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highest possible average

by jain2016 » Sat Mar 26, 2016 9:01 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

OAC

Hi Experts ,

please explain.

Thanks,

SJ
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by DavidG@VeritasPrep » Sat Mar 26, 2016 9:09 am
jain2016 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

OAC

Hi Experts ,

please explain.

Thanks,

SJ
To start, we've got 7 distinct integers, 2m is the largest, and m is the median. So we've got: ___,___,___,m,___,___, 2m

If the terms are all integers and distinct, and we want to maximize the average of the set, the largest the terms between 2m and m can be are 2m -1 and 2m-2. And the largest the terms less than m can be are m -1, m -2, and m-3. Now we have the following set:

m -3, m -2, m-1, m, 2m -2, 2m-1, 2m. Let's find the average

Sum: 10m - 9
Average: (10m - 9)/7
Answer: C
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by [email protected] » Sat Mar 26, 2016 9:21 am
Hi jain2016,

Here's a discussion of the various Tactics and 'math' that you can use to answer this question:

https://www.beatthegmat.com/integer-m-t280595.html

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by Brent@GMATPrepNow » Sat Mar 26, 2016 9:22 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
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 jain2016 » Sat Mar 26, 2016 10:59 pm
So, we get: 2, 3, 4, 5, 8, 9, 10
The average = (2 + 3 + 4 + 5 + 8 + 9 + 10)/7 = 41/7

Hi Brent ,

Thanks for your reply.

Just a quick question. Is there any specific reason to take 2,3,4,7,8 these as integers. Can we also take other integers value other than the above?

Please explain.

Many thanks for your reply.

SJ
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by Brent@GMATPrepNow » Sun Mar 27, 2016 6:19 am
jain2016 wrote:
So, we get: 2, 3, 4, 5, 8, 9, 10
The average = (2 + 3 + 4 + 5 + 8 + 9 + 10)/7 = 41/7

Hi Brent ,

Thanks for your reply.

Just a quick question. Is there any specific reason to take 2,3,4,7,8 these as integers. Can we also take other integers value other than the above?

Please explain.

Many thanks for your reply.

SJ
Hi SJ,

There are other possible sets that meet the given conditions.
For example, let's see what happens if we let m = 8.
So, when we arrange all 7 values in ASCENDING order, 8 is the MEDIAN: _ _ _ 8 _ _ _
Since all values in set S are equal to or less than 2m, the BIGGEST possible value is 16.
So, we get: _ _ _ 5 _ _ 16

At this point, we are tying to MAXIMIZE the other values AND make sure all are DISTINCT.
So, we get: 5, 6, 7, 8, 14, 15, 16
The average = (5 + 6 + 7 + 8 + 14 + 15 + 16)/7 = 71/7

Now plug m = 8 into the answer choices to see which one yields an average of 71/7

.
.
.
C) 10m/7 - 9/7. So, we get: 10(8)/7 - 9/7 = 71/7 BINGO!!

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by jain2016 » Mon Mar 28, 2016 3:25 am
So, we get: _ _ _ 5 _ _ 16
I think there should be 8 instead of 5 right?

At this point, we are tying to MAXIMIZE the other values AND make sure all are DISTINCT.
So, we get: 5, 6, 7, 8, 14, 15, 16
The average = (5 + 6 + 7 + 8 + 14 + 15 + 16)/7 = 71/7
What I understood is that if I can fulfil the given conditions, then rest no. will be any distinct right?

For e.g. m = 10 , Since all values in set S are equal to or less than 2m, the BIGGEST value is 20

so I can pick rest no. will be any distinct right?

Please advise and correct.

Many thanks in advance.

SJ
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by GMATGuruNY » Mon Mar 28, 2016 3:38 am
jain2016 wrote:What I understood is that if I can fulfil the given conditions, then rest no. will be any distinct right?

For e.g. m = 10 , Since all values in set S are equal to or less than 2m, the BIGGEST value is 20

so I can pick rest no. will be any distinct right?
To maximize the average, we must make each of the 7 distinct integers AS BIG AS POSSIBLE.

Here is my solution:

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 jain2016 » Mon Mar 28, 2016 7:15 am
Hi ,

Thank you so much sir all clear.

Thanks,

SJ
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