a range of 25 and a median of 25

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Topic: a range of 25 and a median of 25
PostWed Apr 08, 2009 6:01 am
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A set of 15 different integers have a range of 25 and a median of 25. What is greatest possible integer that could be in this set?
A. 32
B. 37
C. 40
D. 43
E. 50


OA D
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PostWed Apr 08, 2009 6:22 am
median: 25=> 7 integers on the left of 25 are <25 and 7 on the right of the median are >25.

to maximize the largest integer, we must minimize the smallest integer. this is possible if the numbers are consecutive integers in increasing order on the left of the median (as the integers must be all different)

so the smallest integer possible is 18.
since range is 25, largest possible integer in the set is 18+25=43

hence, D
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PostTue Jun 02, 2009 7:02 pm
Why can't we just make the highest 50 and all the rest of the 14 integers = 25. That way we can have 50-25 = 25?

Am I missing something?
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anksgupta
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PostTue Jun 02, 2009 8:46 pm
Umar82 wrote:
Why can't we just make the highest 50 and all the rest of the 14 integers = 25. That way we can have 50-25 = 25?

Am I missing something?
All the integers need to be different as per the question
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vpmba2009
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PostWed Jun 03, 2009 11:06 am
We have 15 number, x1, x2....x15
We need to find x1 and x15
We know that the median is x8=25 => x1= 25-7 = 18
We know that x15 -x1 = 25 (*)
Replace x1 = 18 into (*) we have x15 - 18 = 25
Thus x15 = 25 + 18 = 43
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Nigogo
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PostWed Nov 18, 2009 12:16 am
vpmba2009 wrote:
We have 15 number, x1, x2....x15
We need to find x1 and x15
We know that the median is x8=25 => x1= 25-7 = 18
We know that x15 -x1 = 25 (*)
Replace x1 = 18 into (*) we have x15 - 18 = 25
Thus x15 = 25 + 18 = 43
can anybody explain me why we are assuming that 7 numbers to the left from the median should be starting with 18??? not 1, 2, 3, etc. ??? Thus we can easily maximize the largest integer in the set.
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