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Median Problem;HELP GMAT Tomorrow

Expert replies
by soni_pallavi » Sat Nov 24, 2012 12:08 am
Q1) 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

AnsD
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Source: — Problem Solving |

by Anurag@Gurome » Sat Nov 24, 2012 3:25 am
soni_pallavi wrote:Q1) 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

AnsD
Median of a set of 15 different integers will be the 8th integer of the series when arranged according to their values.

Now, Largest - Smallest = Range => Largest = (Smallest + Range)
As the range is fixed, we can maximize largest number by maximizing the smallest number.

Maximum possible value of the smallest integer in the set is (25 - 7) = 18, as all the terms are different and 25 is the 8th term.

Hence, greatest possible integer in the set = (18 + Range) = (18 + 25) = 43

The correct answer is D.
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by GMATGuruNY » Sat Nov 24, 2012 3:31 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
Correct answer 43
Please explain how?
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 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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