a range of 25 and a median of 25

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a range of 25 and a median of 25

by sanju09 » Wed Apr 08, 2009 5:01 am
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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by scoobydooby » Wed Apr 08, 2009 5:22 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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by Umar82 » Tue Jun 02, 2009 6:02 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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by anksgupta » Tue Jun 02, 2009 7:46 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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by vpmba2009 » Wed Jun 03, 2009 10:06 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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by Nigogo » Tue Nov 17, 2009 11:16 pm
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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by Brent@GMATPrepNow » Sun May 15, 2011 2:55 pm
sanju09 wrote: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
Let's tackle this one step at a time.

First, we have 15 different integers.
We can let these 15 spaces represent the 15 numbers written in ascending order: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

If the median is 25, we can add this as the middle value: _ _ _ _ _ _ _ 25 _ _ _ _ _ _ _
Notice that 7 of the remaining numbers must be greater than 25 and the other 7 remaining number must be less than 25.

Since, we are told that the range is 25, we know that the greatest number minus the smallest number = 25

Now notice two things:
1) Once we know the value of the smallest number, the value of the greatest number is fixed.
For example, if the smallest number were 10, then the greatest number would have to be 35 in order to have a range of 25
Similarly, if the smallest number were 12, then the greatest number would have to be 37 in order to have a range of 25

2) If we want to maximize the value of the greatest number, we need to maximize the value of the smallest number.

So, how do we maximize the value of the smallest number in the set?
To do this, we must maximize each of the 7 numbers that are less than the median of 25.

Since the 15 numbers are all different, the largest values we can assign to the numbers less than the median of 25 are as follows:
18 19 20 21 22 23 24 25 _ _ _ _ _ _ _ (this maximizes the value of the smallest number)

If 18 is the maximum value we can assign to the smallest number, and if the range of the 15 numbers is 25, then greatest number must equal 43 (since 43 - 18 = 25)

So, the numbers are as follows: 18 19 20 21 22 23 24 25 _ _ _ _ _ _ 43 (the missing numbers don't really matter here)

This means the answer is D

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Brent
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