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# There are 6 numbers, A,B,C,D,E,F and their average is 42. Is

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Manager
Joined: 30 Mar 2007
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There are 6 numbers, A,B,C,D,E,F and their average is 42. Is [#permalink]  02 Jun 2007, 10:22
There are 6 numbers, A,B,C,D,E,F and their average is 42. Is the median less than the average?
A. A,B,C,D,E are at least 41.
B. F=47
VP
Joined: 08 Jun 2005
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statement 1

If A,B,C,D,E is at least 41 and the avarege is 42.

trying two options will give us:

A=41
B=41
C=41
D=41
E=41
F=47

Avarege = 42 & Median = 41

A=41
B=41
C=42
D=42
E=43
F=43

Avarege = 42 & Median = 42

insufficient

statement 2

insufficient

statement 1&2

knowing that F=47 - we can know that Avarege = 42 & Median = 41

sufficient

Manager
Joined: 07 Jan 2007
Posts: 59
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Adding to the above analysis, combing the two statements,

if F=47, then the average of remaining numbers A,B,C,D,E is (6*42-47)/5 = 41

stmt 1 says, "A,B,C,D,E are at least 41". This is possible only if all of the numbers A,B,C,D,E are 41.
VP
Joined: 08 Jun 2005
Posts: 1147
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buffetfollower wrote:
Adding to the above analysis, combing the two statements,

if F=47, then the average of remaining numbers A,B,C,D,E is (6*42-47)/5 = 41

stmt 1 says, "A,B,C,D,E are at least 41". This is possible only if all of the numbers A,B,C,D,E are 41.

That's what I meant ! excellent explanation buffetfollower.

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