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Each of the 45 boxes on shelf J weighs less than each of the
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26 Aug 2014, 06:56
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Expert Reply
If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order; If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.
Set J+K has 89 terms, which means that the median would be the 45th term, so the heaviest box on shelf J (as each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K, when arranged in ascending order we'll have j1, j2, ..., j45, k1, k2, ..., k44).
(1) The heaviest box on shelf J weighs 15 pounds --> \(j_{45}=15\)--> directly gives us the answer. Sufficient.
(2) The lightest box on shelf K weighs 20 pounds --> \(k_1=20\) --> just tells us that median is less than 20. Not sufficient.
Re: Each of the 45 boxes on shelf J weighs less than each of the
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31 Dec 2014, 11:35
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Hi jhasac,
DS questions on the Official GMAT are often based on patterns/concepts that you know, but are presented in ways that you're not used to thinking about.
If you were told you that there were 89 values in a group and you were asked to find the median of the group, what would you do?
You'd... 1) Put the numbers in order from least to greatest. 2) Look at the 45th number - even if there are duplicates (since that's the one in "the middle")
Those rules don't change, even though the question is worded in a quirky way. Maybe there's a way to figure out the 45th term WITHOUT putting ALL the numbers in a row first though....
We're told that EACH of the 45 boxes on shelf J weigh LESS than EACH of the 44 boxes on shelf K. This means that the 44 boxes on shelf K would be the "heaviest 44" numbers in the group; by extension, the "heaviest" value on shelf J MUST be the median. With that realization, the question is really asking "what is the heaviest value on shelf J?" From here, getting to the correct answer doesn't require much work.
The "take-away" from all of this is that by spending a little bit of time thinking about the "setup" in the prompt and doing some work to "rewrite" the question can make answering the question considerably easier (and save you some time).
Re: Each of the 45 boxes on shelf J weighs less than each of the
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15 Oct 2015, 03:33
Bunuel wrote:
If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order; If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.
Set J+K has 89 terms, which means that the median would be the 45th term, so the heaviest box on shelf J (as each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K, when arranged in ascending order we'll have j1, j2, ..., j45, k1, k2, ..., k44).
(1) The heaviest box on shelf J weighs 15 pounds --> \(j_{45}=15\)--> directly gives us the answer. Sufficient.
(2) The lightest box on shelf K weighs 20 pounds --> \(k_1=20\) --> just tells us that median is less than 20. Not sufficient.
How can we derive the median from a large or very large set of numbers. E.g. the median from a set of consecutive integers from 1 to 2000? Is it simply dividing 2000 / 2 = 1000 to arrive at 1000 and 1001 which both are the middle numbers? Then divide the sum of those by two?
Do you have more on this concept? If you have a big set of numbers with odd # of members in it, e.g. 1 - 2003 consecutive integers. How to you arrive at the exact median of this set? Can you say that by dividing 2003/2 = 1001.5 and rounding to the next integer 1002 is the median of the set?
Re: Each of the 45 boxes on shelf J weighs less than each of the
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15 Oct 2015, 05:17
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Kudos
reto wrote:
Bunuel wrote:
If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order; If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.
Set J+K has 89 terms, which means that the median would be the 45th term, so the heaviest box on shelf J (as each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K, when arranged in ascending order we'll have j1, j2, ..., j45, k1, k2, ..., k44).
(1) The heaviest box on shelf J weighs 15 pounds --> \(j_{45}=15\)--> directly gives us the answer. Sufficient.
(2) The lightest box on shelf K weighs 20 pounds --> \(k_1=20\) --> just tells us that median is less than 20. Not sufficient.
How can we derive the median from a large or very large set of numbers. E.g. the median from a set of consecutive integers from 1 to 2000? Is it simply dividing 2000 / 2 = 1000 to arrive at 1000 and 1001 which both are the middle numbers? Then divide the sum of those by two?
Do you have more on this concept? If you have a big set of numbers with odd # of members in it, e.g. 1 - 2003 consecutive integers. How to you arrive at the exact median of this set? Can you say that by dividing 2003/2 = 1001.5 and rounding to the next integer 1002 is the median of the set?
