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05 Mar 2021, 01:15
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Sets S and T contain an equal number of elements, all of which are positive integers. x is the median of S and y is the average (arithmetic mean) of T. Is x > y ?
(1) The sum of S is greater than the sum of T
(2) S consists of consecutive even numbers and T consists of consecutive odd numbers
(1) The sum of S is greater than the sum of T
(2) S consists of consecutive even numbers and T consists of consecutive odd numbers
08 Mar 2021, 01:30
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Sets S and T contain an equal number of elements, all of which are positive integers. x is the median of S and y is the average (arithmetic mean) of T. Is x > y ?
(1) The sum of S is greater than the sum of T
(2) S consists of consecutive even numbers and T consists of consecutive odd numbers
(1) The sum of S is greater than the sum of T
(2) S consists of consecutive even numbers and T consists of consecutive odd numbers
Sets S and T contain an equal number of elements, all of which are positive integers. x is the median of S and y is the average (arithmetic mean) of T. Is x > y ?
(1) The sum of S is greater than the sum of T.
If S = {2} (median = x = 2) and T = {1} (average = y = 1), then x > y;
If S = {1, 1, 10} (median = x = 1) and T = {1, 2, 3} (average = y = 2), then x < y.
Not sufficient.
(2) S consists of consecutive even numbers and T consists of consecutive odd numbers.
If S = {2, 4} (median = x = 3) and T = {1, 3} (average = y = 2), then x > y;
If S = {2, 4} (median = x = 3) and T = {3, 5} (average = y = 4), then x < y.
Not sufficient.
(1)+(2) In an evenly spaced set (for example, in a set of consecutive even or odd integers, or in a set of any other arithmetic progression) mean = median. So, the mean of S equals to the median of S. So, we need to compare the mean of S to the mean of T. Now, {mean} = {sum of the elements}/{number of elements}.
We also know that sets S and T contain an equal number of elements and the sum of S is greater than the sum of T, hence {sum of the elements in S}/{number of elements} > {sum of the elements in T}/{number of elements}. Sufficient.
Answer: C.
General Discussion
05 Mar 2021, 09:31
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Since the sets have equal numbers of elements, Statement 1 tells us the mean of S is greater than the mean of T. We need to compare the median of S with the mean of T, so Statement 1 is not sufficient. If we want an example where the mean of S is greater than the mean of T, but the median of S is smaller than the mean of T, the sets could be:
S = 1, 2, 3, 4, 1000000000000000
T = 98, 99, 100, 101, 102
Statement 2 is not sufficient because we have no idea how big any of the elements are.
Statement 2 does, however, guarantee that the sets are both 'equally spaced', and in any equally spaced list, the mean and median are equal. So if from Statement 1, S has a greater mean than T, then the median of S must also be greater than the mean of T, and the answer is C.
S = 1, 2, 3, 4, 1000000000000000
T = 98, 99, 100, 101, 102
Statement 2 is not sufficient because we have no idea how big any of the elements are.
Statement 2 does, however, guarantee that the sets are both 'equally spaced', and in any equally spaced list, the mean and median are equal. So if from Statement 1, S has a greater mean than T, then the median of S must also be greater than the mean of T, and the answer is C.
