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The average (arithmetic mean) of the positive integers x, y, and z is

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The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 30 Jul 2018, 21:59
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A
B
C
D
E

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  25% (medium)

Question Stats:

66% (00:54) correct 34% (00:54) wrong based on 526 sessions

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

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New post 30 Jul 2018, 22:02
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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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 30 Jul 2018, 22:06
2

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.

Answer: B
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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 30 Jul 2018, 23:48
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

The best answer is B.
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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 31 Jul 2018, 08:18
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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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 17 Aug 2018, 22:04
Bunuel so zero isn't counted as an integer? This always confuses me.
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Re: The average (arithmetic mean) of the positive integers x, y, and z is  [#permalink]

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New post 17 Aug 2018, 22:44
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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.

Check for more below threads:
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Hope it helps.
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Re: The average (arithmetic mean) of the positive integers x, y, and z is   [#permalink] 17 Aug 2018, 22:44
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