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Set W is made up of positive numbers. One number is removed,

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Set W is made up of positive numbers. One number is removed, [#permalink]

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07 Jul 2006, 08:47
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Set W is made up of positive numbers. One number is removed, and the remaining numbers comprise set V. Is the mean of the numbers in V equal to the mean of the numbers in W?

(1) All numbers in W are integers.
(2) The mean of the numbers in W is 17.5.
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07 Jul 2006, 10:22
MA, how are you getting (B)?

I know it is not (A). I am pretty confident it is not (B). (Just pick 3 numbers for W with average 17.5 and experiment with removing one. There infinite number of possibilities.)

I am guessing it is not (C) -- too much work.

So this would lead me to (E)
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07 Jul 2006, 10:22
MA, how are you getting (B)?

I know it is not (A). I am pretty confident it is not (B). (Just pick 3 numbers for W with average 17.5 and experiment with removing one. There infinite number of possibilities.)

I am guessing it is not (C) -- too much work.

So this would lead me to (E)
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07 Jul 2006, 11:14
would go for (C)

I - insufficient - Ifall no.s are same, then mean can be same.. or if no.s are different, then anything is possible.

II. insufficient - mean is 17.5.. it may be possible to have all no.s as 17.5, which would yield same mean for V/W .. or no.s can be differnet, yielding to dif. mean.

Combining,
we know that no.s are integers and not all integers are same .. So, If we remove one integer from the set, it shd yield a diff. mean
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07 Jul 2006, 13:25
sgrover wrote:
would go for (C)

Combining,
we know that no.s are integers and not all integers are same .. So, If we remove one integer from the set, it shd yield a diff. mean

Still not convinced this proves it. What if you find a set with all but one integers the same that yields 17.5 as average? If you remove any one number from (assuming there more than 3 numbers in the set) the set the median would not change. Is it true that there would be no combination that can give you mean of 17.5 again? Hard to believe, there are so many numbers out there...
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07 Jul 2006, 14:03
v1rok wrote:
sgrover wrote:
would go for (C)

Combining,
we know that no.s are integers and not all integers are same .. So, If we remove one integer from the set, it shd yield a diff. mean

Still not convinced this proves it. What if you find a set with all but one integers the same that yields 17.5 as average? If you remove any one number from (assuming there more than 3 numbers in the set) the set the median would not change. Is it true that there would be no combination that can give you mean of 17.5 again? Hard to believe, there are so many numbers out there...

sorry guys. i was rushing to guess B thinking that all numbers in set w are integers. it should be C.

we know from 1 and 2 that w has all integers and the number of integers is even. for ex: 10, 15, 20, 25. the sum is 70 and avg = 17.5. if we take out one integer, the avg of the remaining integers wont be the same. it is only possible with numbers.

So its C.
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07 Jul 2006, 14:20
Great work, guys/gals!

The answer is indeed (C), as explained above!
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07 Jul 2006, 15:24
sgrover wrote:
would go for (C)

I - insufficient - Ifall no.s are same, then mean can be same.. or if no.s are different, then anything is possible.

II. insufficient - mean is 17.5.. it may be possible to have all no.s as 17.5, which would yield same mean for V/W .. or no.s can be differnet, yielding to dif. mean.

Combining,
we know that no.s are integers and not all integers are same .. So, If we remove one integer from the set, it shd yield a diff. mean

Aren't you assuming that all the integers are different ?? As I understand, unelss mentioned that all integers are different,we have consider a situation where they could be same..
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07 Jul 2006, 15:28
If the integers were all the same, the average of the set would be precisely that integer, not 17.5
07 Jul 2006, 15:28
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