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by GmatTakerNo.1 » Mon May 03, 2010 10:58 am
In the sequence of 7 positive integers 2, 3, k, r, 7, 7, 11 if the average equals the median, which is one of the members in the sequence. What is the average?
(1) r=6
(2) k=r

The answer is B, but I got A, because if r is 6 you have the median, and because median=average you have the average as well.
Am I wrong?
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by iamseer » Mon May 03, 2010 11:24 am
average=median=value of one of the member element.

from statement 1:
IF r = 6, we have no information about k. Therefore we cannot find the average. NOT SUFFICIENT

from statement 2:
k=r
therefore average is (30+2k)/7. This greater than 2 or 3 for sure.
therefore (30+2k)/7 = k or 7 or 11
(30+2k)/7 = 7 not possible, since k is integer
(30+2k)/7 = 11 not possible, since k is integer

(30+2k)/7 = k gives 5k=30, k=6=r
So set is 2,3,6,6,7,7,11 .... median is 6, average is 6

answer B
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by GmatTakerNo.1 » Mon May 03, 2010 11:45 am
But the question also says that the median is equal to the average. And because the numbers are ordered from small to big ones, actually it´s clear that k is not bigger than r. So if r is 6, then the median is 6 which implies due to the question that the average is 6 as well.
Where is the flaw in my idea?

Can you maybe explain your way of solution a little bit more in detail?
Why do you check if (30+2k)/7 is 7 or 11?
And where do you get the term (30+2k)/7?

I have my Gmat tommorow and a strange feeling that I can use the knowledge to solve this ;)
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by liferocks » Mon May 03, 2010 6:31 pm
the average of the series is (30+r+k)/7 or 4+(2+r+k)/7

from 1 we get the average is 4+(2+6+k)/7 or 5+(1+k)/7
now for k=6 average is 6 and median 6...
now for k=13 average is 7 and median 7..

Not sufficient to conclude what is avarage

From 2
average 4+(2+k+k)/7 or 4+2*(1+k)/7
since the set is of integers median=average must be integer ..hence 1+k is a multiple of 7
for k=7n-1
when n=1,k=6 and median=average= 6
when n>1,k>=13 hence median 7..and average>=8
hence only solution is k=r=median=average=6
condition 2 is sufficient

ans B
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by ssuarezo » Wed May 05, 2010 10:26 am
GmatTakerNo.1 wrote:In the sequence of 7 positive integers 2, 3, k, r, 7, 7, 11 if the average equals the median, which is one of the members in the sequence. What is the average?
(1) r=6
(2) k=r

The answer is B, but I got A, because if r is 6 you have the median, and because median=average you have the average as well.
Am I wrong?
Me too, anyone please tell me what's wrong with this: (I get D !! )

From the quest, I simplified to (Avg=Med): (30 + k)/6 = r --> since r is both the median and avg, and assuming the list is sorted (seems so). Maybe this is the mistake. If sorted:

Stm 1: r=6, and applied to (30 + k)/6 = 6, then k = 6 Suff.
Stm 2: k=r, and applied to (30 + r)/6 = r, then r =k=6 Suff

Maybe I dont understand the question: which is one of the members in the sequence, but I do got the average.

Please help !
Silvia
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by liferocks » Wed May 05, 2010 5:55 pm
ssuarezo wrote:
GmatTakerNo.1 wrote:In the sequence of 7 positive integers 2, 3, k, r, 7, 7, 11 if the average equals the median, which is one of the members in the sequence. What is the average?
(1) r=6
(2) k=r

The answer is B, but I got A, because if r is 6 you have the median, and because median=average you have the average as well.
Am I wrong?
Me too, anyone please tell me what's wrong with this: (I get D !! )

From the quest, I simplified to (Avg=Med): (30 + k)/6 = r --> since r is both the median and avg, and assuming the list is sorted (seems so). Maybe this is the mistake. If sorted:

Stm 1: r=6, and applied to (30 + k)/6 = 6, then k = 6 Suff.
Stm 2: k=r, and applied to (30 + r)/6 = r, then r =k=6 Suff

Maybe I dont understand the question: which is one of the members in the sequence, but I do got the average.

Please help !
Silvia
here you have considered r as median which is not necessary correct.
if k=2 and r=1 then the series will be

r k 2 3 7 7 11-->median 3
don't consider that the series is in ascending order unless it is specified.
"If you don't know where you are going, any road will get you there."
Lewis Carroll
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by kstv » Wed May 05, 2010 6:22 pm
(1) r = 6 so average = (36+k)/7
(36+k)/7 = median so (36+k)/7 hass to be a positive integer
the possible values of k are 6,13 etc
if k = 6,13 then the median & mean will be 6,7 respectively
so insuff
IMO B
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by ssuarezo » Thu May 06, 2010 9:17 am
liferocks wrote:
ssuarezo wrote: Me too, anyone please tell me what's wrong with this: (I get D !! )

From the quest, I simplified to (Avg=Med): (30 + k)/6 = r --> since r is both the median and avg, and assuming the list is sorted (seems so). Maybe this is the mistake. If sorted:

Stm 1: r=6, and applied to (30 + k)/6 = 6, then k = 6 Suff.
Stm 2: k=r, and applied to (30 + r)/6 = r, then r =k=6 Suff

Maybe I dont understand the question: which is one of the members in the sequence, but I do got the average.

Please help !
Silvia
here you have considered r as median which is not necessary correct.
if k=2 and r=1 then the series will be

r k 2 3 7 7 11-->median 3
don't consider that the series is in ascending order unless it is specified.
Ok LR, if not specifically indicated that they are in ascending order, they are not (worse case).
Thanks for your patience.
Silvia
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