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Expert replies
by j_shreyans » Fri Nov 07, 2014 11:19 pm
Hi Guys ,

I have query but i dont remember the options and the OA.

Pls help me on this.

There are 15 set of distinct integer , Median is 25 and the range is 25. What is the greatest possible value?
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Source: — Problem Solving |

by [email protected] » Fri Nov 07, 2014 11:32 pm
Hi J_shreyans,

When it comes to maximizing or minimizing a value in a group of numbers, you have to think about what the other numbers would need to be to accomplish your goal.

Here, we have a group of 15 DISTINCT (meaning DIFFERENT) integers with a median of 25 and a RANGE of 25. That range will dictate how large the largest value can be.

With a median of 25, we know that 7 numbers are LESS than 25 and 7 numbers are GREATER than 25:

_ _ _ _ _ _ _ 25 _ _ _ _ _ _ _

To maximize the largest value, we need to maximize the smallest value. Here's how we can do it:

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

With 18 as the smallest value, and a range of 25, the largest value would be 43.

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by Brent@GMATPrepNow » Sat Nov 08, 2014 2:12 am
Here's the complete question:
A set of 15 different integer has a median of 25 & 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
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 [spoiler]43 = D[/spoiler]

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by GMATGuruNY » Sat Nov 08, 2014 9:09 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?
32
37
40
43
50
Range = biggest - smallest.
Thus:
25 = biggest - smallest.
Smallest = biggest - 25.

We can plug the answer choices into the equation above.
Since we need the greatest possible integer that could be in the set, we should start with the greatest answer choice.

Answer choice E: 50
Smallest = 50-25 = 25.
Since all of the integers must be different, the smallest integer cannot be equal to the median.
Eliminate E.

Answer choice D: 43
Smallest = 43-25 = 18.
Thus, the 7 integers below the median of 25 could be {18,19,20,21,22,23,24}.
The 6 integers between 25 (the median) and 43 (the biggest) could be any 6 different integers between 25 and 43.
This works.

The correct answer is D.
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by Mathsbuddy » Mon Nov 10, 2014 9:32 am
The median here is in the 7th position (because 7 = (N+1)/2).
Therefore the largest value that the minimum value could be is 25(median) - 7 = 18
Maximum value = 18 + range(25) = 43
Last edited by Mathsbuddy on Tue Nov 11, 2014 5:21 am, edited 1 time in total.
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by Matt@VeritasPrep » Mon Nov 10, 2014 10:21 am
Since there are 15 different integers, 7 of them will be below the median and 7 of them will be above the median.

The largest number is tied to the smallest number -- it must be 25 greater -- so we need the smallest number to be as large as it can be.

Hence we'll have the seven integers less than the median be equal to the seven integers preceding 25, which means those seven integers are 18, 19, 20, 21, 22, 23, and 24.

The smallest integer is thus 18, making the largest integer 18+25, or 43.

50 is clearly impossible, as it's 25 greater than the MEDIAN, not the least term. Thus D must be right!
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