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Mike, Scott, Jim, Kate, and Pete each have a different numbe
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17 Nov 2019, 04:34
17 Nov 2019, 04:34
Expert Reply
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Question Stats:
72% (01:19) correct
27% (01:21) wrong based on 58 sessions
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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?
Indicate all such people.
A. Mike
B. Scott
C. Jim
D. Kate
E. Pete
Kudos for the right answer and solution.
Indicate all such people.
A. Mike
B. Scott
C. Jim
D. Kate
E. Pete
Kudos for the right answer and solution.
Re: Mike, Scott, Jim, Kate, and Pete each have a different numbe
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23 Nov 2019, 16:24
23 Nov 2019, 16:24
Can you please explain.
I do not understand how could Pete have the median number of assignments since Pete has the smallest number of assignments.
I try to order them from the person holding the smallest number of assignments to the biggest number of assignment
Pete - Scott - Jim - Mike - Kate
I would probably go for B, C, and D
I do not understand how could Pete have the median number of assignments since Pete has the smallest number of assignments.
I try to order them from the person holding the smallest number of assignments to the biggest number of assignment
Pete - Scott - Jim - Mike - Kate
I would probably go for B, C, and D
Re: Mike, Scott, Jim, Kate, and Pete each have a different numbe
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23 Nov 2019, 17:32
23 Nov 2019, 17:32
1
Expert Reply
\(P < K\)
\(K > M\)
\(M> J\)
\(J > S\)
The question is NOT a must be but a could be
From the chain above
\(P < J < M < K\)
So K is on one extreme. However, we do not precise where S stands
We could have \(P < S < J < M < K\)
OR
\(S < P < J < M < K\) for instance.
In any case what could be is that the median could be formed by P , J, and M
Hope now is more clear
\(K > M\)
\(M> J\)
\(J > S\)
The question is NOT a must be but a could be
From the chain above
\(P < J < M < K\)
So K is on one extreme. However, we do not precise where S stands
We could have \(P < S < J < M < K\)
OR
\(S < P < J < M < K\) for instance.
In any case what could be is that the median could be formed by P , J, and M
Hope now is more clear
