Set S contains seven distinct integers

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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

How will i solve this kind of problem?

OA C

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by GMATGuruNY » Tue Nov 14, 2017 3:51 am
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
Let the 7 distinct integers be a, b, c, m, d, e, and f, such that a<b<c<m<d<e<f.
Let m = 10.
To MAXIMIZE the average, we must maximize the values of a, b, c, d, e and f.
The greatest possible values for a, b and c are 7, 8 and 9.
Since none of the integers can be greater than 2m=20, the greatest possible values for d, e and f are 18, 19, and 20.
Thus, the greatest possible average = (7+8+9+10+18+19+20)/7 = 91/7. This is our target.

Now plug m=10 into the answers to see which yields our target of 91/7.
Only C works:
10m/7 - 9/7 = (10*10)/7 - 9/7 = 91/7.

The correct answer is C.
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by Brent@GMATPrepNow » Tue Nov 14, 2017 5:55 am
lheiannie07 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

How will i solve this kind of problem?

OA C
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: C

Cheers,
Brent
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by Scott@TargetTestPrep » Wed Dec 13, 2017 1:08 pm
lheiannie07 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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by BTGmoderatorDC » Wed Jan 10, 2018 9:19 pm
Scott@TargetTestPrep wrote:
lheiannie07 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
Thanks a lot!

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by BTGmoderatorDC » Wed Jan 10, 2018 9:19 pm
Brent@GMATPrepNow wrote:
lheiannie07 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

How will i solve this kind of problem?

OA C
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: C

Cheers,
Brent
Thanks a lot!