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25 Sep 2019, 13:16
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Set S={-8,-17,-23,-2,4,16,x,y}. If x and y are integers, is the median of set S positive?
1) The average of elements of set S is 0
2) The range of set S is less than 40
1) The average of elements of set S is 0
2) The range of set S is less than 40
Updated on: 25 Sep 2019, 20:38
Originally posted by CareerGeek on 25 Sep 2019, 20:32.
Last edited by CareerGeek on 25 Sep 2019, 20:38, edited 1 time in total.
Last edited by CareerGeek on 25 Sep 2019, 20:38, edited 1 time in total.
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Asad
Median is average of 4th & 5th terms
As we can see in ascending order,
4th term is a negative integer,
5th term has to be derived
(1) The average of elements of set S is 0
Average = 0
—> Sum/8 = 0
—> Sum = 0
—> -25 + 5 + x + y = 0
—> x + y = 30
Possible values of (x, y) =
(0, 30) —> Median = (-2+0)/2 = -1, negative
or
(12, 18) —> Median = (-2+4)/2 = 2, positive
Insufficient
(2) The range of set S is less than 40
—> Range possible = 39, since Range of constant number of the set = 16 - (-23) = 39
x, y can take any values from -23 to 16
—> Median can be positive or negative
Insufficient
Combining (1) & (2),
x + y = 30, Range = 39 and maximum value can not exceed 16
—> Possible values of (x, y)
= (14, 16) or (15, 15)
—> Median is average of -2 & 4 = 1, Always positive
Sufficient
IMO Option C
Pls Hit kudos if you like the solution
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25 Sep 2019, 20:35
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Set S={-8,-17,-23,-2,4,16,x,y}. If x and y are integers, is the median of set S positive?
Depends on the value of x and y..
(A) If both x and y are positive, the median will be {(-2)+4}/2=1, hence positive
(B) In all other cases , the median will be negative.... When even one is negative or zero irrespective of the value of other...
1) The average of elements of set S is 0
so (-8)+(-17)+(-23)+(-2)+4+16+x+y=0....x+y=30
x=40, y=-10, median = -2-8/2=-5
x=15,y=15, median = \(\frac{{(-2)+4}}{2}=1\)
Insuff
2) The range of set S is less than 40
x and y can be anywhere between -23 and 16, so insuff
Combined..
Both x and y have to be positive as x+y=30 and the max value of x and y is <40-23 or <17
Hence the median is always 1
Suff
C
Depends on the value of x and y..
(A) If both x and y are positive, the median will be {(-2)+4}/2=1, hence positive
(B) In all other cases , the median will be negative.... When even one is negative or zero irrespective of the value of other...
1) The average of elements of set S is 0
so (-8)+(-17)+(-23)+(-2)+4+16+x+y=0....x+y=30
x=40, y=-10, median = -2-8/2=-5
x=15,y=15, median = \(\frac{{(-2)+4}}{2}=1\)
Insuff
2) The range of set S is less than 40
x and y can be anywhere between -23 and 16, so insuff
Combined..
Both x and y have to be positive as x+y=30 and the max value of x and y is <40-23 or <17
Hence the median is always 1
Suff
C
