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# DS: Average

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Manager
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03 Oct 2006, 09:14
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If M is a positive odd integer, what is the average of a certain set of M integers?

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

2) The median of the set of integers is 33.
Manager
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03 Oct 2006, 09:15
OA is E. Someone explain why it isn't C. Since we know that it is a consecutive set and the median is 33, why isn't the average 33. Since in a consecutive set, the median is equal to the mean.
Director
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03 Oct 2006, 12:50
Your reasoning does not account for the case where the set contains -ve numbers. In such a case, the median may be 33, but the average (mean) may be lower.
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03 Oct 2006, 20:26
Futuristic wrote:
Your reasoning does not account for the case where the set contains -ve numbers. In such a case, the median may be 33, but the average (mean) may be lower.

This is not right. If they are consecutive, then even if there are negatives, the average will still be 33.

Additionally, we don't really consider negative multiples.

I think it is still C, until someone gives me another explanation.
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03 Oct 2006, 21:50
I agree. I also think it should be C, even if we were to consider negative integers
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29 Feb 2008, 04:48

Question clearly says M is +ve as well as Odd. So there is no question of -ve numbers.
Moreover having M as odd does not mean that Set contains odd number of M intergers.

Now Using Statement 1:

It tells us set is consecutive multiples of 3. But it does not tell us how many numbers are there in the set.

Now Using Statement 2:

It tells us median of the set is 33. But this again does not tells us how many numbers are there in the set. So for following 2 sets (32,33,34) as well as (31,32,33,34,35). Moreover if there is even number of terms in series then answer can be quite different.

Now Combine both 1 and 2

Again no such proof as how many terms and whether number of Ms in set is odd or even, which can affect the average. So I will go with E.
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29 Feb 2008, 07:02
govinam wrote:

If M is a positive odd integer, what is the average of a certain set of M integers?

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

2) The median of the set of integers is 33.

no of integers in a set = m, which is +ve odd odd. in fact number of integers cannot be -ve. so it must be +ve.

1: m could be 1,3, 5, 7, or 9 or so on. lets say m is 3.
3 integers could be 2, 5, and 10 or 4, 8, and 11. so not suff.

2: since m is odd and median is 33:
again, lets say m is 3.

3 integers could be 2, 33, and 34 or 31, 33, and 35. so not suff.

togather, no matter whether m is 3 or 5 or 7 or 9, since the median is 33 and m integers are consecutive, the mean of m integers are also 33.
so suff and it is C.
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29 Feb 2008, 13:01
bmwhype2 wrote:
govinam wrote:
OA is E. Someone explain why it isn't C. Since we know that it is a consecutive set and the median is 33, why isn't the average 33. Since in a consecutive set, the median is equal to the mean.

OA is C

Maybe the OP thinks the OA is E, but the OA is C? I can't see E being possible here.
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29 Feb 2008, 13:17
I'm for C as well.

Stmt 1:
insuff.
3 6 9 or 6 9 12 or 6 9 12 15 18 etc.

Stmt 2:
Suff.
30 33 36 or 27 30 33 36 39 etc.
In all cases the average will always be 33, even if you go into negative numbers.
Ex. -3+69= 66, 66/2=33
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29 Feb 2008, 13:24
if m is odd..isnt the avg=median? if thats the case isnt statement 2 sufficient???
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29 Feb 2008, 13:31
fresinha12 wrote:
if m is odd..isnt the avg=median? if thats the case isnt statement 2 sufficient???

only if the elements of set are consecutive, which we do not know from 2.

again, suppose: m (no of elements in the set) = 3.
if the elements are: 1, 33, and 34 ......... no.
if the elements are: 31, 33, and 35 ...... yes.
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Re: DS: Average   [#permalink] 29 Feb 2008, 13:31
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