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11 Nov 2022, 08:33
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:20) correct
32%
(01:04)
wrong
based on 38
sessions
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Not Attempted Yet
If the two sets have an equal number of numbers, is the mean of set Q lower than the mean of set P?
(1) Set Q consists of consecutive even integers and set P of consecutive odd integers.
(2) The median of Q is higher than the mean of P
(1) Set Q consists of consecutive even integers and set P of consecutive odd integers.
(2) The median of Q is higher than the mean of P
Updated on: 11 Nov 2022, 13:14
Originally posted by gmatophobia on 11 Nov 2022, 10:07.
Last edited by gmatophobia on 11 Nov 2022, 13:14, edited 1 time in total.
Last edited by gmatophobia on 11 Nov 2022, 13:14, edited 1 time in total.
Corrected Typo
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Given
Question
\(Mean_Q < Mean_P\)
Statement 1
Set Q consists of consecutive even integers and set P of consecutive odd integers.
We don't know the starting or the ending values of the P and Q.
Case 1
P = {2, 4, 6}
Q = {101, 103, 105}
Is \(Mean_Q < Mean_P\) -- No
Case 1
P = {2233, 2235, 2237}
Q = {102, 104, 106}
Is \(Mean_Q < Mean_P\) -- Yes
The statement is not sufficient and we can eliminate A and D
Statement 2
The median of Q is higher than the mean of P
Just as in Statement 1, the number could lie anywhere.
Case 1
Q = {-10000, 50, 51}
P = { 10, 11, 12}
Is \(Mean_Q < Mean_P\) -- Yes
Case 2
Q = {49, 50, 51}
P = { 10, 11, 12}
Is \(Mean_Q < Mean_P\) -- No
This statement is not sufficient as well and we can eliminate B.
Combined
We know that the number of terms in each set is same. We are given that median of Q is higher than the mean of P, hence in set Q all the terms on the left of the median will have a value greater than the mean of set P. As the number of terms is same, and the difference between terms is constant, we cannot adjust the excess value of Q in any way in set P.
Hence the mean of P will be less than the mean of Q
Is \(Mean_Q < Mean_P\) -- No
We have a definite answer.
IMO C
- The number of terms in both the sets are same.
Question
\(Mean_Q < Mean_P\)
Statement 1
Set Q consists of consecutive even integers and set P of consecutive odd integers.
We don't know the starting or the ending values of the P and Q.
Case 1
P = {2, 4, 6}
Q = {101, 103, 105}
Is \(Mean_Q < Mean_P\) -- No
Case 1
P = {2233, 2235, 2237}
Q = {102, 104, 106}
Is \(Mean_Q < Mean_P\) -- Yes
The statement is not sufficient and we can eliminate A and D
Statement 2
The median of Q is higher than the mean of P
Just as in Statement 1, the number could lie anywhere.
Case 1
Q = {-10000, 50, 51}
P = { 10, 11, 12}
Is \(Mean_Q < Mean_P\) -- Yes
Case 2
Q = {49, 50, 51}
P = { 10, 11, 12}
Is \(Mean_Q < Mean_P\) -- No
This statement is not sufficient as well and we can eliminate B.
Combined
We know that the number of terms in each set is same. We are given that median of Q is higher than the mean of P, hence in set Q all the terms on the left of the median will have a value greater than the mean of set P. As the number of terms is same, and the difference between terms is constant, we cannot adjust the excess value of Q in any way in set P.
Hence the mean of P will be less than the mean of Q
Is \(Mean_Q < Mean_P\) -- No
We have a definite answer.
IMO C
11 Nov 2022, 13:16
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kapil1995
I see there was a typo. I didn't even think of the values before typing them as a part of explanation
, all I knew was that the statement was not sufficient as we don't know at what point in space the values lie.Thanks for pointing that. Corrected in the explanation. Kudos for the observation !
