|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Updated on: 22 May 2021, 13:27
Originally posted by vaivish1723 on 10 May 2009, 12:12.
Last edited by Bunuel on 22 May 2021, 13:27, edited 2 times in total.
Last edited by Bunuel on 22 May 2021, 13:27, edited 2 times in total.
Edited the question and added the OA
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:09) correct
29%
(02:11)
wrong
based on 1501
sessions
History
Date
Time
Result
Not Attempted Yet
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.
(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.
Every tricky question should trigger a follow up quiz.
Build that habit with
GMAT Club Forum Quiz →.
17 Jan 2010, 17:01
Kudos
Bookmarks
vaivish1723
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.
17 Feb 2011, 13:22
Kudos
Bookmarks
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.
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.
