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avg of m integers

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Source: — Data Sufficiency |
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by heshamelaziry » Mon Nov 09, 2009 11:25 pm
IMO E.

if the set has 3,6 the average is 3. if the set has 3,6,9 the average is 6. Stetmnt 1 ---> insuficient.

when 33 is the median, many multiples of 3 above and below 33 could be sufficient and the average will differ; range unknown. ----> insufficient.

Combo:

No new info can be deduced.
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by okigbo » Mon Nov 09, 2009 11:33 pm
IMO the answer is C

For a consecutive set of integers, median=mean. You need both statements.

Let me know if I am wrong.
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by heshamelaziry » Mon Nov 09, 2009 11:46 pm
You are absoulutly correct. I am sufferering from memory loss :oops:
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by vivekjaiswal » Tue Nov 10, 2009 12:45 am
okigbo wrote:IMO the answer is C

For a consecutive set of integers, median=mean. You need both statements.

Let me know if I am wrong.
Even i think its C...
Thanks for reminding this important relationship between mean and median of an equally spaced set.
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by xcusemeplz2009 » Tue Nov 10, 2009 1:02 am
IMO C

M CAN BE ANYTHING 1,3,5,7

STATEMNET1) M CAN BE 3 6 9 OR 3 6 9 12 15 ......
HENCE AVG IS NOT CONISITENT

STATMENT2) M CAN BE 32 33 34 OR
1 33 100

AVG ARE DIFF..

COMB...
M CAN BE 27 30 33 36 39
OR 30 33 36

LET ANY NO CASES
MED AND MEAN IS SAME 33

HENCE C
It does not matter how many times you get knocked down , but how many times you get up
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