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.
avg of m integers
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heshamelaziry
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heshamelaziry
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vivekjaiswal
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Even i think its C...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.
Thanks for reminding this important relationship between mean and median of an equally spaced set.
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xcusemeplz2009
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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
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
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