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DS_Median, mode_plzz explain how...

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Director
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DS_Median, mode_plzz explain how... [#permalink] New post 05 Sep 2007, 11:58
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D
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 05 Sep 2007, 12:42
IrinaOK wrote:
ds


Got A.

(1) (a-b+b-a+a+b)/3 = a+b
=> 0 = 2(a+b)
This means that a+b = 0, or a = -b
Therefore, pick some numbers...(a,b)
(0,0) => median =0
(-1,1) => median =0
(1,-1) => median =0
Median will always equal to zero.
SUFFICIENT.

(2) We know that range = 2b
This means that a-b is minimum and a+b is maximum since a+b-a+b = 2b. However, it is impossible to find out b-a since we don't know the value of a and b.
INSUFFICIENT.
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 05 Sep 2007, 20:23
bkk145 wrote:
IrinaOK wrote:
ds


Got A.

(1) (a-b+b-a+a+b)/3 = a+b
=> 0 = 2(a+b)
This means that a+b = 0, or a = -b
Therefore, pick some numbers...(a,b)
(0,0) => median =0
(-1,1) => median =0
(1,-1) => median =0
Median will always equal to zero.
SUFFICIENT.

(2) We know that range = 2b
This means that a-b is minimum and a+b is maximum since a+b-a+b = 2b. However, it is impossible to find out b-a since we don't know the value of a and b.
INSUFFICIENT.


hmm......do we need to find the value of medain, mean or range.

for me it should be D.
1: mean should be medain. suff....
2: if 2b is range, lowest and highest values are (a - b) and (a+b) respectively. so (b-a) is the median. also suff...

but 1 and 2 give different answers. so seems something not like OG type/standard question.
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 05 Sep 2007, 22:44
bkk145 wrote:
IrinaOK wrote:
ds


Got A.

(1) (a-b+b-a+a+b)/3 = a+b
=> 0 = 2(a+b)
This means that a+b = 0, or a = -b
Therefore, pick some numbers...(a,b)
(0,0) => median =0
(-1,1) => median =0
(1,-1) => median =0
Median will always equal to zero.
SUFFICIENT.

(2) We know that range = 2b
This means that a-b is minimum and a+b is maximum since a+b-a+b = 2b. However, it is impossible to find out b-a since we don't know the value of a and b.
INSUFFICIENT.


Great explanation. Thanks. I go with A as well. The answer cannot be D as Fistail said because though we know that the median is (b-a) from stat 2, we do not know the value of either a or b making it impossible to find the median.
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 [#permalink] New post 05 Sep 2007, 23:17
Agree that media is 0. The mean of S is [(a-b)+(b-a)+(a+b)]/3 = (a+b)/3
But we are told that the mean of S is (a+b). Therefore,

a+b = (a+b)/3, which holds only if a+b = 0 or a = -b (1)

Taking (1) into consideration, we have that S = (-2b, 2b, 0) and therefore, the median will be always 0.
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 06 Sep 2007, 06:32
GK_Gmat wrote:
bkk145 wrote:
IrinaOK wrote:
ds


Got A.

(1) (a-b+b-a+a+b)/3 = a+b
=> 0 = 2(a+b)
This means that a+b = 0, or a = -b
Therefore, pick some numbers...(a,b)
(0,0) => median =0
(-1,1) => median =0
(1,-1) => median =0
Median will always equal to zero.
SUFFICIENT.

(2) We know that range = 2b
This means that a-b is minimum and a+b is maximum since a+b-a+b = 2b. However, it is impossible to find out b-a since we don't know the value of a and b.
INSUFFICIENT.


Great explanation. Thanks. I go with A as well. The answer cannot be D as Fistail said because though we know that the median is (b-a) from stat 2, we do not know the value of either a or b making it impossible to find the median.



Do agree with GK_Gmat, though 2b is a range we don't know whether a+b is higher or a-b (there is no values of a and b, they could be positive or negative numbers).

1) is sufficient because a=-b making median 0 IrinaOK explained above.

Ans: A
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 [#permalink] New post 06 Sep 2007, 07:21
St1:
(a+b)/3 = a+b
a+b = 3(a+b) --> only way is if a+b = 0. So a = -b.
Now, a-b = -2b, a+b = 0, b-a = 2b. So the median is 0 (or a+b). Sufficient.

St2:
Range = 2b

Can either be (a+b) - (a-b) = 2b or (b-a)-(a-b) = 2a. More than one possibility. Insufficient.

Ans A
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 06 Sep 2007, 09:02
Fistail wrote:
bkk145 wrote:
IrinaOK wrote:
ds


Got A.

(1) (a-b+b-a+a+b)/3 = a+b
=> 0 = 2(a+b)
This means that a+b = 0, or a = -b
Therefore, pick some numbers...(a,b)
(0,0) => median =0
(-1,1) => median =0
(1,-1) => median =0
Median will always equal to zero.
SUFFICIENT.

(2) We know that range = 2b
This means that a-b is minimum and a+b is maximum since a+b-a+b = 2b. However, it is impossible to find out b-a since we don't know the value of a and b.
INSUFFICIENT.


hmm......do we need to find the value of medain, mean or range.

for me it should be D.
1: mean should be medain. suff....
2: if 2b is range, lowest and highest values are (a - b) and (a+b) respectively. so (b-a) is the median. also suff...

but 1 and 2 give different answers. so seems something not like OG type/standard question.


Question is asking for specific value. Here median must be specific number. I made the same mistake.
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 06 Sep 2007, 09:05
IrinaOK wrote:
ds


the OA is A

ywilfred,

Thanks for explnations. It is clear as usuall!!
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Re: DS_Median, mode_plzz explain how... [#permalink] New post 06 Sep 2007, 12:15
IrinaOK wrote:
ds



hmmm from S1 I got b=-a. and then a=0, so b=0. So im guessin this is suff.


S2: I don't really know, but i don't see how this is suff.


On the real GMAT. i dunno if i could have gotten this in 2min. i got in 3...

My ans is A. I rephrashed S1 like this.

a-b+b-a+a/3=a+b. cancel out common terms. we get a+b/3=a+b

so a+b=3a+3b which comes out to -2b=2a b=-a. plug it back in to the equation. I get 2a=0. so a must be 0. since -a=0. b must be 0.

From this i said A must be suff.

But i dunno how right I am on this.
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 [#permalink] New post 07 Sep 2007, 08:10
I didn't get why we are looking for a "specific" numerical value for the median. The question doesn't state that directly. Thanks
  [#permalink] 07 Sep 2007, 08:10
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