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The arithmetic mean (average) of a set of 10 numbers is 10.

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The arithmetic mean (average) of a set of 10 numbers is 10. [#permalink] New post 26 Sep 2007, 07:07
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C
D
E

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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.
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 [#permalink] New post 26 Sep 2007, 07:16
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.
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 [#permalink] New post 26 Sep 2007, 09:31
E

Although half of the numbers are less than 10 in wht order are they placed. it can be 1-5 or 2-6 so on.
mode 10 still not suff.
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Re: DS : Mean/Median/Mode [#permalink] New post 26 Sep 2007, 11:04
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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 [#permalink] New post 26 Sep 2007, 11:21
I get the same - C - using the same rationale as GK_GMAT. What's the OA?
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 [#permalink] New post 26 Sep 2007, 12:07
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 [#permalink] New post 26 Sep 2007, 13:02
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
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 [#permalink] New post 26 Sep 2007, 16:15
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
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 [#permalink] New post 26 Sep 2007, 16:24
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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 [#permalink] New post 26 Sep 2007, 17:02
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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 [#permalink] New post 26 Sep 2007, 17:59
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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 [#permalink] New post 26 Sep 2007, 18:32
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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 [#permalink] New post 26 Sep 2007, 18:36
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 [#permalink] New post 10 May 2008, 23:09
****makes no sense so Deleted*****

Last edited by vshaunak@gmail.com on 11 May 2008, 05:54, edited 1 time in total.
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Re: DS : Mean/Median/Mode [#permalink] New post 11 May 2008, 05:43
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 [#permalink] New post 11 May 2008, 05:52
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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Re: [#permalink] New post 12 May 2008, 12:05
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.
Re:   [#permalink] 12 May 2008, 12:05
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