## The average (arithmetic mean) of four distinct positive inte

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### The average (arithmetic mean) of four distinct positive inte

by guerrero » Tue Jun 11, 2013 1:24 pm

00:00

A

B

C

D

E

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The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

12
14
15
16
17

OAB

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by Brent@GMATPrepNow » Tue Jun 11, 2013 1:40 pm
guerrero wrote:The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

12
14
15
16
17

OAB
Let the 4 different positive integers be A, B, C, and D such that A < B < C < D

The average (arithmetic mean) of four distinct positive integers is 10
So, (A+B+C+D)/4 = 10
This means A+B+C+D = 40

The average of the smaller two of these four integers is 8
So, the average of A and B is 8
In other words, (A+B)/2 = 8
So, A+B = 16

Since we already know that A+B+C+D = 40, we can replace A+B with 16 to get:
16+C+D = 40
So, C + D = 24

We want to maximize the value of D. To do this, we need to minimize the value of C.
Also, since B < C, we want to minimize the value of B.

Since A+B = 16, the smallest possible value of B is 9.
So, we get A = 7
B = 9
So, C = 10 is the smallest we can make C
This means that D = [spoiler]14[/spoiler]

Cheers,
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by GMATGuruNY » Wed Jun 12, 2013 2:58 am
guerrero wrote:The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

12
14
15
16
17

OAB
Sum = number * average.

Since the average of the 4 integers is 10, their sum = 4*10 = 40.

In ascending order, let the 4 integers = A+B+C+D.
Since A+B+C+D = 40, we get:
D = 40 - (A+B+C).
To MAXIMIZE the value of D, we must MINIMIZE the value of A+B+C.

To minimize the value of C, we must minimize the value of B.
Since the average of A and B is 8, their sum = 2*8 = 16.
Thus, the least option B is 9:
A+B = 7+9.
Thus, the least option for C is 10.

Thus:
Greatest possible value for D = 40 - (A+B+C) = 40 - (7+9+10) = 14.

The correct answer is B.
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by faraz_jeddah » Thu Jun 13, 2013 12:17 am
Brent@GMATPrepNow wrote:
guerrero wrote:The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

12
14
15
16
17

OAB
Let the 4 different positive integers be A, B, C, and D such that A < B < C < D

The average (arithmetic mean) of four distinct positive integers is 10
So, (A+B+C+D)/4 = 10
This means A+B+C+D = 40

The average of the smaller two of these four integers is 8
So, the average of A and B is 8
In other words, (A+B)/2 = 8
So, A+B = 16

Since we already know that A+B+C+D = 40, we can replace A+B with 16 to get:
16+C+D = 40
So, C + D = 24

We want to maximize the value of D. To do this, we need to minimize the value of C.
Also, since B < C, we want to minimize the value of B.

Since A+B = 16, the smallest possible value of B is 9.
So, we get A = 7
B = 9
So, C = 10 is the smallest we can make C
This means that D = [spoiler]14[/spoiler]

Cheers,
Brent
Brent how can we assume A < B < C < D
The question does not tell us that they are arranged in ascending order.

The 4 numbers could be - 1 15 7 17 which would make 17 the max value.

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by Brent@GMATPrepNow » Thu Jun 13, 2013 4:32 am
faraz_jeddah wrote: Brent how can we assume A < B < C < D
The question does not tell us that they are arranged in ascending order.

The 4 numbers could be - 1 15 7 17 which would make 17 the max value.
We're told that the 4 integers are distinct, which means no two values are the same.
So, one of them will be the smallest, one of them will be the second smallest, one of them will be the second biggest, one of them will be the biggest. To make things easier to discuss, I named these values A, B, C and D.

The example you give, {1 15 7 17} does not meet the condition that "If the average of the smaller two of these four integers is 8." In your example, the two smallest integers are 1 and 7, so their average is 4.

Cheers,
Brent
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by faraz_jeddah » Thu Jun 13, 2013 10:20 am
Thanks.

I over looked that segment.

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### Re: The average (arithmetic mean) of four distinct positive inte

by Scott@TargetTestPrep » Mon Jun 29, 2020 8:28 am
guerrero wrote:
Tue Jun 11, 2013 1:24 pm
The average (arithmetic mean) of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum possible value of the largest integer?

12
14
15
16
17

OAB
Solution:

Since the sum of the four distinct integers is 10 x 4 = 40 and the sum of the two smallest integers is 8 x 2 = 16, the sum of the largest two integers is 40 - 16 = 24. Since we want the maximum possible value of the largest integer, we can let the second largest integer be as small as possible. The second integer can’t be 9; otherwise, the sum of the two smallest integers would be at most 7 + 8 = 15 (recall that all the integers are distinct). However, if the second smallest integer is 10 (and the two smallest integers are 7 and 9), the largest integer will then be 24 - 10 = 14.