of terms in T}, let's say N. (1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is \(Mean\{S\}>Mean\{T\}\)? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}. (2) \(Sum\{S\}>Sum\{T\}\). Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1). (1)+(2) From (1) question became is \(Mean\{S\}>Mean\{T\}\)? > As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is \(\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}\) true? > Is \(Sum\{S\}>Sum\{T\}\)? This is exactly what is said in statement (2) that \(Sum\{S\}>Sum\{T\}\). Hence sufficient. Answer: C. Hi Bunuel, In the above Question, it says there are same number of positive integers but does not explicitly rule out negative numbers or fractions etc. After combining 2 statements and taking the case of negative numbers we can have the following as well Let us say S= 2,0,2,4,6 and 8, Median will be 3 T= 1,1,3,5 and7, Median will be 3 Clearly there can be 2 cases and ans can be E. I think the question needs to mention that the list contains only positive integers Please confirm
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”



Intern
Joined: 15 Jul 2014
Posts: 9

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
04 Dec 2014, 20:07
hey Bunuel.
in your explanation for (1) and (2) combined, you derived a relationship for the Means of the two sets from statement (1). can you please explain how you got that. Statement (1) only tells us that one set has even consecutive integers and one odd. so how do you know that Mean of S > Mean of T?



Math Expert
Joined: 02 Sep 2009
Posts: 58437

Lists S and T consist of the same number of positive
[#permalink]
Show Tags
05 Dec 2014, 04:33
kritiu wrote: hey Bunuel.
in your explanation for (1) and (2) combined, you derived a relationship for the Means of the two sets from statement (1). can you please explain how you got that. Statement (1) only tells us that one set has even consecutive integers and one odd. so how do you know that Mean of S > Mean of T? From (1) we don't know whether \(Mean\{S\}>Mean\{T\}\). From (1) the question became "is \(Mean\{S\}>Mean\{T\}\)?"" Meaning that based on the information given in (1), the original question "is \(Median\{S\}>Mean\{T\}\)?" could be rephrased "is \(Mean\{S\}>Mean\{T\}\)?". Please reread the solution. Hope it helps.
_________________



Manager
Joined: 22 Aug 2014
Posts: 136

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
03 Apr 2015, 04:12
Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T.[/quote] Hey Bunuel, PLease explain this.When I take below sets I get answer E. Set S=(6,8,10)...Median=8 Set T=(1,3,5)...Mean=4.5 Median of S>mean of t If Set s=(2,4,6)...Median=4 Set T=(1,3,5)...Mean=4.5 Here,Median of S IS NOT GREATER THAN mean of T



Math Expert
Joined: 02 Sep 2009
Posts: 58437

Lists S and T consist of the same number of positive
[#permalink]
Show Tags
03 Apr 2015, 04:40
ssriva2 wrote: Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T. Hey Bunuel, PLease explain this.When I take below sets I get answer E. Set S=(6,8,10)...Median=8 Set T=(1,3,5).. .Mean=4.5Median of S>mean of t If Set s=(2,4,6)...Median=4 Set T=(1,3,5)... Mean=4.5Here,Median of S IS NOT GREATER THAN mean of T The median/mean of {1, 3, 5} is 3, not 4.5.
_________________



Manager
Joined: 22 Aug 2014
Posts: 136

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
03 Apr 2015, 05:05
Oh mean is 3,that was indeed a silly question!



Director
Affiliations: CrackVerbal
Joined: 03 Oct 2013
Posts: 563
Location: India

