If m is a positive odd integer, what is the average (arithmetic mean) of a certain set of m integers

(1) The integers in the set are consecutive multiples of 3. Clearly insufficient. Consider: {0, 3, 6} and {3, 6, 9}. Not sufficient.

From this statement, though we can deduce that since the set is evenly spaced then its mean equals to its median.

(2) The median of the set of integers is 33. Also, insufficient. Consider: {31, 33, 35} and {31, 33, 133}. Not sufficient.

(1)+(2) From (1) we have that mean=median and from (2) we have that meidan=33, thus mean=median=33. Sufficient.

Answer: C.
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in a set of consecutive numbers the mean is equal to the median. hence, C)...

Statement 1 says

multiples of 3

so it can be 3,6,9 or 3,6,9,12,15

different mean's insufficient


Statement 2 says

33 is the middle value but we have no other info insufficient

Combined we know that 33 is the middle value so there are 10 terms on either side of 33

1st value is 3
11th value is 33
21st value is 63


---------33-------

So we can find out mean = median = 33 sufficient

Given : m is a positive Odd Integer

Question : Arithmetic Mean of a set of m Integers = ?

Statement 1: The integers in the set are consecutive multiples of 3.

If the Integers are set of Consecutive Multiples of 3 then They are definitely equally spaced (in Arithmetic Progression)

CONCEPT: Terms in Arimetic Progression (Equally spaced) have
Mean = Media = Middle term = Average of First and Last Term of the set

But since no information about first and Last term or median of the set or the number of terms is known so
NOT SUFFICIENT

Statement 2: The median of the set of integers is 33.
The media may be mean ONLY if the terms are equally spaced and are in Arithmetic progression But since it's unknown whether the terms in the set are equally spaced or not, Hence,
NOT SUFFICIENT

Combining the Two statements:
Statement 1 mentions that terms in the set are on Arithmetic Progression
Statement 2 mentions that Median = 33
We know that Mean = Median when terms are in AP. hence,
SUFFICIENT

Answer: option C
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Answer and solution apart, is there a contradiction between question stem and (1)?
m is a positive odd integer
Clearly, then such sets as {0, 3, 6} and {3, 6, 9} are not possible
Further, since sets are of consecutive multiples of 3, it can have only 1 element e.g. {3} or {9}
I'm only interested in knowing whether my interpretation is right since I agree with the solution given by you. Thanks in advance for your comments!

No, your interpretation is not right.

Given: m is a positive odd integer.
Question: what is the average of a set of m integers.

So, we need to know the average of a set which has m integers in it (the number of element in the set is m, which is a positive odd integer). It does NOT mean that the set consists of positive odd integers.
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(1) The integers in the set are consecutive multiples of 3.
{3,6,9,12,15} or {3,6} or {-99,-96,-93,-90} ==> insufficient
(2) The median of the set of integers is 33
{1,23,33,55,99} or {1,2,3,4,5,} or {-88,33,1098}==> Insuficient

Marge both:- Sufficient
{first term=3,..........middle term=33,...........last term=63}

so the Median is 33; since the set contains consecutive multiples of 3 therefore mean = median = 33

ANSWER IS C
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M is a +ve Odd Integer {1,3,5,7,9,11...} What is Avg of {M}?
Rephrase the Question as What is M?

ADBCE Grid:-
1) M:- {1,3,5,7,9,11..}
M:-{3,6} {3,6,9,12,15}.... Insuff to Find the Average

2) Median of set is 33. Insuff

Using 1 and 2
M( 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,....66,69}
Median is 33.
Since M is +ve odd integer,M should be odd.
If M is odd, then the middle integer will be the median

So if Median is 33, then M should be 21. { 3,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63}

Hence C is Suff
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