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

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

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02 Jun 2007, 10:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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
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02 Jun 2007, 12:10
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

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02 Jun 2007, 12:50
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.
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02 Jun 2007, 13:07
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.

02 Jun 2007, 13:07
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