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
03 Apr 2015, 23:37
Hi All, Given: List S and T have same no. of elements. Median of S > average of T? Statement 1 is insufficient: Whenever it is consecutive median is always equal to mean. Let S =2 ,4 ,6 median is 4. Let T = 1,3,5 median(mean) is 3. Answer to the question is YES. Let S =2 ,4 ,6 median is 4. Let T = 3,5,7 median(mean) is 5. Answer to the question is NO. So not sufficient. Statement 2 is insufficient: Sum of the integers in S is greater than T(here the list not necessarily be consecutive). Let S =1 ,2 ,3 median is 2. Let T = 0,1,2 median(mean) is 1. Answer to the question is YES. Let S =1 ,5 ,20 median is 5. Let T = 5,6,7 median(mean) is 6. Answer to the question is NO. So not sufficient. Together it is sufficient. Since the S and T are consecutive even and odd and Sum of S is greater than T. Then median of S is always greater than mean(median) of T . So answer is together sufficient .
_________________
 CrackVerbal Prep Team For more info on GMAT and MBA, follow us on @AskCrackVerbal Register for the Free GMAT Kickstarter Course : http://bit.ly/2DDHKHqRegister for our Personal Tutoring Course : https://www.crackverbal.com/gmat/personaltutoring/ Join the free 4 part GMAT video training series : http://bit.ly/2DGm8tR



Current Student
Joined: 22 Sep 2016
Posts: 159
Location: India
GPA: 4

Lists S and T consist of the same number of positive
[#permalink]
Show Tags
Updated on: 06 Aug 2017, 06:33
Bunuel wrote: vaivish1723 wrote: Lists S and T consist of the same number of positive integers. Is the median of the integers in S greater than the average (arithmetic mean) of the integers in T? (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. (2) The sum of the integers in S is greater than the sum of the integers in T. OA is Pl explain Q: is \(Median\{S\}>Mean\{T\}\)? Given: {# of terms in S}={# of terms in T}, let's say N. (1) From this statement we can derive that as set S and set T are evenly spaced their medians equal to their means. So from this statement question becomes is \(Mean\{S\}>Mean\{T\}\)? But this statement is clearly insufficient. As we can have set S{2,4,6} and set T{21,23,25} OR S{20,22,24} and T{1,3,5}. (2) \(Sum\{S\}>Sum\{T\}\). Also insufficient. As we can have set S{1,1,10} (Median{S}=1) and set T{3,3,3} (Mean{T}=3) OR S{20,20,20} (Median{S}=20) and T{1,1,1} (Mean{T}=1). (1)+(2) From (1) question became is \(Mean\{S\}>Mean\{T\}\)? > As there are equal # of term in sets and mean(average)=(Sum of terms)/(# of terms), then we have: is \(\frac{Sum\{S\}}{N}>\frac{Sum\{T\}}{N}\) true? > Is \(Sum\{S\}>Sum\{T\}\)? This is exactly what is said in statement (2) that \(Sum\{S\}>Sum\{T\}\). Hence sufficient. Answer: C. I'm sorry, but I disagree. The question says that " the lists S and T consist of SAME NUMBER OF POSITIVE INTEGERS". Now you just know that the number of positive integers are same on both lists. What if there are some negative integers as well? or infact zero? I didn't interpret the question to say that there are same number of terms.
_________________
Desperately need 'KUDOS' !!
Originally posted by rekhabishop on 04 Aug 2017, 18:32.
Last edited by rekhabishop on 06 Aug 2017, 06:33, edited 1 time in total.



Manager
Joined: 12 Sep 2016
Posts: 64
Location: India
GPA: 3.15

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
05 Aug 2017, 12:27
We cannot assume that set A and set B do not contain negative numbers. Either the question needs to be edited to "same number of integers" or the OA needs to be changed to E.



Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 832

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
05 Aug 2017, 14:06
darn wrote: We cannot assume that set A and set B do not contain negative numbers. Either the question needs to be edited to "same number of integers" or the OA needs to be changed to E. We actually can assume that! The question itself says "S and T consist of the same number of positive integers." If it said " contain the same number of positive integers", your argument would be correct. But when a GMAT problem tells you that a set 'consists of' something, you know they've told you about everything that's in the set. Since the question uses the word 'consist' and says 'positive integers', you know that both sets consist of only positive integers.
_________________



NonHuman User
Joined: 09 Sep 2013
Posts: 13241

Re: Lists S and T consist of the same number of positive
[#permalink]
Show Tags
21 Jan 2019, 04:26
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Lists S and T consist of the same number of positive
[#permalink]
21 Jan 2019, 04:26






