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03 Aug 2016, 10:25
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Zaraki22
Bunuel
enigma123
A set of 25 different integers has a median of 50 and a range of 50. What is the greatest possible integer that could be in this set?
(A) 62
(B) 68
(C) 75
(D) 88
(E) 100
Any idea how to solve this question please?
(A) 62
(B) 68
(C) 75
(D) 88
(E) 100
Any idea how to solve this question please?
Consider 25 numbers in ascending order to be \(x_1\), \(x_2\), \(x_3\), ..., \(x_{25}\).
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 \(x_{13}=50\);
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 \(50=x_{25}-x_{1}\) --> \(x_{25}=50+x_{1}\);
We want to maximize \(x_{25}\), hence we need to maximize \(x_{1}\). Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\) and thus the maximum value of \(x_{25}\) is \(x_{25}=38+50=88\).
The set could be {38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 88}
Answer: D.
Hope it's clear.
Having all lower 13 as 50, can still lead a distribution with a median of 50 and a range of 50, whereby the answer will be E. Any thoughts?
No, Question says "set of 25 different integers"
15 Dec 2016, 16:14
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Excellent Official Question.
Here is what i did in this one=>
Median => 13th element = 50
Now range =50.
Let a and b be the minimum value and maximum value in set.
As range is given=> To maximise b => We must maximise a.
As the integers involved are different and all the element to the left of the median must be less than or equal to it=>
50
49
48
47
46
45
44
43
42
41
40
39
38
Hence the maximum value of a will be 38
Now b-38=50(Range)
Hence b=50+38=88
Hence D
Note=>If the question didn't mention that all integers are different that a would have been 50 and b would have been 100
Here is what i did in this one=>
Median => 13th element = 50
Now range =50.
Let a and b be the minimum value and maximum value in set.
As range is given=> To maximise b => We must maximise a.
As the integers involved are different and all the element to the left of the median must be less than or equal to it=>
50
49
48
47
46
45
44
43
42
41
40
39
38
Hence the maximum value of a will be 38
Now b-38=50(Range)
Hence b=50+38=88
Hence D
Note=>If the question didn't mention that all integers are different that a would have been 50 and b would have been 100
28 Sep 2017, 11:56
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enigma123
I think Bunuel , you meant "We want to maximize \(x_{25}\), hence we need to minimize \(x_{1}\). I still don't understand how did you get the below:
Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\)
Since all integers must be distinct then the maximum value of \(x_{1}\) will be \(median-12=50-12=38\)
How did we assue the least number to be 50-12? why it can't be 0 or 1 or 2 or 34 for that matter?
