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mean/median ds

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Director
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mean/median ds [#permalink] New post 13 Jan 2006, 04:55
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
A list contains 3 diferent numbers. Does the median of the 3 numbers equal the arithmetic mean of the 3 numbers?

1) The range of the 3 numbers is equal to twice the difference betwenn the greatest number and the median

2) the sum of the 3 numbers is equal to 3 times one of the numbers
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 [#permalink] New post 13 Jan 2006, 08:50
As far as I can see it's D

1)

The only way the range is equal to twice the difference betwenn the greatest number and the median is when the three numbers are equal.

Consider some ex.

[123] Median=2 Range=1, => out
[125] M=2 R=4 => out
[222] M=2 R=0 fits

Sufficient

2)
[-1,0,1] fits
[222] also
[125]nope
[-2,0,3]nope
Sufficient

This is a stupid method and it's inaccurate too, maybe I've overseen a set.
Hope you provide a more elegant way to answer this.
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 [#permalink] New post 13 Jan 2006, 09:11
allabout wrote:
As far as I can see it's D

1)

The only way the range is equal to twice the difference betwenn the greatest number and the median is when the three numbers are equal.

Consider some ex.

[123] Median=2 Range=1, => out
[125] M=2 R=4 => out
[222] M=2 R=0 fits

Sufficient

2)
[-1,0,1] fits
[222] also
[125]nope
[-2,0,3]nope
Sufficient

This is a stupid method and it's inaccurate too, maybe I've overseen a set.
Hope you provide a more elegant way to answer this.


allabout, The question asks whether median = mean.

It should be A?

Let {x1,x2,x3} be the numbers

Q: Is x2 = (x1+x2+x3)/3 ? or
is x2 = (x1+x3)/2 ?

(1) Given x3-x1 = 2(x3-x2)
or x2 = (x1+x3)/2
SUFFICIENT.
(2) INSUFFICIENT, since the sum can equal any of the numbers x1,x2 or x3.

Hence A.
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 [#permalink] New post 13 Jan 2006, 09:19
giddi77 wrote:
allabout wrote:
As far as I can see it's D

1)

The only way the range is equal to twice the difference betwenn the greatest number and the median is when the three numbers are equal.

Consider some ex.

[123] Median=2 Range=1, => out
[125] M=2 R=4 => out
[222] M=2 R=0 fits

Sufficient

2)
[-1,0,1] fits
[222] also
[125]nope
[-2,0,3]nope
Sufficient

This is a stupid method and it's inaccurate too, maybe I've overseen a set.
Hope you provide a more elegant way to answer this.


allabout, The question asks whether median = mean.

It should be A?

Let {x1,x2,x3} be the numbers

Q: Is x2 = (x1+x2+x3)/3 ? or
is x2 = (x1+x3)/2 ?

(1) Given x3-x1 = 2(x3-x2)
or x2 = (x1+x3)/2
SUFFICIENT.
(2) INSUFFICIENT, since the sum can equal any of the numbers x1,x2 or x3.

Hence A.


Don't know; the sum of the three numbers is equal to three times one number. Do you know a set that fullfills this condition (2) but in which the mean isn't equal to the median?
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 [#permalink] New post 13 Jan 2006, 10:59
allabout wrote:

Don't know; the sum of the three numbers is equal to three times one number. Do you know a set that fullfills this condition (2) but in which the mean isn't equal to the median?


Hmm. I understand it now sir/madam(?) THANK YOU!!

It will work only if the numbers are of the form you mentioned.
(-n,0,n) or (n,n,n). Hence SUFFICIENT.

Therefor D.

Another classic case of "Answer Biasing" as duttsit would like to call it. I just rejected (2) because I didn't like it :wall
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 [#permalink] New post 13 Jan 2006, 12:32
D it is.

Say numbers are x,y,z and these are in ascending order, So median here is y

Range = z-x

St1:

Range = 2 * (z-y)

so z-x = 2z-2y
-x = z-2y
0 = z+x -2y
y = z+x -2y + y
i.e 3y = x+y+z
y = (x+y+z)/3

So median = mean SUFF

St2:
this have three cases

1. x+y+z = 3x
x + (x +a) + (x+b) = 3x
so a + b = 0 (Remember that a and b are +ve)
This means all must be equal to x i.e mean = median
2. x+y+z = 3y
(y-a) + y + y(+b) = 3y
so b-a = 0 i.e a = b (Remember that a and b are +ve)
This means all must be equal to y i.e mean = median

3. x+y+z = 3z
(z-a) + (z-b) + z = 3z
so -a-b = 0 i.e a +b = 0 (Remember that a and b are +ve)
This means all must be equal to z i.e mean = median

SUFF
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 [#permalink] New post 13 Jan 2006, 13:12
En contraire, I'm sorry giddi that I haven't made it clear enough.

dahiyas approach is clearer and intuitive. One should strive always for clarity even when the time ticks.

By the way, I'm male.
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 [#permalink] New post 14 Jan 2006, 00:46
I would go with D

The nos are a, b, c

i) c-a=2(c-b)=>b=(a+c)/2 =>which is AP which means mean=median
so SUFFICIENT
ii) a+b+c=2a/2b/2c=>Again an AP so SUFFICIENT

So ans is D
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 [#permalink] New post 14 Jan 2006, 02:40
the oa is D

great explanation guys

btw how long did it take you to solve?
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 [#permalink] New post 15 Jan 2006, 00:51
Close to 1.5 min..
The real trick was to know that the series is an AP..Once you get that its harly a matter of 2 sec
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 [#permalink] New post 15 Jan 2006, 23:06
I also got D. Could not fail the second statement. Looks like 3 times one number always means 3 time median.

Last edited by Bhai on 15 Jan 2006, 23:31, edited 1 time in total.
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 [#permalink] New post 15 Jan 2006, 23:25
joemama142000 wrote:
what is an AP?


AP= Arithmetic Progression

example
5 +10 +15 + 20 + 25 +... + 100

AP formula
S = n/2 ((2a + (n-1)*d) = 20/2 (5*2 + 19*5) = 1050
  [#permalink] 15 Jan 2006, 23:25
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