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16 Jan 2022, 10:52
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C
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Difficulty:
65%
(hard)
Question Stats:
55% (02:18) correct
45%
(01:52)
wrong
based on 94
sessions
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In one list of N consecutive integers, the smallest value is S and the median is M. In a second list of \(n\) consecutive integers, the smallest value is \(s\) and the median is \(m\). What is the value of M −\(m\)?
1) N −\(n\) = 4
2) S −\(s\) = 2
1) N −\(n\) = 4
2) S −\(s\) = 2
20 Jan 2022, 21:24
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nislam
1) N −\(n\) = 4
This tells us that the first has \(4\) elements more than the second. Obviously we have no idea of the elements of the sets HENCE INSUFF.
2) S −\(s\) = 2
We know that the smallest element of the first set is two more than the smallest element in second set .
So we could have :
N \(=( 3, 4, 5,6.....)\)
n\(= (1, 2,3,4.....)\)
or
N \(=( 5, 6, 7,8.....)\)
n \( =(3, 4,5,6.....)\)
or
etc. There could be many other combinations
INSUFF.
1+2
N \(=( 3,4,5,6,7,8,9) \)-> Median(M) \(= 6\)
n \(=( 1,2,3)\)-> Median \((m)= 2\)
M-m\(= 4 \)
N\(=(7,8,9,10,11,12,13,14)\) M\(= 10.5\)
n\(= (5,6,7,8)\) m\(= 6.5 \)
M-m\(= 4\)
So it doesn't matter how many elements are there in \(n \) or which is the first element in \(n \) , we know set \(N \) will always have \(s+2\) as the first element and \(n+4\) no. of elements. Hence \(M-m\) remains constant.
Hence both together are SUFF.
Ans C
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