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16 Sep 2014, 00:56
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C
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Difficulty:
55%
(hard)
Question Stats:
57% (01:32) correct
43%
(02:03)
wrong
based on 110
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Set \(S\) consists of consecutive multiples of 3, while set \(T\) consists of consecutive multiples of 6. Is the median of set \(S\) greater than the median of set \(T\)?
(1) The smallest element in both sets is 6.
(2) Set \(T\) has twice as many elements as set \(S\).
(1) The smallest element in both sets is 6.
(2) Set \(T\) has twice as many elements as set \(S\).
16 Sep 2014, 00:56
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Official Solution:
Set \(S\) consists of consecutive multiples of 3, while set \(T\) consists of consecutive multiples of 6. Is the median of set \(S\) greater than the median of set \(T\)?
(1) The least element in both sets is 6.
If \(S=\{6, 9\}\) and \(T=\{6, 12\}\), then the answer is NO. However, if \(S=\{6, 9, 12, 15\}\) and \(T=\{6, 9\}\), then the answer is YES. Not sufficient.
(2) Set \(T\) has twice as many elements as set \(S\).
If \(S=\{6, 9\}\) and \(T=\{60, 66, 72, 78\}\), then the answer is NO. However, if \(S=\{6, 9\}\) and \(T=\{-78, -72, -66, -60\}\), then the answer is YES. Not sufficient.
(1)+(2) Given that set \(T\) contains twice as many elements as set \(S\), and the smallest element in both sets is 6, the median of set \(T\) will always be greater than the median of set \(S\). For example, we can have \(S=\{6\}\) and \(T=\{6, 12\}\), or \(S=\{6, 9\}\) and \(T=\{6, 12, 18, 24\}\), or \(S=\{6, 9, 12\}\) and \(T=\{6, 12, 18, 24, 30, 36\}\), and so on. In each case, the median of set \(T\) is greater than the median of set \(S\). Therefore, the answer to the question is NO. The combination of statements (1) and (2) is sufficient.
Answer: C
Set \(S\) consists of consecutive multiples of 3, while set \(T\) consists of consecutive multiples of 6. Is the median of set \(S\) greater than the median of set \(T\)?
(1) The least element in both sets is 6.
If \(S=\{6, 9\}\) and \(T=\{6, 12\}\), then the answer is NO. However, if \(S=\{6, 9, 12, 15\}\) and \(T=\{6, 9\}\), then the answer is YES. Not sufficient.
(2) Set \(T\) has twice as many elements as set \(S\).
If \(S=\{6, 9\}\) and \(T=\{60, 66, 72, 78\}\), then the answer is NO. However, if \(S=\{6, 9\}\) and \(T=\{-78, -72, -66, -60\}\), then the answer is YES. Not sufficient.
(1)+(2) Given that set \(T\) contains twice as many elements as set \(S\), and the smallest element in both sets is 6, the median of set \(T\) will always be greater than the median of set \(S\). For example, we can have \(S=\{6\}\) and \(T=\{6, 12\}\), or \(S=\{6, 9\}\) and \(T=\{6, 12, 18, 24\}\), or \(S=\{6, 9, 12\}\) and \(T=\{6, 12, 18, 24, 30, 36\}\), and so on. In each case, the median of set \(T\) is greater than the median of set \(S\). Therefore, the answer to the question is NO. The combination of statements (1) and (2) is sufficient.
Answer: C
13 Sep 2016, 15:25
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i think answer should be E, negative numbers are also the multiple, example -9 is multiple of 3
set T contains 13 elements all negatives
example ( -13*3, -12*3, -11*3 ........-1*3 ) so median is -7*3= -21
set S contains 06 elements- all positive integers, multiple of 3
example (3*1, 3*2, .......3*6) so median will be 10.5 so now median of S greater than median of T
if we take T set starting from 21 ( 21, 24,27,30..........57) set T will have much higher MEDIAN than S
set T contains 13 elements all negatives
example ( -13*3, -12*3, -11*3 ........-1*3 ) so median is -7*3= -21
set S contains 06 elements- all positive integers, multiple of 3
example (3*1, 3*2, .......3*6) so median will be 10.5 so now median of S greater than median of T
if we take T set starting from 21 ( 21, 24,27,30..........57) set T will have much higher MEDIAN than S
