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integer m

by j_shreyans » Wed Oct 29, 2014 10:58 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 Brent@GMATPrepNow » Wed Oct 29, 2014 11:39 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

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by Brent@GMATPrepNow » Wed Oct 29, 2014 11:48 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

Cheers,
Brent
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by [email protected] » Wed Oct 29, 2014 11:50 am
Hi j_shreyans,

I'm a big fan of TESTing VALUES for this type of question (as Brent showed in his explanation). This question can also be solved algebraically:

We're told that the 7 numbers in Set S have to be DISTINCT (meaning they have to be different). With a median of M, we have....

_ _ _ M _ _ _

We're told that all values are less than or equal to 2M; we're asked for the HIGEST possible average of the group of numbers. This means that we need to MAXIMIZE every value in the group (while still using DISTINCT values). The values would be:

M-3, M-2, M-1, M, 2M-2, 2M-1, 2M

Adding these 7 values gives us a total of:

10M - 9

The average would be:

(10M - 9)/7 = 10M/7 - 9/7

Final Answer: C

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by YTarhouni » Fri Aug 25, 2017 5:57 pm
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by Matt@VeritasPrep » Wed Aug 30, 2017 5:45 pm
Suppose we have no restrictions. We'd maximize everything:

* Any number on the left side of the median = m
* Any number on the right side of the median = the maximum = 2m

We've got three on each side of the median, giving us:

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

Average = Sum/# = (2m + 2m + 2m + m + m + m + m) / 7 = 10m/7

But we know the integers are distinct, so we've got to make everything just a tiny bit smaller:

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

Same formula: (m - 3 + m - 2 + m - 1 + m + 2m - 2 + 2m - 1 + 2m) / 7 => (10m - 9)/7

and we're set!
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