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Math Expert
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Re M1527
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16 Sep 2014, 00:56
Official Solution: (1) The least element in both sets is 6. If \(S=\{6\}\) and \(T=\{6\}\) then the answer is NO but if \(S=\{6, 9\}\) and \(T=\{6\}\) then the answer is YES. Not sufficient. (2) Set T contains twice as many elements as set \(S\). If \(S=\{6\}\) and \(T=\{6, 12\}\) then the answer is NO but if \(S=\{6\}\) and \(T=\{0, 6\}\) then the answer is YES. Not sufficient. (1)+(2) Since \(T\) contains twice as many elements as set \(S\) and the least element in both sets is 6 then the median of \(T\) will always be more than the median of \(S\), so the answer to the question is NO. For example we can have that \(S=\{6\}\) and \(T=\{6, 12\}\) or \(S=\{6, 9\}\) and \(T=\{6, 12, 18, 24\}\) or \(S=\{6, 9, 12\}\) and \(T=\{6, 12, 18, 24, 30, 36\}\) ... As you can see in each case the median of \(T\) is more than the median of \(S\). Sufficient. Answer: C
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Re: M1527
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07 Apr 2015, 10:49
Can we consider only one element in a set and evaluate further if it is specifically mentioned in the question that the set has consecutive multiples/numbers.???? Because for the question to be justified there should be at least two elements in the set.



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Re: M1527
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13 Sep 2016, 15:25
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



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Re: M1527
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14 Sep 2016, 01:10
vijaisingh2001 wrote: 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 Notice that (1) says that the smallest element in either set is 6. So, when considering statements together we cannot take negative values.
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Re: M1527
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21 Sep 2016, 15:39
language is playing trick here,
if sentence 1 is as mentioned at below, E will be the right answer, one S makes the the difference
(1) The least elementS in either set is 6.
whenever we read least, it is always the numbers of elements in the set
smallest would have been more appropriate THAN LEAST, looks like this is a question of sentence correction more than the question of DS



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Re: M1527
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29 Jun 2017, 22:24
Bunuel wrote: Set \(S\) consists of consecutive multiples of 3 and set \(T\) consists of consecutive multiples of 6. Is the median of set \(S\) larger than the median of set \(T\)?
(1) The least element in either set is 6.
(2) Set \(T\) contains twice as many elements as set \(S\). Shouldn't we use 'both' instead of 'either' in statement in (1). Either gives an impression that least element in either of the set is 6. Or am i over thinking this?



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Re: M1527
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29 Jun 2017, 23:56



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Re: M1527
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30 Jun 2017, 00:14



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Re: M1527
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30 Jun 2017, 02:56
Bunuel wrote: FYI. The way it was written was also correct. Here is an official guide question with the same usage of "either of the two": https://gmatclub.com/forum/ifrandta ... 43303.htmlThank you Bunuel. I had a look at the question. I feel the way it is worded makes it unambiguous (at least to my eyes) since it uses keyword "two" > Phrase: The tens digit of r is less than either of the other two digits of r. Not that I wanted to contest the wording of this question, just wanted to clarify my doubt in case I encounter similar wording in future.



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Re: M1527
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30 Jun 2017, 03:01










