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GMAT PREP q3

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GMAT PREP q3 [#permalink] New post 27 Aug 2006, 03:29
QC42: GMAT PREP q
Please explain your ans
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Re: GMAT PREP q3 [#permalink] New post 27 Aug 2006, 04:11
shinewine wrote:
QC42: GMAT PREP q
Please explain your ans


Here it goes...

Statement 1 is insuff. It only talks about S nothing about T.

Statement 2 is insuff. It talks about the sum, nothing about the median.

Taking these statements together, this is what i get.

S: median is 0.
integers are consecutive.
Therefore set S = {-2,-1,0,1,2}

T: same sum as S. S sums upto 0.
Therefore T sums upto 0.
integers are consecutive.
Therefore set S = {-3,-2,-1,0,1,2,3}

Median of Set S and Set T is 0. Therefore, equal.

Answer C.
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Re: GMAT PREP q3 [#permalink] New post 27 Aug 2006, 05:53
SuperHumanAmit wrote:
shinewine wrote:
QC42: GMAT PREP q
Please explain your ans


Here it goes...

Statement 1 is insuff. It only talks about S nothing about T.

Statement 2 is insuff. It talks about the sum, nothing about the median.

Taking these statements together, this is what i get.

S: median is 0.
integers are consecutive.
Therefore set S = {-2,-1,0,1,2}

T: same sum as S. S sums upto 0.
Therefore T sums upto 0.
integers are consecutive.
Therefore set S = {-3,-2,-1,0,1,2,3}

Median of Set S and Set T is 0. Therefore, equal.

Answer C.


i think Statement II is sufficient :

as i can thnk of only one case in which can furnish same sum for 5 and 7 consecutive intergers..
This when we have digits on either side of zero..as u ve correctly pointed out.. {-3,-2,-1,0,1,2,3}, {-2,-1,0,1,2}


can you think of some other case in whch sum of 7 and 5 conse. no.s is equal?
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 [#permalink] New post 27 Aug 2006, 06:14
If S={s,s+1,...,s+4} T={t,t+1,...,t+6}

For s+(s+1)+...(s+4)=t+(t+1)+...(t+6)

5s+10=7t+21

5s=7t+11

s=(7t+11)/5

if t=2, s=5
if t=7, s=12

So {5,6,7,8,9} and {2,3,4,5,6,7,8} have the same sum, but not the same median!
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 [#permalink] New post 27 Aug 2006, 07:05
Textbook style kevin. (E) is it!
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 [#permalink] New post 27 Aug 2006, 08:27
kevincan wrote:
If S={s,s+1,...,s+4} T={t,t+1,...,t+6}

For s+(s+1)+...(s+4)=t+(t+1)+...(t+6)

5s+10=7t+21

5s=7t+11

s=(7t+11)/5

if t=2, s=5
if t=7, s=12

So {5,6,7,8,9} and {2,3,4,5,6,7,8} have the same sum, but not the same median!


thank you kevin..
i was almost thr..i calculated till 5s=7t+11...
.. but i could'nt get the appropriate values of s and t,
.
there is so much trial and error.. how did u click upon these values ? is thr any shortcut..
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 [#permalink] New post 27 Aug 2006, 08:51
gmatt73,
why is it E.. taken together, i would think C ..
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 [#permalink] New post 27 Aug 2006, 18:39
gk3.14 wrote:
gmatt73,
why is it E.. taken together, i would think C ..


Consider two example sets

T={ -1, -2,0,1,2} -->median =0

S1={-3,-2, -1, 0,1,2,3} --> median=0

S2={ -5,-2,0,1,1,2,3} ---> median not equal to zero,

However, the sum of each of the sets, s1, s2 = sum of set T.
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 [#permalink] New post 27 Aug 2006, 18:49
aspirations:

but they have to be consecutive numbers.. so S2 has to be consecutive..that example does not work..
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 [#permalink] New post 27 Aug 2006, 22:07
gk3.14 wrote:
aspirations:

but they have to be consecutive numbers.. so S2 has to be consecutive..that example does not work..

Exactly.

Answer should be C.
St1: S = {-2,-1,0,1,2}. No info about T.

St2:
Median equal for S = {-2,-1,0,1,2} T = {-3,-2,-1,0,1,2,3}
Median not equal for S = (5,6,7,8,9} T = {2,3,4,5,6,7,8}: INSUFF

Together:
S = {-2,-1,0,1,2} T = {-3,-2,-1,0,1,2,3} must be true.
Medians are equal.
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 [#permalink] New post 28 Aug 2006, 06:34
gk3.14 wrote:
aspirations:

but they have to be consecutive numbers.. so S2 has to be consecutive..that example does not work..


Agreed, thanks for the catch, its C.
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 [#permalink] New post 04 Sep 2006, 08:39
I think it's B

S= {a,a+1,a+2,a+3,a+4}
T = {b,b+1,b+2,b+3,b+4,b+5,b+6)

Ok the way i see it is that it's a yes or no DS

1) is insuff for obvious reasons

Ok now it gets a bit tricky
We have to find it median of S = median of T

Inshort If a+2 = b+3 => a = b + 1

2) Says the sum of S (5a+10) = sum of T (7b + 16)

We get 5a + 10 = 7b + 16

Implying a = (7b + 6)/5 and not equal to b + 1 . Therefore B

Is there anything wrong with this method?
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 [#permalink] New post 04 Sep 2006, 08:49
jaypshah wrote:
I think it's B

S= {a,a+1,a+2,a+3,a+4}
T = {b,b+1,b+2,b+3,b+4,b+5,b+6)

Ok the way i see it is that it's a yes or no DS

1) is insuff for obvious reasons

Ok now it gets a bit tricky
We have to find it median of S = median of T

Inshort If a+2 = b+3 => a = b + 1

2) Says the sum of S (5a+10) = sum of T (7b + 16)

We get 5a + 10 = 7b + 16

Implying a = (7b + 6)/5 and not equal to b + 1 . Therefore B

Is there anything wrong with this method?


going by ur calculation a=1.4b+1.2

how do u know that 0.4b is not equal to -0.2?

I think, C is the answer
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 [#permalink] New post 04 Sep 2006, 11:26
becasue B is an integer
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 [#permalink] New post 04 Sep 2006, 11:41
what does integer have to do with any of this?

the answer shud b C...

jaypshah wrote:
becasue B is an integer
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 [#permalink] New post 04 Sep 2006, 11:53
good work guys. I myself picked B at first. Now I understand it

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 [#permalink] New post 04 Sep 2006, 22:15
missed "consecutive" :beat .... C it is ... what would I do if I do this the day after tomorrow :-(((
  [#permalink] 04 Sep 2006, 22:15
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