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Harish Dorai
 
 

A set of 15 different integers has a median of 25

by Harish Dorai Tue Jul 31, 2007 11:30 am

A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?

A) 32
B) 37
C) 40
D) 43
E) 50
amrinder
 
 

by amrinder Tue Jul 31, 2007 12:05 pm

Ans is D i.e.43

Assume first value A and Last value x

Range is 25=x-A Also Median =25

Set of different integers
Only one value satisfy this 43-18

18,19,20,21,22,23,24,25
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by dbernst Thu Aug 02, 2007 11:27 am

Harish,

sometimes, when the algebraic solution is not obvious, it can be advantageous to roll up your sleeves and attack the problem with "brute force." In this case, the median is 25, so the 8th number in ascending order must be 25. Moreover, the range is 25, so the difference between the smallest and largest numbers must be 25.

Because this is a "could" problem that asks for the "greatest possible integer," let's attack the largest integers first.

E) 50: If 50 is greatest number than 25 must be smallest (as the range is 25). This, obviously, cannot yield a median of 25.

D) 43: If 43 is greatest number than 18 must be smallest (as the range is 25). Now, just list numbers to check whether 25 can be the 8th number in ascending order. 18, 19, 20, 21, 22, 23, 24, 25... As 25 is the 8th number, the correct answer is D.

-dan

A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?

A) 32
B) 37
C) 40
D) 43
E) 50
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Re: A set of 15 different integers has a median of 25

by borhan11 Fri Sep 23, 2011 1:43 am

The solution doesn't seem to work.

If you consider

18,19,20,21,22,23,24,25

and then go to:

26,27,28,29,30,31,43

The average of these numbers comes to 25.7 and not 25.

So how can 43 be the solution? Am I missing something?
700+
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Re: A set of 15 different integers has a median of 25

by 700+ Sun Sep 25, 2011 8:20 am

borhan11 wrote:and then go to:

26,27,28,29,30,31,43

The average of these numbers comes to 25.7 and not 25.

So how can 43 be the solution? Am I missing something?

There are 15 Integers. 25 is the median.
The numbers after 25 i.e the next 6 integers (7th integer is 43), need not be consecutive. It could be 6 different integers which is greater than 25 and the sum should add to
[(25 x 15) - (18+19+20+21+22+23+24+25) - (43)]

(25 x 15) - Total Sum of all the integers
(18+19+20+21+22+23+24+25) - Sum of first eight integers
(43) - 15th or last integer

Hope I was able to clear your doubt.
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Re: A set of 15 different integers has a median of 25

by RonPurewal Thu Oct 06, 2011 5:47 am

borhan -- nice 4-year thread bump.

borhan11 wrote:The average of these numbers comes to 25.7 and not 25.


this is your problem: you're using the arithmetic mean (average) of the numbers, but the problem actually wants you to use the median.
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Re: A set of 15 different integers has a median of 25

by scotty Tue Feb 14, 2012 2:49 pm

i actually arrived at 50 as my final answer. could somebody please explain to me why my logic is wrong?

here is my thought process going through the problem:

out of the 15 integers in the set, i let 14 of those integers be 25, and i let the last integer be 50. thus, the range of the set is 50-25=25, and 25 technically is the median of the set, as 25 will also be the 8th term in the set (because there are fourteen of them).
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Re: A set of 15 different integers has a median of 25

by RonPurewal Fri Feb 17, 2012 5:59 am

scotty, per the problem statement, all of the integers have to be different. so, having multiple 25's is not ok.
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Re: A set of 15 different integers has a median of 25

by s.pratibha14 Sat Feb 18, 2012 7:10 am

I also have a same kind of data sufficiency doubt.

What is the median of 15 integers?
(1) Exactly seven of the numbers are greater than 7.
(2) Exactly seven of the numbers are less than 7

plz help me to solve this question with explanation.
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Re: A set of 15 different integers has a median of 25

by jnelson0612 Sun Feb 19, 2012 10:28 pm

s.pratibha14 wrote:I also have a same kind of data sufficiency doubt.

What is the median of 15 integers?
(1) Exactly seven of the numbers are greater than 7.
(2) Exactly seven of the numbers are less than 7

plz help me to solve this question with explanation.


Hi,
Is this a question from the GMAT prep tests? If so, please post it in a new thread with the first few words as the thread title. We'll then be happy to help you. Thanks!
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Re:

by NNK314 Sun Jul 17, 2016 8:01 am

Totally understand the working backwards strategy..
However, I tried that during the test and when I got to the second choice, i realized the numbers would be 18, 19, 20...25... so I started wondering that the question didn't really say that the integers would be consecutive.. that threw me off and I tried all choices and eventually spent too much time on this. :roll:

dbernst wrote:Harish,

sometimes, when the algebraic solution is not obvious, it can be advantageous to roll up your sleeves and attack the problem with "brute force." In this case, the median is 25, so the 8th number in ascending order must be 25. Moreover, the range is 25, so the difference between the smallest and largest numbers must be 25.

Because this is a "could" problem that asks for the "greatest possible integer," let's attack the largest integers first.

E) 50: If 50 is greatest number than 25 must be smallest (as the range is 25). This, obviously, cannot yield a median of 25.

D) 43: If 43 is greatest number than 18 must be smallest (as the range is 25). Now, just list numbers to check whether 25 can be the 8th number in ascending order. 18, 19, 20, 21, 22, 23, 24, 25... As 25 is the 8th number, the correct answer is D.

-dan

A set of 15 different integers has a median of 25 and a range of 25. What is the greatest possible integer that could be in this set?

A) 32
B) 37
C) 40
D) 43
E) 50
RonPurewal
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Re: Re:

by RonPurewal Sat Jul 23, 2016 1:38 am

the goal is to make the top number as big as possible.

in order to do that, all of the OTHER numbers need to be "crunched together" as tightly as possible. if these numbers have to be different integers, then the "tightest" you can get is to make them consecutive integers.