Thanks
It does not matter how big a set is. The rules are: for a set of values (integers/non integers), median of n values will be
= (n+1)/2 th term if n = odd
or
= average of n/2 th and (n/2+1)th terms if n = even
example, for a set with 10 values, the median will be = average of the 5th and the 6th terms
while for a set with 11 values, the median will be = (11+1)/2 = 6th term in the set
Do make sure that the set is either in ascending or descending order.
For the questions you have asked, look below
1-2000, you have 2000 integers = even 'n' and thus the median will = average of 1000th + 1001st term = (1000+1001)/2 = 2001/2 = 1000.5
For 1-2003, you have 2003 integers = odd 'n' and thus the median = middlemost term = (2003+1)/2 th term = 2004/2 th term = 1002 nd term = 1002
Re: Each of the 45 boxes on shelf J weighs less than each of the
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14 Sep 2017, 09:45
Expert Reply
udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?
(1) The heaviest box on shelf J weighs 15 pounds. (2) The lightest box on shelf K weighs 20 pounds.
We are given that each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K and need to determine the median weight of the 89 boxes.
We may recall that when the boxes are ordered from least weight to greatest, the box with the median weight is in the (89 + 1)/2 = 90/2 = 45th position. Thus, the box with the median weight is on shelf J and it is the heaviest box on shelf J.
Statement One Alone:
The heaviest box on shelf J weighs 15 pounds.
Since the heaviest box on shelf J weighs 15 pounds, the 45th box weighs 15 pounds. Since the 45th box is the median of the boxes, the median weight is 15 pounds. Statement one alone is sufficient to answer the question.
Statement Two Alone:
The lightest box on shelf K weighs 20 pounds.
Since the lightest box on shelf K weighs 20 pounds, when the boxes on both shelves are ordered from least to greatest weight, the 46th box weighs 20 pounds. However, we do not have enough information to determine the weight of the 45th box, i.e., the median weight.
Re: Each of the 45 boxes on shelf J weighs less than each of the
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16 Sep 2017, 06:22
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Top Contributor
udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?
(1) The heaviest box on shelf J weighs 15 pounds. (2) The lightest box on shelf K weighs 20 pounds.
Target question:What is the median weight of 89 boxes on these shelves?
Given: Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on the shelf K. Let J1 be the weight of the lightest box on shelf J. Let J2 be the weight of the 2nd lightest box on shelf J. . . . Let J45 be the weight of the heaviest box on shelf J. Let K1 be the weight of the lightest box on shelf K. Let K2 be the weight of the 2nd lightest box on shelf K. etc.
So, the given information tells us that J1 < J2 < J3 < ... <J45 < K1 < K2 < ...< K44 Since the 89 boxes are now arranged in ascending order (according to weight), the median weight will be the weight of the middle box. That is, the median weight, will be the weight of the 45th box. So, we can REPHRASE the target question.... REPHRASED target question:What is the weight of box J45?
Statement 1: The heaviest box on shelf J weighs 15 pounds. PERFECT! Box J45 is the heaviest box on shelf J So, box J45 weighs 15 pounds. Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT
Statement 2: The lightest box on shelf K weighs 20 pounds. This tells us that box K1 weighs 20 pounds. This does not help us find the weight of box J45 Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT
Re: Each of the 45 boxes on shelf J weighs less than each of the
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20 Dec 2017, 17:40
Expert Reply
Hi All,
This is a great 'concept' question, meaning that you don't really have to do any math to answer it as long as you understand the concepts involved.
We're told that EACH of the 45 boxes on shelf J weighs LESS than EACH of the 44 boxes on shelf K. We're asked for the MEDIAN weight of the 89 boxes.
To start, the 'median box' will be the 45th box. Since each of the boxes on shelf J weigh LESS than each of the boxes on shelf K, the 'heaviest' box on shelf J will be the median. If we have that one value, then we can answer the question.
1) The heaviest box on shelf J weighs 15 pounds.
This Fact gives us exactly what we need to answer the question. Fact 1 is SUFFICIENT
2) The lightest box on shelf K weighs 20 pounds.
Unfortunately, this Fact doesn't tell us anything about the weights of the boxes on shelf J. Maybe the median is 19 pounds, but it could just as easily be 18 pounds, 17 pounds, 16.5 pounds, etc. Fact 2 is INSUFFICIENT
Re: Each of the 45 boxes on shelf J weighs less than each of the
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05 Sep 2021, 15:41
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