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Set A consists of integers {3, -8, Y, 19, -6}

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Set A consists of integers {3, -8, Y, 19, -6}

by sachin_yadav » Wed Jun 04, 2014 3:57 am
Set A consists of integers {3, -8, Y, 19, -6} and Set B consists of integers {K, -3, 0, 16, -5, 9}. Number L represents the median of Set A, number M represents the mode of set B, and number Z = LM. If Y is an integer greater than 21, for what value of K will Z be a divisor of 26?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2

OA is C

It took me 4 mins to solve this question. Please advice any faster method.

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by GMATGuruNY » Wed Jun 04, 2014 9:05 am
The intent of the problem seems to be that Z = L^M, as shown in red below:
sachin_yadav wrote:Set A consists of integers {3, -8, Y, 19, -6} and Set B consists of integers {K, -3, 0, 16, -5, 9}. Number L represents the median of Set A, number M represents the mode of set B, and number Z = L^M. If Y is an integer greater than 21, for what value of K will Z be a divisor of 26?
(A) -2
(B) -1
(C) 0
(D) 1
(E) 2
The MODE of a set is the value that appears the GREATEST NUMBER OF TIMES.
For Set B to have a mode -- for one of the values of Set B to appear more than any other value in set B -- K must be equal to one of the 5 values already present:
-3, 0, 16, -5, 9.
Of the 5 answer choices, only C -- K=0 -- is included in the list above.

The correct answer is C.
Last edited by GMATGuruNY on Thu Jun 05, 2014 10:36 am, edited 1 time in total.
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by sachin_yadav » Wed Jun 04, 2014 9:35 am
Mitch,

Thank you. Quick one :D

Regards
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by sidelinesk » Thu Jun 05, 2014 10:32 am
But then Z becomes 0, and how can 0 be a divisor?
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by GMATGuruNY » Thu Jun 05, 2014 10:41 am
sidelinesk wrote:But then Z becomes 0, and how can 0 be a divisor?
I believe that the posted problem was transcribed incorrectly.
The intent of the problem is not that Z = LM but that Z = L^M.
If the mode of set B is 0, then M=0, implying that Z = L� = 1.
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