Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Mar 2009
Posts: 296

Lists S and T consist of the same number of positive integer
[#permalink]
Show Tags
Updated on: 04 Jun 2013, 04:47
Question Stats:
62% (02:07) correct 38% (01:55) wrong based on 419 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 10 May 2009, 12:12.
Last edited by Bunuel on 04 Jun 2013, 04:47, edited 1 time in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 49964

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
17 Feb 2011, 13:22
Original question is: 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?Q: is \(Median\{S\}>Mean\{T\}\)? Given: {# of terms in S}={# of terms in T}, let's say \(N\). (1) The integers in S are consecutive even integers, and the integers in T are consecutive odd integers. 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) The sum of the integers in S is greater than the sum of the integers in T. \(Sum\{S\}>Sum\{T\}\). 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 16 Apr 2009
Posts: 206
Schools: Ross

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
11 May 2009, 13:00
vaivish1723 wrote: Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T? 1. S is consecutive even integers and T is consecutive odd integers. 2. The sum of S is greater than sum of T. Please help. We'll use simple sets of 5 numbers (once we find out both conditions): S: [E, E+2, E+4, E+6, E+8]  we want median from Set S T: [O, O+2, O+4, O+6, O+8]  we want mean from Set T (1) Clearly not enough; 1 set could start with very small numbers & the other very big (& vice versa); (2) If we don't know statement 1, we can pick any numbers for both sets, and adjust each's mean & median to whatever we want, also not enough. (combined): We can make a few sets using both conditions combined: i.e. S: [2, 4, 6, 8, 10]; T: [1, 3, 5, 7, 9] Median of S = 6, mean of T = 5; in this case, answer  yes. S: [2, 4, 6, 8]; T: [1, 3, 5, 7] Median of S = 5, mean of T = 4; yes again. S: [2, 4, 6]; T: [1, 3, 5] Median of S = 4, mean of T = 3; yes So, now we can pretty much be sure we can answer this question reliably; so answer (C).  My opinion: It can be S: [2, 4, 6, 8]; T: [1, 3, 5, 7] or S: [ 6, 8,9,10]; T: [1, 3, 5, 7] too As the question states only about the same number of positive integers and the Stat(1) says S is consecutive even integers and T is consecutive odd integers.does that mean both S and T should be consecutive?hence I would go with 'E'. Correct me ,if I am wrong.
_________________
Keep trying no matter how hard it seems, it will get easier.



Manager
Joined: 16 Apr 2009
Posts: 206
Schools: Ross

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
11 May 2009, 13:28
Quote:  My opinion: It can be S: [2, 4, 6, 8]; T: [1, 3, 5, 7] or
S: [2, 4, 6, 8]; T: [7,9, 11, 13] too as the question states only about the same number of positive integershence I would go with 'E'.
Correct me ,if I am wrong.
Under the second condition, the sum of S must be greater than the sum of T, therefore the sets above do not qualify. S: [ 6, 8,10,12]; T: [1, 3, 5, 7] If this is the case,what would be the answer? Will the answer be E?
_________________
Keep trying no matter how hard it seems, it will get easier.



Senior Manager
Joined: 08 Jan 2009
Posts: 302

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
18 May 2009, 00:35
S and T have the same number of integers say n . Is Median of S > Median of T?
1. S is consecutive even integers and T is consecutive odd integers. So S could be 2,4,6 or 111112,1111114. So Insufficient. 2. The sum of S is greater than sum of T. If the sets are equally spaced then, Mean = Median. sum of S > sum of T then (sum of S)/n > (sum of T)/n If the set are not, Then S = (1,1,1,1,1) T=(1,2,1,2,3) here sum of S > sum of T but Median of S = Median of T
Together
Yes we know S and T are equally sapced. So as seen in Statement two yes. So Ans . C



Intern
Joined: 08 Jul 2009
Posts: 28

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
13 Oct 2009, 04:57
This can be solved the following ( Hope it is right) without pluging numbers. First: it is given that the two sets have same number of elements Stmt 1: Since each set has consecutive numbers then for each set the median will equal the mean. That's for odd and even consecutive sets. Ex: S { 1,2,3,4} mean= 2,5. Median = 2,5 T{ 1,2,3,4,5} mean = 3 and Median= 3 however it is Ins. since it gives us nothing else to help in determining the requested Stmt 2: since the sum of the first set is bigger than the second, it means the first has bigger mean however nothing is given about the median.. Together, yes the set with bigger mean has bigger median and that's applicable on set 2.. please correct me if there's anything wrong



Senior Manager
Joined: 03 Nov 2005
Posts: 337
Location: Chicago, IL

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
13 Oct 2009, 21:23
vaivish1723 wrote: Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T?
1. S is consecutive even integers and T is consecutive odd integers. 2. The sum of S is greater than sum of T.
Please help. ST. (1) two possible scenarios S{2,4,6} and T{1,3,5} or S{2,4,6} and T{3,5,7} both satsify the conditions under statement (1). So, st. 1 is insufficient. ST. (2) (insufficient) Look at the examples 1. S{1,3,5} and t{0,0,8) and S(10,40,60), while T{ 0, 45, 54}. St. 2 is also insufficient. Combining ST 1 and St. 2, we can conclude that the median of S is greater thean the mean of T.
_________________
Hard work is the main determinant of success



