In one list of N consecutive integers, the smallest value is S and

Question

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

stne · Answer

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

rockstar09rulez · Answer

(C)

Decoding the question:
First list: S........M........S+(N-1)  
Second list: s........m........s+(n-1)

Evenly spaced sets: mean=median=(first term + last term) / 2
M = 2S+N-1 / 2
m = 2s+n-1 / 2
M-m = (2(S-s) + (N-n))/2

As long as we have S-s and N-n, we can find M-m
Statement 1 and 2 together provide both values. Hence, C is the answer.

sujoykrdatta · Answer

Obviously, from individual statements that provide information about the number of integers or the smallest integer, we cannot determine any relationship between their medians. Combining: List#1 has N consecutive integers while list#2 has n N - n = 4 Since the numbers are consecutive integers: The values in list#1 will be S, S+1, S+2 ... S+(N-1) => Median, M = (1st + last term)/2 = /2 = S + N/2 - 1/2 The values in list#2 will be s, s+1, s+2 ... s+(n-1) => Median, m = (1st + last term)/2 = /2 = s + n/2 - 1/2 Note: In an AP, median = mean = (1st term + last term)/2 Thus, M - m = S - s + (N/2 - n/2) = 2 + 2 = 4 - Sufficient Answer: C Recorded GMAT course https://topmate.io/sujoy_kumar_datta/1762791 Live one-on-one classes Whatsapp: https://wa.me/+919433063089

In one list of N consecutive integers, the smallest value is S and

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