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Math Expert V
Joined: 02 Sep 2009
Posts: 59725
The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 68% (00:53) correct 32% (00:54) wrong based on 594 sessions

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The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

NEW question from GMAT® Quantitative Review 2019

(PS00986)

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Manager  S
Joined: 01 Jan 2018
Posts: 78
Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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The smallest positive integer is 1 and 2nd smallest 2.
So, 3*3 = 3+x.
Hence, x= 6. B is the correct choice.

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e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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4

Solution

Given:
• The average (arithmetic mean) of the positive integers x, y, and z is 3
• Also, x < y < z

To find:
• The greatest possible value of z

Approach and Working:
As the average is 3,
• Then sum of all 3 = 3 * 3 = 9

Also, x < y < z and each are positive integers.

To maximise z, we should take minimum value of x and y
• Minimum possible x = 1
• Minimum possible y = 2
• Therefore, maximum possible z = 9 – (1 + 2) = 6

Hence, the correct answer is option B.

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Joined: 31 Oct 2013
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Concentration: Accounting, Finance
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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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Bunuel wrote:
The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

NEW question from GMAT® Quantitative Review 2019

(PS00986)

Given

$$\frac{x + y + z}{3} = 3$$

x + y+ z =9

In order to maximize z we must minimize x and y. Note: x<y<z and all are positive integers. So the minimum value we have is 1.

1 + 2 + 6 = 9

Board of Directors D
Status: QA & VA Forum Moderator
Joined: 11 Jun 2011
Posts: 4834
Location: India
GPA: 3.5
Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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Bunuel wrote:
The average (arithmetic mean) of the positive integers x, y, and z is 3. If x < y < z, what is the greatest possible value of z ?

A. 5
B. 6
C. 7
D. 8
E. 9

NEW question from GMAT® Quantitative Review 2019

(PS00986)

$$x + y + z = 9$$

We have to find the max value of z is minimizing the value of x & y

Since, $$x < y < z$$, minimizing x & y will be -

x < y < z = 1 < 2 < 6 where x + y + z = 9, Thus Answer must be (B)
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Manager  B
Joined: 20 Jun 2018
Posts: 72
Location: India
Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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Bunuel so zero isn't counted as an integer? This always confuses me.
Math Expert V
Joined: 02 Sep 2009
Posts: 59725
Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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1
1
Shbm wrote:
Bunuel so zero isn't counted as an integer? This always confuses me.

0 is an integer but it's neither positive nor negative.

ZERO.

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

ALL YOU NEED FOR QUANT ! ! !

Hope it helps.
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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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_________________ Re: The average (arithmetic mean) of the positive integers x, y, and z is   [#permalink] 30 Nov 2019, 07:05
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