Set T contains more than one element. Is the median of set T greater than its mean?
(1) Set S has positive range.
(2) The elements of the set are not consecutive integers
Mean Vs Median
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- karthikpandian19
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(1) Set S has positive range.karthikpandian19 wrote:Set T contains more than one element. Is the median of set T greater than its mean?
(1) Set S has positive range.
(2) The elements of the set are not consecutive integers
Clearly insufficient
(2) The elements of the set are not consecutive integers
Mean and median for {2,4,6,8,10} are the same
Mean and media for {1.7.10.103} are not the same
insufficient
Combining two statements
Still insufficient
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(1) Set T has a positive range means T can have any range of numbers from 1 till infinity; NOT sufficient.karthikpandian19 wrote:Set T contains more than one element. Is the median of set T greater than its mean?
(1) Set S has positive range.
(2) The elements of the set are not consecutive integers
(2) The elements of the set are not consecutive integers implies it could be any range of numbers from 1 till infinity; NOT sufficient.
Combining (1) and (2), again is NOT sufficient to find if the median of set T is greater than its mean.
The correct answer is E.
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Let's give a simpler explanation
Remember 1 thing: If the median is closer to the smaller value/values mean will be greater than the median and median is closer to the bigger values median will be greater than
Statement 1:
Range is +ive
S= {0,5}--mean = median
S= {0,1,5}--median<mean
S= {0,3,5}--median>mean
So INSUFFICIENT
Statement 2:
Elements are not consecutive
S= {5,5}--mean = median
S= {1,3,6}--mean >median
S= {1,5,7}--median > mean
INSUFFICIENT
Combining also ..INSUFFICIENT
(E) is the answer
Remember 1 thing: If the median is closer to the smaller value/values mean will be greater than the median and median is closer to the bigger values median will be greater than
Statement 1:
Range is +ive
S= {0,5}--mean = median
S= {0,1,5}--median<mean
S= {0,3,5}--median>mean
So INSUFFICIENT
Statement 2:
Elements are not consecutive
S= {5,5}--mean = median
S= {1,3,6}--mean >median
S= {1,5,7}--median > mean
INSUFFICIENT
Combining also ..INSUFFICIENT
(E) is the answer