Intern
Joined: 01 Sep 2009
Posts: 33

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
14 Oct 2009, 12:50
C
Median = Mean for consec odd/even #s
If sum of S is greater than sum of T, then median and mean of S are greater than median and mean of T



Director
Joined: 25 Aug 2007
Posts: 799
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
15 Apr 2010, 05:55
vannu wrote: vaivish1723 wrote: Lists S and T consist of the same number of positive integers, Is the median of S is greater than mean of T?
1. S is consecutive even integers and T is consecutive odd integers. 2. The sum of S is greater than sum of T.
Please help.
We'll use simple sets of 5 numbers (once we find out both conditions):
S: [E, E+2, E+4, E+6, E+8]  we want median from Set S T: [O, O+2, O+4, O+6, O+8]  we want mean from Set T
(1) Clearly not enough; 1 set could start with very small numbers & the other very big (& vice versa); (2) If we don't know statement 1, we can pick any numbers for both sets, and adjust each's mean & median to whatever we want, also not enough.
(combined): We can make a few sets using both conditions combined:
i.e. S: [2, 4, 6, 8, 10]; T: [1, 3, 5, 7, 9] Median of S = 6, mean of T = 5; in this case, answer  yes. S: [2, 4, 6, 8]; T: [1, 3, 5, 7] Median of S = 5, mean of T = 4; yes again. S: [2, 4, 6]; T: [1, 3, 5] Median of S = 4, mean of T = 3; yes
So, now we can pretty much be sure we can answer this question reliably; so answer (C).
 My opinion: It can be S: [2, 4, 6, 8]; T: [1, 3, 5, 7] or
S: [ 6, 8,9,10]; T: [1, 3, 5, 7] too As the question states only about the same number of positive integers and the Stat(1) says S is consecutive even integers and T is consecutive odd integers.does that mean both S and T should be consecutive?hence I would go with 'E'.
Correct me ,if I am wrong. I think if u opted for E then u r write. My opinion: It can be S: [2, 4, 6, 8]; T: [1, 3, 5, 7]  Median can be 4 or 6 which is still insufficientor S: [ 6, 8,9,10]; T: [1, 3, 5, 7] Yes Median (S) > Avg (T) We dont know the number of elements in two set. If the number of elements is odd then yes we can find the relationship by combining both 1 and 2 statements while if it is even then we cant. So, E must be OA. In either case the Median of set S will be more than avg of set T.
_________________
Want to improve your CR: http://gmatclub.com/forum/crmethodsanapproachtofindthebestanswers93146.html Tricky Quant problems: http://gmatclub.com/forum/50trickyquestions92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/keyfundamentalsofgrammerourcruciallearningsonsc93659.html



Manager
Status: I am Midnight's Child !
Joined: 04 Dec 2009
Posts: 110
WE 1: Software Design and Development

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
17 Feb 2011, 12:49
ykaiim wrote: I think if u opted for E then u r write. My opinion: It can be S: [2, 4, 6, 8]; T: [1, 3, 5, 7]  Median can be 4 or 6 which is still insufficient or
S: [ 6, 8,9,10]; T: [1, 3, 5, 7] Yes Median (S) > Avg (T) We dont know the number of elements in two set. If the number of elements is odd then yes we can find the relationship by combining both 1 and 2 statements while if it is even then we cant. So, E must be OA.
In either case the Median of set S will be more than avg of set T. S: [2, 4, 6, 8]  Median = (4+6)/2 = 5T: [1, 3, 5, 7]  Mean = (1+7)/2 = 4Hence, here also Median of S > Mean of T . So, Answer should be C only.
_________________
Argument : If you love long trips, you love the GMAT. Conclusion : GMAT is long journey.
What does the author assume ? Assumption : A long journey is a long trip.
GMAT Club Premium Membership  big benefits and savings



Retired Moderator
Joined: 16 Nov 2010
Posts: 1439
Location: United States (IN)
Concentration: Strategy, Technology

Re: Mean and median of sets.. please help
[#permalink]
Show Tags
17 Feb 2011, 23:13
From 1 we can two sets as : 6 4 2  2 4 6  {S} 5 3 1  1 3 5  {T} From 2 we can have two sets as : 1 1 1 22 25  5 5 5 5 5  {S} 1 1 3 4 5  1 1 1 1 1  {T} So insufficient From 1 and 2, sufficient, so C.
_________________
Formula of Life > Achievement/Potential = k * Happiness (where k is a constant)
GMAT Club Premium Membership  big benefits and savings



Director
Joined: 29 Nov 2012
Posts: 788

Re: Lists S and T consist of the same number of positive integer
[#permalink]
Show Tags
29 Jul 2013, 06:01
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. So as per the translation of this statement ( in red) it means that the set could be Random example irrespective of the statements would be T(1,2,3,4, 5,6,7) S( 2,3,4,1,2,3) It doesn't mean all of them are positive right?
_________________
Click +1 Kudos if my post helped...
Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/
GMAT Prep software What if scenarios http://gmatclub.com/forum/gmatprepsoftwareanalysisandwhatifscenarios146146.html



NonHuman User
Joined: 09 Sep 2013
Posts: 8372

Re: Lists S and T consist of the same number of positive integer
[#permalink]
Show Tags
17 Jan 2018, 07:11
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: Lists S and T consist of the same number of positive integer &nbs
[#permalink]
17 Jan 2018, 07:11






