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Set S contains seven distinct integers. The median of set S is the [#permalink]

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18 Nov 2009, 21:35

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44% (02:01) correct
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Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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18 Nov 2009, 21:51

kairoshan wrote:

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ? m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m

10m/7 -9/7

lets consider m = 7 and set as [4,5,6,7,12,13,14] all distinct and will give highest possible average

1. m is median --> x x x m x x x 2. 2m is the maximum value. x x x m x x 2m 3. because integers are distinct, we should find as large integers as we can under above restrictions:

x x (m-1) m x (2m-1) 1m (m-3) (m-2) (m-1) m (2m-2) (2m-1) 2m Sum = 10m-9 Average = 10/7m-9/7

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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19 Nov 2009, 06:20

walker wrote:

1. m is median --> x x x m x x x 2. 2m is the maximum value. x x x m x x 2m 3. because integers are distinct, we should find as larger integers as we can under above restrictions:

x x (m-1) m x (2m-1) 1m (m-3) (m-2) (m-1) m (2m-2) (2m-1) 2m Sum = 10m+9 Average = 10/7m-9/7

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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27 May 2010, 01:53

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m _________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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29 May 2010, 02:11

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m

mean of 7 numbers = (Sum of 7 numbers)/7 To find the highest mean we need to maximise the numerator.

since m in the median and we've 7 numbers so m will take the 4th position i.e. the Set S must have(for max avg)

m, m, m, m, 2m, 2m, 2m

Max avg mean = (m + m + m + m + 2m + 2m + 2m)/7 = 10m/7

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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30 May 2010, 04:34

OA is C

But why cant the set be

m,m,m,m,2m,2m,2m

it is a set and m can be the median, and average will be more than

m-3,m-2,m-1,m,2m-2,2m-1,2m

Amiman wrote:

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m

mean of 7 numbers = (Sum of 7 numbers)/7 To find the highest mean we need to maximise the numerator.

since m in the median and we've 7 numbers so m will take the 4th position i.e. the Set S must have(for max avg)

m, m, m, m, 2m, 2m, 2m

Max avg mean = (m + m + m + m + 2m + 2m + 2m)/7 = 10m/7

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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30 May 2010, 12:44

BlueRobin wrote:

OA is C

But why cant the set be

m,m,m,m,2m,2m,2m

it is a set and m can be the median, and average will be more than

m-3,m-2,m-1,m,2m-2,2m-1,2m

Amiman wrote:

[highlight]Set S contains seven distinct integers[/highlight]. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m

mean of 7 numbers = (Sum of 7 numbers)/7 To find the highest mean we need to maximise the numerator.

since m in the median and we've 7 numbers so m will take the 4th position i.e. the Set S must have(for max avg)

m, m, m, m, 2m, 2m, 2m

Max avg mean = (m + m + m + m + 2m + 2m + 2m)/7 = 10m/7

Note that "Set S contains seven distinct integers".

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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30 May 2010, 13:09

Fistail wrote:

BlueRobin wrote:

OA is C

But why cant the set be

m,m,m,m,2m,2m,2m

it is a set and m can be the median, and average will be more than

m-3,m-2,m-1,m,2m-2,2m-1,2m

Amiman wrote:

[highlight]Set S contains seven distinct integers[/highlight]. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

m 10m/7 10m/7 – 9/7 5m/7 + 3/7 5m

mean of 7 numbers = (Sum of 7 numbers)/7 To find the highest mean we need to maximise the numerator.

since m in the median and we've 7 numbers so m will take the 4th position i.e. the Set S must have(for max avg)

m, m, m, m, 2m, 2m, 2m

Max avg mean = (m + m + m + m + 2m + 2m + 2m)/7 = 10m/7

Note that "Set S contains seven distinct integers".

Yeah thanks for point zzzzzzzzzzzz that out, i am awake now. _________________

Re: Set S contains seven distinct integers. The median of set S is the [#permalink]

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01 Dec 2014, 10:04

walker wrote:

1. m is median --> x x x m x x x 2. 2m is the maximum value. x x x m x x 2m 3. because integers are distinct, we should find as large integers as we can under above restrictions:

x x (m-1) m x (2m-1) 1m (m-3) (m-2) (m-1) m (2m-2) (2m-1) 2m Sum = 10m-9 Average = 10/7m-9/7

So, C

Had the integers not been distinct would it have been like this: m m m m 2m 2m 2m And the median would still have been m trying to clarify an imp concept P.S. I'm a technology handicap so please dont beat your head at my posts' misplacements and redundancies

1. m is median --> x x x m x x x 2. 2m is the maximum value. x x x m x x 2m 3. because integers are distinct, we should find as large integers as we can under above restrictions:

x x (m-1) m x (2m-1) 1m (m-3) (m-2) (m-1) m (2m-2) (2m-1) 2m Sum = 10m-9 Average = 10/7m-9/7

So, C

Had the integers not been distinct would it have been like this: m m m m 2m 2m 2m And the median would still have been m trying to clarify an imp concept P.S. I'm a technology handicap so please dont beat your head at my posts' misplacements and redundancies

Yes, that's correct.

Set S contains seven distinct integers. The median of set S is the integer m, and all values in set S are equal to or less than 2m. What is the highest possible average (arithmetic mean) of all values in set S ?

A. m B. 10m/7 C. 10m/7 – 9/7 D. 5m/7 + 3/7 E. 5m

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order; If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

So median of S, which contains seven terms is 4th term when arranged in ascending order;: \(median=4th \ term=m\).

Now, to maximize the mean we should maximize the terms. As numbers in S are distinct integers and the highest number in S could be equal to \(2m\), then maximum values of the terms would be: \(m-3\), \(m-2\), \(m-1\), \(median=m\), \(2m-2\), \(2m-1\), \(2m\).

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