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The arithmetic mean (average) of a set of 10 numbers is 10.
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26 Sep 2007, 08:07
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The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10.
2) The mode of the set of numbers is 10. == Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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I am sure that it can't be (A). Here is why :
Median = (5th + 6th)/2
From (1), we know that 5th < 10, but we know nothing about 6th.
Case 1 : 5th = 9, 6th = 10, median = 19/2
Case 2 : 5th = 9, 6th = 11, median = 10
Therefore (1) is insufficient.



VP
Joined: 09 Jul 2007
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E
Although half of the numbers are less than 10 in wht order are they placed. it can be 15 or 26 so on.
mode 10 still not suff.



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Re: DS : Mean/Median/Mode
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26 Sep 2007, 12:04
coldweather999 wrote: The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10. 2) The mode of the set of numbers is 10.
I'm getting C. Knowing both Stat 1 and Stat 2, we can say that the median value will never be 10.
Median = (5th term + 6th term)/2
Stat 1:
set = 5, 5, 5, 5, 5, 15, 15, 15, 15, 15
In this case median = 10.
set = 4, 4, 4, 4, 9, 15, 15, 15, 15, 15
median is not equal to 10.
Insuff
Stat 2:
set = 10 times 10.
Median = 10
set = 3, 4, 4, 5, 9, 10, 10, 10, 20, 25
median is not equal to 10
Insuff.
Stat 1 & 2:
Since exactly half the values are under 10 then the 5th term has to be less than 10. Since the mode is 10, the 6th term has to be 10.
(10+number less than 10)/2 has to be less than 10.
Suff.



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I get the same  C  using the same rationale as GK_GMAT. What's the OA?



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Bluebird wrote: I get the same  C  using the same rationale as GK_GMAT. What's the OA?
I also got (C) using the same logic as GK_GMAT...(i didn't assume the numbers to be whole numbers since it's not mentioned anywhere)...but the OA is (E)



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Re: DS : Mean/Median/Mode
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26 Sep 2007, 14:02
coldweather999 wrote: The arithmetic mean (average) of a set of 10 numbers is 10. Is the median value of the same set also equal to 10?
1) Exactly half of the numbers are less than 10. 2) The mode of the set of numbers is 10.
C for me.
(1)
1,2,3,4,5,15,16,17,18,19 gives you median of 10
1,2,3,4,8,15,16,17,18,22 gives you median more than 10
INSUFFICIENT
(2)
If all numbers are 10, then median =10
1,2,3,4,5,6,10,10, large, large gives median not equal to 10
INSUFFICIENT
Together, you can have
1,2,3,4,5,10,10,20,22,23
gives you median of not 10
Say a<10
You know that median = (10+a)/2
Since a is always less than 10, the median will never equals to 10.
SUFFICIENT



VP
Joined: 28 Dec 2005
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why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10



VP
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pmenon wrote: why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10
The question said exactly half the terms (5 terms) are less than 10. The mode is 10. This means the sixth term must be 10.



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bkk145 wrote: pmenon wrote: why does the 6th term have to be 10 ?
It could be 11, and the 5th term could be 9, which give median of 10
Or, the 6th term could be 25, and the 5th term could be 3, therefore giving a median of 15
Just because the mode is 10, I dont think it means that the 6th term has to be 10 The question said exactly half the terms (5 terms) are less than 10. The mode is 10. This means the sixth term must be 10.
Mode refers to the value in a set that is most repeated, right ? If that is the case, how does it mean that the sixth term has to be 10 ?
As long as 10 is the most repeated number in the set, thats all that matters right ?
Forgive me ! lol



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St1:
The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5
The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10.
Insufficient.
St2:
The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10
The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5
Insufficient.
St1 and St2:
The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5
The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5.
Insufficient.
Ans E



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ywilfred wrote: St1: The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5 The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10. Insufficient.
St2: The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10 The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5 Insufficient.
St1 and St2: The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5 m The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5. Insufficient.
Ans E
how does it make C insufficient?
If we take both sets together 10 can never be median..question solved.



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rishi2377 wrote: ywilfred wrote: St1: The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5 The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10. Insufficient.
St2: The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10 The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5 Insufficient.
St1 and St2: The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5 m The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5. Insufficient.
Ans E how does it make C insufficient? If we take both sets together 10 can never be median..question solved.
my bad.... it should be C. I thought the question was asking for the value of the median in which case the answer would be E.



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Re: DS : Mean/Median/Mode
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Updated on: 11 May 2008, 06:54
****makes no sense so Deleted*****



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Re: DS : Mean/Median/Mode
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11 May 2008, 06:43
vshaunak@gmail.com wrote: Should be 'E'.
S1 + S2 not suff
case1  set = {7,8,9,10,10,16} mean=median=mode=10 and 3 elements are smaller than 10. case2 set={1,2,3,4,10,10,10,40} mean=10, mode=10, median=7 and 4 elements and smaller than 10. C) case1  median will be (9 + 10)/2 = 9.5 <10



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Re: DS : Mean/Median/Mode
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11 May 2008, 06:52
tarkumar wrote: vshaunak@gmail.com wrote: Should be 'E'.
S1 + S2 not suff
case1  set = {7,8,9,10,10,16} mean=median=mode=10 and 3 elements are smaller than 10. case2 set={1,2,3,4,10,10,10,40} mean=10, mode=10, median=7 and 4 elements and smaller than 10. C) case1  median will be (9 + 10)/2 = 9.5 <10 Oops..wrong numbers.... yes it should be 'C'



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ywilfred wrote: St1: The set could be {0,0,0,0,1,10,10,10,10,59}. Mean = 10, Median = 5.5 The set could be {0,0,0,0,0,20,20,20,20,20}. Mean = 10, Median = 10. Insufficient.
St2: The set could be {10,10,10,10,10,10,10,10,10,10}. Mode = Mean = Median = 10 The set could be {1,2,3,4,5,6,10,10,10,49}. Mode = 10. Mean = 10. Median = 5.5 Insufficient.
St1 and St2: The set could be {1,2,3,4,5,10,10,10,10,45}. Mode = Mean = 10. Median = 7.5 The set could eb {0,0,0,0,0,10,10,10,10,60}. Mode = Mean = 10. Median = 5. Insufficient.
Ans E Quick question. In you example St1 and St2. Wouldn't the mode be 0 as it's used 5 times and 10 is used 4 times.



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Re: The arithmetic mean (average) of a set of 10 numbers is 10.
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17 Sep 2017, 03:16
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Re: The arithmetic mean (average) of a set of 10 numbers is 10. &nbs
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