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Set S

by atulmangal » Fri May 06, 2011 4:59 am
Set S consists of five consecutive integers, and set T consists of seven consecutive integers. Is the median of the numbers in set S equal to the median of the numbers in set T?

(1) The median of the numbers in Set S is 0.
(2) The sum of the numbers in set S is equal to the sum of the numbers in set T.
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Source: — Data Sufficiency |
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by vineeshp » Fri May 06, 2011 6:43 am
OA B?

Stmt 1) Not sufficient.
It tells us that Set S is -2,-1,0,1 and 2.
But It doesnt tel us anything about set T.

Stmt 2:
Sums are equal.
For consecutive numbers, such sums can only be equal if their medians are both 0.
Taking an example,
-1, 0 , 1, 2,3 - Sum is 5.
Now to identify a set of 7 that gives sum 5 is impossible.
-1, 0 , 1, 2,3, 4, 5.

This statement is valid only for one set and that is the sets hovering around 0.

Hence B.
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by MAAJ » Fri May 06, 2011 7:01 am
IMO B

Because the sum of 5 CI and the sum of 7 CI can only be equal if they are spread apart from 0.
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by atulmangal » Fri May 06, 2011 8:49 am
Sorry brothers

OA is not B..lets wait for some expert after that i vll reveal the OA
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by clock60 » Fri May 06, 2011 12:55 pm
my try
to me the answer is C
(1) s-2,s-1,s,s+1,s+2 ( 5 consecutive)
t-3,t-2,t-1,t,t+1,t+2,t+3 (7 consecutive)
we need to find does s=t
(1) not suff as it can be
(-2 -1 0 1 2)
(-3 -2 -1 0 1 2 3) here s=t=0 yes
but onther version
(-2 -1 0 1 2)
(1,2,3,4,5,6,7) 0 does not equal to 4 the answer is no

(2) the sum of the numbers are equal in both sets
sum=mean*number of terms
in s it will 5*s and in t it will be 7*t (as in series of consecutive numbers mean=median)
5s=7t when is this st holds true?
if s=t=0 in this case the answer is yes
but if s=7 and t=5 , our series will look as
s(5.6.7.8.9) t(2,3,4,5,6,7,8)
the answer is no,
so 2 st insufficient
from both we no that 5s=7t, and s=0 , it comes that t also =0 so both suff
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by Stendulkar » Fri May 06, 2011 1:30 pm
Hmmm.... Atul is saying that B isnt the answer so my explanation might be wrong anyway...however...my try :


S : {n,n+1,n+2,n+3,n+4}
T : {m, m+1, m+2, m+3, m+4, m+5, m+6}

We need to check if n+2 = m+3... (i)

Statement 1 :

Median in S = 0
n+2 = 0
n = -2

S : { -2,-1,0,1,2}

No connection with T is given. So insufficient.

Statement 2 :

Sum of elements in S = Sum of elements in T

Theremore, 5n + 10 = 7m + 21

5(n+2) = 7 (m+3)

n+2 = (7/5)(m+3)

Compare the above with (i)

therefore the meadians are not matching. Sufficient.

I am getting B. Statement 2 is correct for sure. Atul says that B isnt correct. So I feel only other option is D. For that Statement 1 has to be correct but dnt understand how..
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by clock60 » Fri May 06, 2011 1:59 pm
Stendulkar wrote:Hmmm.... Atul is saying that B isnt the answer so my explanation might be wrong anyway...however...my try :


S : {n,n+1,n+2,n+3,n+4}
T : {m, m+1, m+2, m+3, m+4, m+5, m+6}

We need to check if n+2 = m+3... (i)

Statement 1 :

Median in S = 0
n+2 = 0
n = -2

S : { -2,-1,0,1,2}

No connection with T is given. So insufficient.

Statement 2 :

Sum of elements in S = Sum of elements in T

Theremore, 5n + 10 = 7m + 21

5(n+2) = 7 (m+3)

n+2 = (7/5)(m+3)

Compare the above with (i)

therefore the meadians are not matching. Sufficient.

I am getting B. Statement 2 is correct for sure. Atul says that B isnt correct. So I feel only other option is D. For that Statement 1 has to be correct but dnt understand how..
hi Stendulkar
it seems that you are slightly wrong
you got that
5(n+2)=7(m+3) but this equation holds true if
n=-2, and m=-3, 5*(-2+2)=7(-3+3), 0=0
in this case you sets will be
-2,-1,0,1,2 and
-3,-2,-1,0,1,2,3 here medians are equal =0
or
5(n+2)=7(m+3) also holds true for
n=5, m=2, 5(5+2)=35, and 7(2+3)=35
and sets are
5,6,7,8,9 and
2.3.4.5.6.7.8 in this case medians 7 and 5 are not equal
for this reason st 2 alone insuff
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by Stendulkar » Fri May 06, 2011 2:07 pm
Oops...u are correct...if both the medians are zero then that equation will still hold true...Answer should be E??
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by Stendulkar » Fri May 06, 2011 2:08 pm
This has to be a high difficulty level problem...one that u get when u are cruising at 50 levels on Quant..
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by atulmangal » Fri May 06, 2011 2:32 pm
OA is C guys

@Clock: Good work !!!

I believe its a tough question not because it requires some heavy logic etc but because its a very tricky problem.
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