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# good median Qs for practice :-)

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Senior Manager
Joined: 20 Feb 2007
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good median Qs for practice :-) [#permalink]  18 Mar 2007, 15:51
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?

VP
Joined: 22 Oct 2006
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[#permalink]  19 Mar 2007, 09:49
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
Senior Manager
Joined: 29 Jan 2007
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[#permalink]  19 Mar 2007, 10:07
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
VP
Joined: 22 Oct 2006
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[#permalink]  19 Mar 2007, 11:15
yea 3 looks right , didnt take into account the median and Y as having the same value
Manager
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[#permalink]  19 Mar 2007, 13:43
interesting so the median of 8,9,9 is 9?
Manager
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yes [#permalink]  19 Mar 2007, 14:28
I just checked. The median of 3,9,9 will be 9. Interesting.
Senior Manager
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[#permalink]  19 Mar 2007, 14:48
also mean median and mode of 9,9,9 is 9
VP
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[#permalink]  19 Mar 2007, 14:49
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.
Senior Manager
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[#permalink]  19 Mar 2007, 15:20

Good job Kyatin!!
[#permalink] 19 Mar 2007, 15:20
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# good median Qs for practice :-)

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