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44 and 45 boxes.

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by ska7945 » Thu Aug 28, 2008 10:52 pm
each of the 45 boxes on shelf J weighs less than each of the 44 boxes on the shelf K. what is the median weight of 89 boxes on these shelves?

1)the highest box on shelf J weighs 15 pounds.
2)the highest box on shelf K weighs 20 pounds.


oa A
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Re: 44 and 45 boxes.

by parallel_chase » Thu Aug 28, 2008 11:11 pm
ska7945 wrote:each of the 45 boxes on shelf J weighs less than each of the 44 boxes on the shelf K. what is the median weight of 89 boxes on these shelves?

1)the highest box on shelf J weighs 15 pounds.
2)the highest box on shelf K weighs 20 pounds.


oa A
The answer is A.

Statement I

the highest box on shelf J weighs 15 pounds.

If we arrange all the 89 boxes in ascending or descending order by weight the highest box on shelf J will be the middle value.

median is the middle value therefore the median will be 15

Sufficient.


Statement II
the highest box on shelf K weighs 20 pounds.

If we arrange all the 89 boxes in ascending or descending order by weight the starting value or ending value will be 20, therefore we cannot find the middle value i.e. media.

Insufficient.

Hence A is the answer
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Re: 44 and 45 boxes.

by sudhir3127 » Thu Aug 28, 2008 11:13 pm
parallel_chase wrote:
ska7945 wrote:each of the 45 boxes on shelf J weighs less than each of the 44 boxes on the shelf K. what is the median weight of 89 boxes on these shelves?

1)the highest box on shelf J weighs 15 pounds.
2)the highest box on shelf K weighs 20 pounds.


oa A
The answer is A.

Statement I

the highest box on shelf J weighs 15 pounds.

If we arrange all the 89 boxes in ascending or descending order by weight the highest box on shelf J will be the middle value.

median is the middle value therefore the median will be 15

Sufficient.


Statement II
the highest box on shelf K weighs 20 pounds.

If we arrange all the 89 boxes in ascending or descending order by weight the starting value or ending value will be 20, therefore we cannot find the middle value i.e. media.

Insufficient.

Hence A is the answer
Agree with chase...

45th term will be the middle term.. hence 15 is the median..



A
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by jeffxujian » Fri Aug 29, 2008 1:14 am
IMO A 2 same reasoning as above.
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by moriken » Thu Jan 27, 2011 1:33 am
Don't you have to take it consideration that some of boxes are same weight? I mean, if there are three 15 pounds boxes, you can't decide the median. Or, does the phrase, "the" heaviest box" mean only one 15 pounds box?
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by Anurag@Gurome » Thu Jan 27, 2011 1:52 am
moriken wrote:Don't you have to take it consideration that some of boxes are same weight? I mean, if there are three 15 pounds boxes, you can't decide the median. Or, does the phrase, "the" heaviest box" mean only one 15 pounds box?
"heaviest" means there is only one such box.
If there was more than one box with maximum weight, then it would've been "heavier two/three... boxes"
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by tomada » Fri Jan 28, 2011 11:34 am
I knew what the creator of the question meant, but using "highest" isn't best when describing the weight of something. THe problem should've used the word "heaviest".
I'm really old, but I'll never be too old to become more educated.
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by moriken » Fri Jan 28, 2011 10:46 pm
Anurag@Gurome wrote:
moriken wrote:Don't you have to take it consideration that some of boxes are same weight? I mean, if there are three 15 pounds boxes, you can't decide the median. Or, does the phrase, "the" heaviest box" mean only one 15 pounds box?
"heaviest" means there is only one such box.
If there was more than one box with maximum weight, then it would've been "heavier two/three... boxes"
Thanks a lot. I understand. I hope I don't lose points because of English any more.
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