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# Mike, Scott, Jim, Kate, and Pete each have a different number of assig

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Re: Mike, Scott, Jim, Kate, and Pete each have a different number of assig [#permalink]
Bunuel wrote:
Mike, Scott, Jim, Kate, and Pete each have a different number of assignments this month. Pete has fewer assignments than Kate, Kate has more assignments than Mike, Mike has more assignments than Jim, and Jim has more assignments than Scott. Which of the following could be the person who has the median number of assignments this month for the five people listed?

I. Mike
II. Jim
III. Pete

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

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We have:

K > P

K > M

M > J

J > S

Using the inequalities above, we can determine the number of assignments (except for Pete) is ordered as:

K > M > J > S

If P > M, then the median is M.

If M > P > J, then the median is P.

If J > P, then the median is J.

Thus, we see that the person with the median number of assignments could be Mike, Pete or Jim.

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Re: Mike, Scott, Jim, Kate, and Pete each have a different number of assig [#permalink]
Assuming Mike = M, Jim = J, Kate = K, and Pete = P

Given, K>P; K>M; M>J; J>S

From the above, we can say K>P and K>M>J>S

Listing the possibilities now (in increasing order) to calculate the median:

K-M-J-S-P (Median is J)
K-P-M-J-S (Median is M)
K-M-P-J-S (Median is P)

Hence, we have all three possibilities i.e. J,M and P