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Sets S and T contain an equal number of elements, all of

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Sets S and T contain an equal number of elements, all of  [#permalink]

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New post Updated on: 29 Jul 2014, 08:24
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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

Originally posted by goodyear2013 on 29 Jul 2014, 06:53.
Last edited by Bunuel on 29 Jul 2014, 08:24, edited 1 time in total.
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Re: Sets S and T contain an equal number of elements, all of  [#permalink]

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New post 29 Jul 2014, 08:39
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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.

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.
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Re: Sets S and T contain an equal number of elements, all of  [#permalink]

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New post 16 Mar 2018, 20:13
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Re: Sets S and T contain an equal number of elements, all of &nbs [#permalink] 16 Mar 2018, 20:13
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