Set S contains seven distinct integers. The median of set S

Seven distinct integers, the highest value is equal to twice the median. To achieve the highest possible mean you need the following seven integers:

Set S = \({2m, 2m - 1, 2m - 2, m, m - 1, m - 2, m - 3}\)

\(10m-\frac{9}{7}\)= \(\frac{10m}{7} - \frac{9}{7}\)

Answer is C

To maximize the average we need to maximize all the distinct integers in the set.

the numbers then will be

m-3, m-2, m-1,m,2m-2,2m-1, 2m average is sum of these divided by 7.

\(\frac{(10m-9)}{7} = \frac{10m}{7}- \frac{9}{7}\)

C.

\(Mean=\frac{(m-3)+(m-2)+(m-1)+(m)+(2m-2)+(2m-1)+(2m)}{7} = \frac{10m-9}{7}\)
Answer :C

just assume m = 7 (median)

now since each term is distinct and max is 2m, to achieve max sum ... terms should be 4,5,6,7,12,13,14
mean = 61/7

by options, C is satisfied. so C.

looks like picking numbers won't help and one should solve with algebra

Hi, i got the right answer but i am a bit concern. The question says all all values in set S are equal to or less than 2m. So is that the maximum value should be 2m-1 because it obviouly says the value is less then 2m. Plz help me. I put 2m-1 for the first time and it is not in the showed answer, so i have to take 2m as a maximum value. Thanks

one word that change the whole game . . "distinct"

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

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:

Cheers,
Brent

I used Picking Numbers to solve this question.

M= 4 Median

2M= 8 All values are equal or less than 8

This way, the 7 different integers in Set S are: 1, 2, 3, 4, 6, 7, 8

AVG = \(\frac{sum of terms}{# of terms}\) = \(\frac{1+2+3+4+6+7+8}{7}\) = \(\frac{31}{7}\)

When we plug M=4 in the answer choices, we have to find the match AVG = \(\frac{31}{7}\)

A) 4 not a match

B) \(\frac{40}{7}\) not a match

C) \(\frac{40}{7}\) - \(\frac{9}{7}\) = \(\frac{31}{7}\) that's a match

D) \(\frac{20}{7}\) + \(\frac{3}{7}\) = \(\frac{23}{7}\) not a match

E) 20 not a match


Hence, answer C is the correct choice.


Hope it helps!

Thanks! Alecita

@Bunel,
To maximise the average, the difference between each number should be 1 starting from the highest number, which is 2m. So, I calculated 2m, 2m-1, 2m-2, 2m-3 ( Which is m), 2m-4, 2m-5, 2m-6 and got a different answer. Can you help me understand where I went wrong?

sequence of the terms would be with median i.e 4th term as m
m-3,m-2,m-1,m,2m-2,2m-1,2m
2m is highest term
sum ; 10m-9
avg ; 10m/7-9/7
option C

Total numbers: 7

Median: 4th number: m

The highest number equal to 2m. Therefore the smallest number will be m - 3

Average: \(\frac{[(m - 3) + (m - 2) + (m - 1) + (m ) + (2m - 1) + (2m - 2) + (2m)] }{ 7}\)

=> Average: \(\frac{[10m - 9] }{7}\)

=> Average: \(\frac{10m }{ 7} - \frac{9}{ 7}\)

Answer C

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