Median and Range

[sanjana]

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

OA : D

Can someone please explain this one?

[xcusemeplz2009]

IMO D

set of 15(x1,x2......,x15) int will have median at 8th position i.e x8=25

range=x15-x1 or x15= range+x1=25+x1

x1....x7 will be distinct int and <25( because otherwise med will shift)

in order to get highest value x1 has to be maximum which is 18

therefore x15=43(25+18)

[mehravikas]

Can you please hide the answers using a spoiler..

[spoiler]Answer: D[/spoiler]

Thanks,
Vikas

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

OA : D

Can someone please explain this one?

[Stuart@KaplanGMAT]

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

OA : D

Can someone please explain this one?

The range is set at 25. Therefore, to maximize the biggest number we also need to maximize the smallest number.

The key to this question is the word "different". If we didn't have that restriction, our set could have simply been:

{25, 25, 25, 25, 25, ... 25, 50}

However, since each term must be distinct, we can't use all those 25s.

As xcusemeplz2009 (I can't believe that there were 2008 other xcusemeplzes!) notes, in a set with an odd number of terms the median is the middle term. With 15 terms, that means that the 8th term will be the median.

So, let's count back from 25, our median:

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

That's 8 terms, so the biggest possible smallest term is 18.

The range is 25, so the biggest term in the set is 18 + 25 = 43.