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by j_shreyans » Wed Oct 29, 2014 9:11 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

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by GMATGuruNY » Wed Oct 29, 2014 10:14 am
j_shreyans 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
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 » Wed Oct 29, 2014 11:51 am
j_shreyans 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
Here's an algebraic approach:

When we arrange all 7 values in ASCENDING order, with m as the MEDIAN, we get: _ _ _ m _ _ _
All values in set S are equal to or less than 2m
Since we are trying to MAXIMIZE the average, we'll take 2m as a value in set S
So, we get: _ _ _ m _ _ 2m
At this point, we are tying to MAXIMIZE the other values AND make sure all are DISTINCT.
The 2nd biggest value is 2m - 1. So, we get: _ _ _ m _ 2m-1, 2m
The 3rd biggest value is 2m - 2. So, we get: _ _ _ m, 2m-2, 2m-1, 2m
The remaining values must be less than m.
When MAXIMIZING these values, we get: m-3, m-2, m-1, m, 2m-2, 2m-1, 2m

The average = [(m-3)+(m-2)+(m-1)+(m)+(2m-2)+(2m-1)+(2m)]/7 =
= (10m - 9)/7
= [spoiler]10m/7 - 9/7[/spoiler]

Answer: C

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by Brent@GMATPrepNow » Wed Oct 29, 2014 11:51 am
j_shreyans 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: C

Cheers,
Brent
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