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Set S contains seven distinct integers. The median of set S

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

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New post 19 Sep 2010, 12:37
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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 ?

A. m
B. 10m/7
C. 10m/7 – 9/7
D. 5m/7 + 3/7
E. 5m
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Re: Highest possible all values in set S  [#permalink]

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New post 19 Sep 2010, 12:48
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Orange08 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 ?
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: \(m-3\), \(m-2\), \(m-1\), \(median=m\), \(2m-2\), \(2m-1\), \(2m\).

\(Mean=\frac{(m-3)+(m-2)+(m-1)+m+(2m-2)+(2m-1)+2m}{7}=\frac{10m-9}{7}\).

Answer: C.
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Re: Highest possible all values in set S  [#permalink]

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New post 19 Sep 2010, 12:52
Seven distinct integers, the highest value is equal to twice the median. To achieve the highest possible mean you need the following seven integers:

Set S = \({2m, 2m - 1, 2m - 2, m, m - 1, m - 2, m - 3}\)

\(10m-\frac{9}{7}\)= \(\frac{10m}{7} - \frac{9}{7}\)

Answer is C
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Re: Highest possible all values in set S  [#permalink]

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New post 19 Sep 2010, 15:28
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I had missed reading distinct and made the set as m,m,m,m,2m,2m,2m and came with option B.

Phuf...why they keep such answer choices :(
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 05 Feb 2014, 07:21
To maximize the average we need to maximize all the distinct integers in the set.

the numbers then will be

m-3, m-2, m-1,m,2m-2,2m-1, 2m average is sum of these divided by 7.

\(\frac{(10m-9)}{7} = \frac{10m}{7}- \frac{9}{7}\)

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

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New post 12 Mar 2015, 03:30
Orange08 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 ?

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


\(Mean=\frac{(m-3)+(m-2)+(m-1)+(m)+(2m-2)+(2m-1)+(2m)}{7} = \frac{10m-9}{7}\)
Answer :C
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 13 Mar 2015, 03:27
Orange08 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 ?

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



just assume m = 7 (median)

now since each term is distinct and max is 2m, to achieve max sum ... terms should be 4,5,6,7,12,13,14
mean = 61/7

by options, C is satisfied. so C.
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 24 Feb 2016, 09:42
looks like picking numbers won't help and one should solve with algebra
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 24 Feb 2016, 10:27
Hi, i got the right answer but i am a bit concern. The question says all all values in set S are equal to or less than 2m. So is that the maximum value should be 2m-1 because it obviouly says the value is less then 2m. Plz help me. I put 2m-1 for the first time and it is not in the showed answer, so i have to take 2m as a maximum value. Thanks
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 14 Mar 2016, 06:30
one word that change the whole game . . "distinct"
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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New post 18 Apr 2017, 16:39
Orange08 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 ?

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


Since there are 7 distinct integers, there are 3 integers below the median and 3 above. Furthermore, since we want the largest possible average of these integers, we want the integers to be as large as possible. Thus we can let the largest integer be 2m, the second largest (2m - 1), and the third largest (2m - 2). The fourth largest is the median, so it must be m. The fifth, sixth, and seventh (or the smallest) integers will be (m - 1), (m - 2), and (m - 3), respectively. Thus, the largest possible average is:

[2m + (2m - 1) + (2m - 2) + m + (m - 1) + (m - 2) + (m - 3)]/7

(10m - 9)/7

10m/7 - 9/7

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

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New post 18 Apr 2017, 16:48
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Orange08 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 ?

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


One approach is to TEST a value of m.

Let's say m = 5.
So, when we arrange all 7 values in ASCENDING order, 5 is the MEDIAN: _ _ _ 5 _ _ _
Since all values in set S are equal to or less than 2m, the BIGGEST value is 10.
So, we get: _ _ _ 5 _ _ 10
At this point, we are tying to MAXIMIZE the other values AND make sure all are DISTINCT.
So, we get: 2, 3, 4, 5, 8, 9, 10
The average = (2 + 3 + 4 + 5 + 8 + 9 + 10)/7 = 41/7

Now plug m = 5 into the answer choices to see which one yields an average of 41/7

A) 5 NOPE
B) 10m/7. So, we get: 10(5)/7 = 50/7 NOPE
C) 10m/7 – 9/7. So, we get: 10(5)/7 - 9/7 = 41/7 BINGO!!
D) 5m/7 + 3/7. So, we get: 5(5)/7 + 3/7 = 28/7 NOPE
E) 5m. So, we get: 5(5) = 25 NOPE

Answer:

Cheers,
Brent
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Re: Set S contains seven distinct integers. The median of set S  [#permalink]

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