Problem Solving

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

Cheers,
Brent

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