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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
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C
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29%
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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.
17 Jan 2010, 17:01
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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
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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.
