number game

Condition I:

l+m=34 and given l=4+m

solving we get median to be 15 and largest to be 19.
we have set as {a, b, 15, c, 19}
a=11, b=13, c=17 we will get mean is 15
a=8, a=9 c=16, we will get mean is 67/5 = 13.6 < 15

a=13. b=14, c=16, we will get 77/5 = 15.04 > 15

Hence I is not sufficient

condition II

m-s=10 and l=4+m
hence l-s = 14

so lets say set is {1,b,11,c,15). We can see that b can take max value as 10 and c = 12, 13 , 14. The mean is always less than a median in all these cases.

Lets take another set {10, b, 20, c, 24).
b=19, c=21,22, 23

We can see mean is less than median

Hence B.

Given:

L=M+4

Condition I:

L + M = 34, Therefore L=19 and M=15.
This doesn't tell us where the mean will lie. It is less that 19 is all we know.


Condition II:

S = M - 10


Therefore Mean = (L+X+M+Y+S)/5, where X and Y are the remaining two numbers beside the smallest, largest and the median.

Therefore Mean = (M+4 + M + M-10 + X + Y)/5 = (3M - 6 + X + Y)/5

According to the definition of Median:
M<= X <= L and S<=Y for the set {L, X, M, Y, S},
therefore max value of X=L and max value of Y = M


Plugging these values, Mean <= (3M - 6 + M+4 + M)/5 = (5M - 2)/5
Therefore Mean <= (M - 2/5)

Hence Mean is always less that Median (M)