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DS_Median, mode_plzz explain how... [#permalink]
 05 Sep 2007, 12:58
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Re: DS_Median, mode_plzz explain how... [#permalink]
05 Sep 2007, 13:42
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]
05 Sep 2007, 21:23
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]
05 Sep 2007, 23:44
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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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]
06 Sep 2007, 07:32
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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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]
06 Sep 2007, 10:02
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]
06 Sep 2007, 10:05
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]
06 Sep 2007, 13:15
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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I didn't get why we are looking for a "specific" numerical value for the median. The question doesn't state that directly. Thanks
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