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Range & Median

Expert replies
by Akansha » Thu May 26, 2011 9:37 am
A set of 15 different integers has median of 25 and a range of 25. What is
greatest possible integer that could be in this set?
a. 32
b. 37
c. 40
d. 43
e. 50

OA is D
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Source: — Problem Solving |

by Frankenstein » Thu May 26, 2011 10:00 am
Hi,
Let the 15 integers when placed in increasing order be a1,a2,...a15.
Median is 25. So, a8=25.
Range is 25. So, a15 = a1 + 25
So, for a15 to be maximum, a1 should be maximum.
As all the integers are distinct each of a1 to a8 must be at least one less than the next term.
So, (a1)max+7=a8 =25 => (a1)max = 18.
So, (a15)max = (a1)max+25 = 18+25 =43.

Hence, answer D

Cheers!
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by GMATGuruNY » Thu May 26, 2011 10:10 am
Akansha wrote:A set of 15 different integers has median of 25 and a range of 25. What is
greatest possible integer that could be in this set?
a. 32
b. 37
c. 40
d. 43
e. 50

OA is D
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 biggest answer choice.

Answer choice E: 50
Smallest = 50-25 = 25.
Since all 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 vietle » Thu May 26, 2011 9:34 pm
Could you please elaborate this:
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.

Why not for example choice C?
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by GMATGuruNY » Fri May 27, 2011 1:34 am
vietle wrote:Could you please elaborate this:
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.

Why not for example choice C?
The question asks for the greatest possible integer that could be in the set. Since 43 is greater than all the remaining answer choices, no need to check A, B and C; the correct answer must be D.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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