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good median Qs for practice :-)

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Senior Manager
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good median Qs for practice :-) [#permalink]

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New post 18 Mar 2007, 16:51
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi, it is really a good question. I solved it but I am posting it here for others to practice.

Q: Average weight of 3 boxes is 7 Kg, median weight is 9 Kg. What is the maximum possible weight of the lightest box?

:)

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New post 19 Mar 2007, 10:49
If we are just talking about integers

I say 2

we know from statement X + Y = 12

from median we know since they cannot all be the same 9+9+9 = 27 so we can conclude

Y > 9

X < 9

so to find max X we need to find lowest possible value of Y

That would be 10, therefore X would be 2

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New post 19 Mar 2007, 11:07
My answer is 3.

Since avg is 7 the total is 21
median is 9
so one value is above 9 and other below it.
The value 'below' will be max when value 'above' is min
thats possible when value above is same as median 9
so max possible weight of lightest box is 21 - 9 -9 = 3

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New post 19 Mar 2007, 12:15
yea 3 looks right , didnt take into account the median and Y as having the same value

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New post 19 Mar 2007, 14:43
interesting so the median of 8,9,9 is 9?

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yes [#permalink]

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New post 19 Mar 2007, 15:28
I just checked. The median of 3,9,9 will be 9. Interesting.

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New post 19 Mar 2007, 15:48
also mean median and mode of 9,9,9 is 9

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New post 19 Mar 2007, 15:49
Tuneman wrote:
interesting so the median of 8,9,9 is 9?


The median of the set is literally the number in the middle, given that the numbers are arranged from the smallest to the biggest and the number of integers in the set is odd. If the number of integers is even, to find a median, you will need to divide 2 numbers in the middle by 2.

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New post 19 Mar 2007, 16:20
Answer is 3 :)

Good job Kyatin!!

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  [#permalink] 19 Mar 2007, 16:20
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good median Qs for practice :-)

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