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If 2, x, y, and z are different positive integers whose

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Manager
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If 2, x, y, and z are different positive integers whose [#permalink]

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New post 09 Oct 2008, 05:35
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If 2, x, y, and z are different positive integers whose average (arithmetic mean) is 10, what is the greatest possible value of z ?

A. 10
B. 24
C. 34
D. 36
E. 40

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Re: math--positive integers [#permalink]

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New post 09 Oct 2008, 05:58
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avg = 10 means sum will be 40.

one of them is 2 , so x+y+z = 38

now all are diff. POSITIVE integers , so z has max value when x and y has lowest value: means x= 1, y = 3 , x=y = 4

38-4= 34

Z = 34
IMO C

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Re: math--positive integers [#permalink]

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New post 09 Oct 2008, 07:24
albany09 wrote:
If 2, x, y, and z are different positive integers whose average (arithmetic mean) is 10, what is the greatest possible value of z ?

A. 10
B. 24
C. 34
D. 36
E. 40

If 35 had been an opion i woulkd have surely ended u[p marking that !!!

tricky picks are always gmat qant questions !!! IMO C here
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Re: math--positive integers [#permalink]

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New post 09 Oct 2008, 10:17
albany09 wrote:
If 2, x, y, and z are different positive integers whose average (arithmetic mean) is 10, what is the greatest possible value of z ?

A. 10
B. 24
C. 34
D. 36
E. 40


average = sum/no

sum = avearge * no = 10*4 = 40

x+y+z = 38, z is max when x+y is min

x,y min possibe is 1,3

thus z = 38-4 = 34

answer is C

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Re: math--positive integers [#permalink]

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New post 09 Oct 2008, 13:23
albany09 wrote:
If 2, x, y, and z are different positive integers whose average (arithmetic mean) is 10, what is the greatest possible value of z ?

A. 10
B. 24
C. 34
D. 36
E. 40


(2 + x + y + z) / 4 = 10
2 + x + y + z = 4 x 10 = 40
x + y + z = 40 - 2 = 38 ---------- (1)

To Max(z), we have to Min(x) and Min(y)
Min(x) = 1
Mix(y) = 3 ---------- because x is not equal to y and y cannot be 2

Therefore,
1 + 3 + z = 38 ---------- From (1)
z = 38 - 1 - 3 = 34

The answer is C.

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Re: math--positive integers   [#permalink] 09 Oct 2008, 13:23
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