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31 Dec 2006, 18:50
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set X has 5 integers, and the average is greater than medium; set Y has 7 integers, and the average is greater than medium. integers in X and Y are all different. set Z combined x and y, is the average greater than medium?
(1)medium of set X is greater than medium of set Y
(2)average of set X is greater than average of set Y
(1)medium of set X is greater than medium of set Y
(2)average of set X is greater than average of set Y
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31 Dec 2006, 21:33
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answer is E.
the key to understand this question is to remember that the concept of average and the concept of median are independent from each other. i.e. given a certain set of numbers it is easy to create a variation of it with the same median but different average, or the same average and a different median.
for example - lowering the lowest element of a set will always lower the average and keep the median the same (assuming there are more than 2 elements in the set).
one should also remember, as a rule of thumb, that the average equals to the median if the set is "symmetric"/"evenly spaced". this helps in building examples.
so the question is all about trying to make you think that there is dependency between these concepts. since there isn't - the answer is E.
but hey - I always tell you - if answer is E there should be an example that shows it.... so lets build one.
first we find an example where all statements are true, but the average is not greater than the median.
so the total set would be: 1,2,3,4,5,6,7,8,9,10,11,12. here the median equals the average = 6.5
X= 7,8,9,10,12
Y=1,2,3,4,5,6,11
both sets have their average greater than their median (see how i built it - i took a symmetric set, and made the largest number even larger)
also both the median and average of X are greater than those of Y. hence the sets satisfy st1 and st2.
the second example is even simpler to build:
Y= 1,2,3,4,5,6,10
X=100,101,102,103,140
again X and Y are "legal" sets, and you can see easily tht average and median of X are greater than Y.
also the average of the union is greater than the median of the union.
hence E.
however the main lesson here is that average and median are independent from each other.
the key to understand this question is to remember that the concept of average and the concept of median are independent from each other. i.e. given a certain set of numbers it is easy to create a variation of it with the same median but different average, or the same average and a different median.
for example - lowering the lowest element of a set will always lower the average and keep the median the same (assuming there are more than 2 elements in the set).
one should also remember, as a rule of thumb, that the average equals to the median if the set is "symmetric"/"evenly spaced". this helps in building examples.
so the question is all about trying to make you think that there is dependency between these concepts. since there isn't - the answer is E.
but hey - I always tell you - if answer is E there should be an example that shows it.... so lets build one.
first we find an example where all statements are true, but the average is not greater than the median.
so the total set would be: 1,2,3,4,5,6,7,8,9,10,11,12. here the median equals the average = 6.5
X= 7,8,9,10,12
Y=1,2,3,4,5,6,11
both sets have their average greater than their median (see how i built it - i took a symmetric set, and made the largest number even larger)
also both the median and average of X are greater than those of Y. hence the sets satisfy st1 and st2.
the second example is even simpler to build:
Y= 1,2,3,4,5,6,10
X=100,101,102,103,140
again X and Y are "legal" sets, and you can see easily tht average and median of X are greater than Y.
also the average of the union is greater than the median of the union.
hence E.
however the main lesson here is that average and median are independent from each other.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
