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average value of set > median of set ?

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average value of set > median of set ?

by gmatjeet » Sun Jun 05, 2011 8:02 am
In a set of 5 numbers if the largest number is 3 more than the median, Is the average value of set > median of set ?

1. All the numbers are different.

2. The median is 10 more than the smallest number in the set.

I tried this question and am confused as my answer is coming out to E but on the forum, everyone is giving the answer as B
-> If the median is considered to be always positive, then the answer should be B. But if we consider the median to be either positive or answer then my answer is E. Can someone please help.
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Source: — Data Sufficiency |
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by cans » Sun Jun 05, 2011 8:16 am
a,b,c,d,e numbers in ascending order
e=c+3
(a+b+c+d+e)>5c?? ->a+b+d>3c-3
1. All numbers are different. insufficient. (take a.b small enough and c large enough and d=c,inequality won't hold. but if you take a=b=c=d,0>-3 true)
2. c=a+10
b+d>2c+7
max value of d=c+3 (which is same as largest number)
min value of d=c (same as median)
if we take min d,b>c+7 not true
if we take max d, b>c+4 not true
Thus avg is always less than mean. Sufficeint
IMO B
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by Frankenstein » Sun Jun 05, 2011 8:24 am
Hi,
As you are aware that (1) is not sufficient, I will skip that
From(2): Let the numbers in increasing order be a,b,c,d,e
i.e. c-10,b,c,d,c+3, where c-10<=b<=c and c<=d<=c+3
Minimum value of sum of the numbers is when b=c-10 and d=c
Sum = 5c-17. So mean = c-3.4
Maximum value of sum of the numbers is when b=c and d=c+3
Sum = 5c-4. So, mean is c-0.8
In both the cases mean < median

Hence B
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