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GMAT Diagnostic Test Question 21
Field: word problems (min/max)
Difficulty: 700
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
Field: word problems (min/max)
Difficulty: 700
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
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08 Jul 2009, 06:56
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Consider 11 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{11}\).
The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{6}=25\);
The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{11}-x_{1}\) --> \(x_{11}=50+x_{1}\);
We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).
The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}
Answer: B.
The median of a set with odd number of elements is the middle number (when arranged in ascending or descending order), so the median of given set is \(x_{6}=25\);
The range of a set is the difference between the largest and the smallest numbers of a set, so the range of given set is \(50=x_{11}-x_{1}\) --> \(x_{11}=50+x_{1}\);
We want to maximize \(x_{11}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-5=25-5=20\) and thus the maximum value of \(x_{11}\) is \(x_{11}=50+20=70\).
The set could be {20, 21, 22, 23, 24, 25, 26, 27, 26, 29, 70}
Answer: B.
20 Nov 2013, 12:24
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Why must the numbers in the set be different in order to calculate the median?
Couldn't I arrange a set like the following? S={25, 25, 25, 25, 25, 25, 75, 75, 75, 75, 75}?
(Note that after x6, the follwoing numbers could be anything between 26 and 74)
In this way, the median is still 25, but the máximum value of x would now be 75, making "C" the correct answer.
Thanks for your comments!
D.
Couldn't I arrange a set like the following? S={25, 25, 25, 25, 25, 25, 75, 75, 75, 75, 75}?
(Note that after x6, the follwoing numbers could be anything between 26 and 74)
In this way, the median is still 25, but the máximum value of x would now be 75, making "C" the correct answer.
Thanks for your comments!
D.
