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number game

by vaivish » Tue Sep 02, 2008 10:41 am
Of the 5 numbers, the largest number is 4 greater than the median. Is the mean greater than the median?
(1) The largest number plus the median is 34.
(2) The median minus the smallest number is 10.
A. Statement (1) ALONE is sufficient but Statement (2) ALONE is not sufficient.
B. Statement (2) ALONE is sufficient but Statement (1) ALONE is not sufficient.
C. BOTH Statements TOGETHER are sufficient, but NEITHER Statement alone is sufficient.
D. Each Statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.


Oa is b
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Source: — Data Sufficiency |
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by sushilmore » Tue Sep 02, 2008 11:24 am
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.
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by mayur00 » Tue Sep 02, 2008 11:59 am
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)
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