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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
71% (01:03) correct
29%
(00:35)
wrong
based on 7
sessions
History
Date
Time
Result
Not Attempted Yet
A set of 19 different integer observations have a range 40 and median 40. The least possible value of set could be
A. 2
B. 6
C. 7
D. 8
E. 9
(hii guys, I can understand the source of the question can feel misleading.. I actually mentioned a few sources where I found similar questions so that other test-takers could see that this topic has been discussed in different forms and is worth practicing. My intention was simply to contribute positively to the community by highlighting it as an important topic and a must-try type of question for exam preparation.. )
A. 2
B. 6
C. 7
D. 8
E. 9
(hii guys, I can understand the source of the question can feel misleading.. I actually mentioned a few sources where I found similar questions so that other test-takers could see that this topic has been discussed in different forms and is worth practicing. My intention was simply to contribute positively to the community by highlighting it as an important topic and a must-try type of question for exam preparation.. )
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[random-gprep:1cck4990][/random-gprep:1cck4990]
ankushsambare
Joined: 13 Jun 2022
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08 Jun 2026, 20:17
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Let the 19 different integers be arranged in increasing order:
a1 < a2 < ... < a19
Since there are 19 observations, the median is the 10th term.
So, a10 = 40.
Range = 40
⇒ a19 − a1 = 40
To make the least value a1 as small as possible, we must make the greatest value a19 as small as possible.
Since all numbers are different integers and the 10th term is 40, the smallest possible values after 40 are:
41, 42, 43, 44, 45, 46, 47, 48, 49
Hence,
a19 = 49
Using the range:
49 − a1 = 40
⇒ a1 = 9
Therefore, the least possible value in the set is 9.
Answer: E) 9.
a1 < a2 < ... < a19
Since there are 19 observations, the median is the 10th term.
So, a10 = 40.
Range = 40
⇒ a19 − a1 = 40
To make the least value a1 as small as possible, we must make the greatest value a19 as small as possible.
Since all numbers are different integers and the 10th term is 40, the smallest possible values after 40 are:
41, 42, 43, 44, 45, 46, 47, 48, 49
Hence,
a19 = 49
Using the range:
49 − a1 = 40
⇒ a1 = 9
Therefore, the least possible value in the set is 9.
Answer: E) 9.
08 Jun 2026, 20:52
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Bookmarks
there are 19 different numbers hence the median will be 10th Number of them
x _ _ _ _ _ _ _ _ 40 _ _ _ _ _ _ _ _ (x+40)
For x to be least, we need to reach from median 40 to x+40 so that the increase is minimum
i.e. x _ _ _ _ _ _ _ _ 40 41 42 43 44 45 46 47 48 (49)
i.e. x+40 = 49 hence, x = 9
Answer: Option E
x _ _ _ _ _ _ _ _ 40 _ _ _ _ _ _ _ _ (x+40)
For x to be least, we need to reach from median 40 to x+40 so that the increase is minimum
i.e. x _ _ _ _ _ _ _ _ 40 41 42 43 44 45 46 47 48 (49)
i.e. x+40 = 49 hence, x = 9
Answer: Option E
nischaysetthi
