What si the median of the set if
a) More than half of the numbers are >=1000
b) less than half of the numbers are <=1000
OA is C however, I'm not sure how that comes out.
Median
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wat ever it is !! the end limit of the either inequality matters !! So given space of time !! lotta ifs and but's might arise ..
So why not wait for some more comments to pour in 8) i am sure the tiger tamer( LOGITECH ) want to join in
Vishu
So why not wait for some more comments to pour in 8) i am sure the tiger tamer( LOGITECH ) want to join in
Vishu
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i also a bit sceptical about this. we simply cant assume that if more than half are greater than 1000, exactly the other part ie numbers less than 1000. eg. suppose we have 100 numbers.sumithshah wrote:Doesn't matter if the number of terms are odd or even?
according to stmt 1.
lets say we have 60 numbers are greater than 1000.
according to stmt 2.
30 numbers are less than 1000.
now combining both. Both the statements hold true for the example.
But, since we don't know about the other 10 numbers, we can't comment on the median.
Unless, there is some logic/rule for this...
So, i will go with E.
Please correct me if am wrong.
Sumith, can u pls tell the source?
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This is my first post in this website . Please point out if my logic is wrong.
I agree with Jimmiejaz in this case.
Statement 1:
Let us consider two sets of example for this statement
Example set 1: 987, 990,1000,1001,1002,1003 Median is 1000.5
Example set 2: 987,991,992,999,1000,1001,1002,1003,1004,1006,1007 ,1008,1009,1010Median is 1002.5
Both the cases above hold good for statement one . But we dont have a definitive median.Not enough
Statement 2:
I guess the same two examples hold good for this statement . Not enough.
There is no point in combining here .
Eventually Choice E
I agree with Jimmiejaz in this case.
Statement 1:
Let us consider two sets of example for this statement
Example set 1: 987, 990,1000,1001,1002,1003 Median is 1000.5
Example set 2: 987,991,992,999,1000,1001,1002,1003,1004,1006,1007 ,1008,1009,1010Median is 1002.5
Both the cases above hold good for statement one . But we dont have a definitive median.Not enough
Statement 2:
I guess the same two examples hold good for this statement . Not enough.
There is no point in combining here .
Eventually Choice E
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Sorry I am late
Well
More than half less than half crap!
So lets call the sample size 5 than for fun
Less than half : 2
More than half : 3
1 2 3 4 5
This case 3rd term will be the median. And question does not give any info whether it has odd or even members:
So whatever we choose for the 3rd term which is greater or equal to 1000 will be our MEDIAN
So E is the answer and C is the incorrect OA
Well
More than half less than half crap!
So lets call the sample size 5 than for fun
Less than half : 2
More than half : 3
1 2 3 4 5
This case 3rd term will be the median. And question does not give any info whether it has odd or even members:
So whatever we choose for the 3rd term which is greater or equal to 1000 will be our MEDIAN
So E is the answer and C is the incorrect OA
LGTCH
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I agree with E). If its C then we are missing something big timeWaiting for logitech, cramya
We dont even know if 1000 is in the list(lets say odd number of terms exists after arranging them in ascending order) or adding the middle terms and dividing by 2 gets us to 1000 after again arranging them in ascending order(if there are even number of terms)
Sumit, what's the source?
If the question was half of the numbers are more than 1000 and half of the numbers are less than 1000 and we know 1000 existed on the list then by definition of median it is 1000 which would be D) if it was a DS question. Any takers for this statement?
Median = value with as many other values above as below
Still think its E) and not C) due to the varius if's and but's in this problem!!
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Guys can I assume this ?
a) More than half of the numbers are >=1000
SO the median is will be always on this side , which means Median will be either equal 1000 or greater than 1000 ?
a) More than half of the numbers are >=1000
SO the median is will be always on this side , which means Median will be either equal 1000 or greater than 1000 ?
LGTCH
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true.logitech wrote:Guys can I assume this ?
a) More than half of the numbers are >=1000
SO the median is will be always on this side , which means Median will be either equal 1000 or greater than 1000 ?
but this is just an inequality (median > 1000). that's not the same as actually having the median.
this is definitely (e); there are all kinds of ways this could turn out.
here are just two:
* 87, 1000, 1000 (satisfies both conditions) --> median = 1000
* 87, 2000, 4000 (satisfies both conditions) --> median = 2000
insufficient.
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now, IF BOTH STATEMENTS ARE SUPPOSED TO SAY "MORE THAN", THEN THE ANSWER IS ACTUALLY (C) AS ORIGINALLY GIVEN.
in other words, if more than half the numbers are < 1000 and more than half the numbers are > 1000, then the median must be 1000.
i can explain this in more detail if anyone so desires.
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by the way, ladies and gentlemen, it isn't that hard to type "<" and ">". all you have to do is type normal "<" or ">" with an underline tag.
it's a lot more pleasant to look at than ">=" or, especially, "<=", which looks like it's supposed to be an arrow.
Ron has been teaching various standardized tests for 20 years.
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Hey Ron, thanks for the amazing explanation, but I am still not sure if C is sufficient, because the neither statement stated that the "middle" number is equal to 1000.
So...we can have -20,-10, 9999, 10000, 10001
median=9999
we also can have...
-20,-10,1000,1001,1002
median=1000
Since we can't pinpoint a specific value, shouldn't E be the right answer?
Thank you in advance Ron, I appreciate your comments.
So...we can have -20,-10, 9999, 10000, 10001
median=9999
we also can have...
-20,-10,1000,1001,1002
median=1000
Since we can't pinpoint a specific value, shouldn't E be the right answer?
Thank you in advance Ron, I appreciate your comments.