D01-21

Official Solution:

A set of 11 different integers has a median of 25 and a range of 50. What is the greatest possible integer that could be in this set?

A. 65
B. 70
C. 75
D. 80
E. 85


Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).

The median of a set with an odd number of elements is the middle number when arranged in ascending or descending order. So, in this case, the median of the given set is \(x_{6}=25\):

The range of a set is the difference between the largest and smallest numbers of the set. Therefore, the range of the given set is \(50=x_{11}-x_{1}\), which gives \(x_{11}=50+x_{1}\).

We want to maximize \(x_{11}\), so we need to maximize \(x_{1}\). Since all the integers in the set must be distinct, the maximum value of \(x_{1}\) can be \(median-5=25-5=20\) (we need \(x_1\) as close to the median as possible). Thus, the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).

Therefore, the set could be {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 70}.


Answer: B
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median-5=25-5=20 - why did we subtract 5 from median?

2 things to be noted here

We need to find the largest number possible in the list and that number is x1

Option C,D and E can be ruled out because if median is 25 and range 50 that means number x1 to x5 will be 25 but we are told there are 11 different integers so this case is not possible.Likewise D and E not possible.

Now to maxmimize X11,you need to minimize other numbers..

Thus when you subtract from 5 median to get 20 as the smallest number then largest possible number will be Range (50)+20= 70..

Mind you even 65 can be an option but since we need to find largest integer possible 70 will be the answer
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Hi, how do you know the 11 numbers are consecutive numbers? What if X1 - X6 are all 25, which makes X11 75.

We are given that the set consists of 11 different integers .
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we can solve it by drawing some lines:

_ _ _ _ _ 25 _ _ _ _ _
the range is 50
thus the last one would be x
and the highest one would be y
for the range to be 50, x can be 20 and y 70.
note that if x is lower than 20, then y would be lower.
and x can't be higher than 20 since all the numbers should be different.

If all numbers before 25 are also 25 :

25,25,25,25,25,25,X1,X2 Then also median will be 25 and Max value will be 25+50 = 75

I fell for the very very annoying GMAT Trap which turns this problem in to a 700+ level problem.

The questions mentions

"A set of 11 different integers"

I may be missing a basic point but why shouldn't we consider negative numbers as part of the set ?

We want to find the greatest possible integer in the set. It turns out that the greatest possible integer is 70, and in this case the set is {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}, so no negative integers could be in the set if the greatest integer is 70, median is 25 and the range is 50.
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Let the 11 integers be =>

x1
x2
x3
x4
x5
25
x7
x8
x9
x10
x1+50


Now to maximise last we must maximise the first and minimise others
NOTE -> Different integers.

Hence
x5=24
x4=23
x3=22
x2=21
x1=20
So last term = 20+50 = 70


Smash that B.

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Highest number - least number = Range

11 numbers, \(6^{th}\) number is the median. The range is 50. So the highest number will be 50 higher than the least number. We need to find the least number.

Numbers are different. The median is 25, lets count down for another 5 number 24, 23,22,21,20

So, the highest number will be 20+50=70
The answer is B
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I think this is a high-quality question and I agree with explanation.

We are given Range as 50 , Range = Highest - Lowest in a set. From this we have Highest = 50 + Lowest , We can see inorder for Highest to be maximized , we need Lowest to be maximum too.
But since all numbers are int and median is 25 and total numbers in set 11 , Inorder for lowest number to be maximized we can only have {20,21,22,23,24,25} as first 6 numbers in set ( including median ). Here lowest is 20 , We know Highest = 50 + lowest = 50+20 = 70

even tho I got this question wrong (because I rushed to the answer) I later learned this from it.

- notice the restrictions (integers and different, AND range = 50)
- be careful, we are not told that they are consecutive, but we should come to this conclusion by acknowledging that:
in order to maximize the largest number in the set, we need to minimize the left side of the set (left to the median)WHILE HOLDING THE RESTRICTIONS, there's only one combination that meets these restrictions.