The average (arithmetic mean) of 6 distinct positive integers is 33. T

Question

The average (arithmetic mean) of 6 distinct positive integers is 33. The median of the three largest numbers is 43. What is the difference between the highest and lowest possible median of the 6 numbers?

A. 37
B. 38
C. 39
D. 40
E. 41

A. 37
B. 38
C. 39
D. 40
E. 41

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GMATinsight · Answer

The detailed step-by-step Video solution of the problem is attached here.

Answer: Option B

https://www.youtube.com/watch?v=Ic-4Icc8pv0

lacktutor · Answer

The average (arithmetic mean) of 6 distinct positive integers is 33. The median of the three largest numbers is 43. What is the difference between the highest and lowest possible median of the 6 numbers?

\(a_1+a_2+a_3+a_4+a_5+a_6= 33*6= 198\)
Let's say that \(a_4\),\(a_5\) and \(a_6\) are three largest numbers.
--> \(a_4\) < \(a_5\) < \(a_6\) and \(a_5 = 43\)

The lowest possible median of the 6 numbers:
--> {\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\), \(a_6\)} = {1, 2, 3, 4, 43, 145} (1+2+3+4+43+145= 198)
--> \(\frac{3+4}{2} = 3.5\)

The highest possible median of the 6 numbers:
--> {\(a_1\), \(a_2\), \(a_3\), \(a_4\), \(a_5\), \(a_6\)}= {1, 2, 3, 42, 43, 107} (1+2+3+42+43+107= 198)
--> \(\frac{42+43}{2} = 42.5\)

The difference between them:
\(42.5 - 3.5 = 38\)

Answer (B).

freedom128 · Answer

The lowest possible median of the 6 distinct positive integers is when we have following sequence: 1,2,3,4,43,x 
--> median = (3+4)/2 = 3.5

The highest possible median of the 6 distinct positive integers is when we have following sequence: x1,x2,41,42,43,x
--> median = (41+42)/2 = 41.5

the difference between the highest and lowest possible median of the 6 numbers is 41.5-3.5 = 38

FINAL ANSWER IS (B)

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hiranmay · Answer

The average (arithmetic mean) of 6 distinct positive integers is 33. The median of the three largest numbers is 43. What is the difference between the highest and lowest possible median of the 6 numbers?

A. 37
B. 38
C. 39
D. 40
E. 41

A. 37
B. 38 -->   correct : max(x3+x4)/2 = (41+42)/2 & min(x3+x4)/2 = (3+4)/2, so the difference between the highest and lowest possible median of the 6 numbers = (41+42)/2 - (3+4)/2 = 76/2=38 
C. 39
D. 40
E. 41

fireagablast · Answer

33*6 = 198 --> but this isn't actually important

a-b-c-d-43-e

min when d is lowest
1-2-3-4-43-MAX E --> 4

Max when D highest
1-2-3-42-43-minE --> 42

42-4=38

Kinshook · Answer

Given: 
1. The average (arithmetic mean) of 6 distinct positive integers is 33. 
2. The median of the three largest numbers is 43.

Asked: What is the difference between the highest and lowest possible median of the 6 numbers?

Let the numbers be a, b, c, d, e, f where a<b<c<d<e<f and all are positive integers

1. The average (arithmetic mean) of 6 distinct positive integers is 33. 
a + b + c + d + e + f = 33*6 = 198 (1)

2. The median of the three largest numbers is 43. 
e = 43

Q. Max (c + d)/2 - Min (c + d)/2 = ?

Max (c+d) = 41 + 42 = 83
Max (c+d)/2 = 83/2 = 41.5

For min (c+d)
1,2,3,4,43,f
Min (c+d) = 3 + 4 = 7
Min (c+d)/2 = 7/2 = 3.5

Max (c + d)/2 - Min (c + d)/2 = 41.5 - 3.5 = 38

IMO B

Kinshook · Answer

A link for solution to the above problem on youtube is provided below: - https://youtu.be/dFlBiXLUtAo

Kinshook · Answer

Hi  lacktutor

Please correct the highlighted portion.

lacktutor · Answer

Thank you,  Kinshook .

Posted from my mobile device

chetan2u · Answer

Let the number in ascending order be - A, B, C, D, E, F, where E is 43.
Now average 33 and 2nd largest being 43, gives you ample of room to play around with numbers ..

1) Maximum value.. 
C, D are maximum...so A, B, 41, 42, 43, E..
We can play around with A, B and E to get the average as 33..
Median = (41+42)/2=41.5

2) Minimum value... 
Pretty straight forward....
Least value of A, B, C and D...1, 2, 3, 4, 43, E
You can find E to get the average as 33
Median = (3+4)/2=3.5

Ans - 41.5-3.5=38

B

KarishmaB · Answer

Given: __ __ __ __ 43 __ (all distinct positive integers) Maximise median - We need to maximise 3rd and 4th terms so that their avg is maximum possible. __ __ 41 42 43 44 Avg needs to be 33 so each number effectively should be 33. This brings in an excess of 8 + 9 + 10 + 11 = + 38. Can we set it off by making the two smallest number a total of 38 less than 33? Sure. The two smallest numbers can be 3, 25 or 10, 18 etc. This can be done in many ways. Maximum median = 41.5 Minimise Median - Minimise 3rd and 4th terms 1 2 3 4 43 __ The last number can take as large a value as we need to make the average 33. It will make up for all shortfall. So minimum median is 3.5 Diff = 41.5 - 3.5 = 38 Answer (B) To learn about excess and shortfall, check here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/0 ... eviations/

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The average (arithmetic mean) of 6 distinct positive integers is 33. T

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