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Updated on: 06 Apr 2025, 02:48
Originally posted by EgmatQuantExpert on 11 Jul 2018, 08:02.
Last edited by Bunuel on 06 Apr 2025, 02:48, edited 3 times in total.
Last edited by Bunuel on 06 Apr 2025, 02:48, edited 3 times in total.
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C
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Difficulty:
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Question Stats:
62% (02:01) correct
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A set of 51 different integers has a median of 30 and a range of 60. What is the value of the least possible integer in this set?
A. -10
B. -6
C. -5
D. 5
E. 10
A. -10
B. -6
C. -5
D. 5
E. 10
11 Jul 2018, 08:13
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To find the least possible integer in the set of 51 different integers
The median of the set is 30
=> If the numbers are arranged in ascending order, the 26th integer is 30
We know that all the integers to the right of the median are \(\geq\) median.
Since the set consists of all different integers, the integers to the right of the median are \(>\) median
Lets make the integers to the right of the median greater than median by least possible value i.e 1
=> value at 27th place will be 31(27+4)
=> value at 28th place will be 32(28+4)
...... => value at 51st place will be 55(51+4)
Given range = 60
=> max_possible_integer - lowest_possible_integer = 60
=> 55 - lowest_possible_integer = 60
=> lowest_possible_integer = 55 - 60 = -5
Hence option C
The median of the set is 30
=> If the numbers are arranged in ascending order, the 26th integer is 30
We know that all the integers to the right of the median are \(\geq\) median.
Since the set consists of all different integers, the integers to the right of the median are \(>\) median
Lets make the integers to the right of the median greater than median by least possible value i.e 1
=> value at 27th place will be 31(27+4)
=> value at 28th place will be 32(28+4)
...... => value at 51st place will be 55(51+4)
Given range = 60
=> max_possible_integer - lowest_possible_integer = 60
=> 55 - lowest_possible_integer = 60
=> lowest_possible_integer = 55 - 60 = -5
Hence option C
EgmatQuantExpert
Given set, Say
X={\(x_{min}\),.....(25 members including \(x_{min}\))......,30,.......(25 members including \(x_{max}\)).....,\(x_{max}\)}
---------------------------------------------------(26th term Median) --------------------------------- \((51^{th} term)\)
Given, \(x_{max} - x_{min}=60\) ----(1)
Since number of members after median is 25, hence x(max)=30+25=55 (Note:- all terms of the set are different as given)
So, substituting in (1), we have
\(55-x_{min}=60\)
Or, \(x_{min}=-5\)
Therefore, the value of the least possible integer in this set is -5.
Ans. (C)
