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x, y, and z are consecutive integers, and x < y < z. What is the avera

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x, y, and z are consecutive integers, and x < y < z. [#permalink]

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New post 25 Jan 2010, 20:38
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x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11

(2) The average of y and z is 12.5.


[Reveal] Spoiler:
Can we considered prime numbers as an option of consecutive integers? Also, how can stmt2 be sufficient?

OA is D
[Reveal] Spoiler: OA

Last edited by Bunuel on 27 Jul 2015, 14:54, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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Re: x, y, and z are consecutive integers, and x < y < z. [#permalink]

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New post 26 Jan 2010, 01:34
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joyseychow wrote:
x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11

(2) The average of y and z is 12.5.


[spoiler]Can we considered prime numbers as an option of consecutive integers? Also, how can stmt2 be sufficient?

OA is D[/spoiler]


D.

Stat 1: X=11 this is enough to determine Y=12 and Z=13. You can compute the average after knowing all the values. Suff.

Stat 2: Y and Z has an average of 12.5 and they have to be consecutive numbers. There are no other choices other than Y being 12 and Z being 13, since Y<Z. Any other combination of Y,Z yielding average of 12.5 would not be consecutive integers. (e.g. Y=11, Z=14). With that in mind, X must be 11 and hence sufficient.

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Re: x, y, and z are consecutive integers, and x < y < z. [#permalink]

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New post 26 Jan 2010, 07:22
1. clearly sufficient.

2. take 3 numbers

a-d a a+d

avg =( a-d + a + a+d )/3 = 3a/3 = a

hence we need to know a only... d is 1

now (y+z)/2 = 25/2

a+a+d = 25
2a+1 = 25
a =24/2 => 12

Hence sufficient.

So ans is D
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x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11
(2) The average of y and z is 12.5.

[Reveal] Spoiler:
The answer is apparently D (both statements alone are sufficient). But my concern is with statement 1. According to the solution:

(1) SUFFICIENT: This statement tells us that x is 11. This definitively answers the rephrased question “what is the value of x, y, or z?” To illustrate that this sufficiently answers the original question: since x, y and z are consecutive integers, and x is the smallest of the three, then x, y
and z must be 11, 12 and 13, respectively.


But couldn't x y and z be something like 11, 22, and 33?
Are those not consecutive integers as well? Aren't consecutive integers just integers with a fixed interval or exhibit a fixed pattern?


Thanks in advance for any help answering my question.

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 22 Feb 2010, 22:29
st 1) x = 11, y = 12, z = 13
Sufficient
st 2) y=12, z=13
sufficient
D

glender wrote:
But couldn't x y and z be something like 11, 22, and 33?
Are those not consecutive integers as well? Aren't consecutive integers just integers with a fixed interval or exhibit a fixed pattern?

Thanks in advance for any help answering my question.


11,22,33 are an arithmetic series- meaning when ordered, the difference between 2 adjacent numbers will be equal. Here the difference is 11

1,2,3 - are consecutive integers.
2,4,6 - are also consecutive but they are consecutive even integers.

Consecutive numbers - if the difference of adjacent numbers is the minimum required to get to the next number (when ordered) for an arithmetic series then then numbers are consecutive.
For above, minimum value for a interger to get to next interger is 1.
For above, minimum value for a even interger to get to next even interger is 2.



HTH

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 24 Feb 2010, 10:08
glender wrote:
But couldn't x y and z be something like 11, 22, and 33?
Are those not consecutive integers as well? Aren't consecutive integers just integers with a fixed interval or exhibit a fixed pattern?[/b]

Thanks in advance for any help answering my question.


No, they cant because 11,22 and 33 are not consecutive.

Aren't consecutive integers just integers with a fixed interval or exhibit a fixed pattern -- the fixed interval is +1

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 25 Feb 2010, 10:28
When I see consecutive integers I pick something like x,x+1,x+2.
Here y=x+1, z=x+2

Average = x+y+z /3 = x+x+1+x+2 /3 = x+1

Statement 1 : Suff alone. If I know x, I can find the average.
Statement 2 : Is Suff alone. if I know average of y and z that is x+1+x+2 / 2, then I can find x and so the average of x,y,z.

So both statements are alone sufficient to ge the answer.

It is D.

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 08 Oct 2014, 22:51
Bunuel wrote:

Tough and Tricky questions: Statistics



x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11
(2) The average of y and z is 12.5.


D.

1) x=11,y=12,z=13
sufficient.

2) let numbers be a-1,a,a+1
=> (a + (a+1))/2 = 12.5
=> a = 12
=> x=11,y=12,z=13
sufficient.
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Illegitimi non carborundum.

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 02 Nov 2014, 17:54
Hi -- can we not assume that these are negative consecutive integers?

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 03 Nov 2014, 01:39
russ9 wrote:
x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11
(2) The average of y and z is 12.5.

Hi -- can we not assume that these are negative consecutive integers?


The stem says that x, y, and z are consecutive integers, (1) says that x = 11 and (2) says that the average of two consecutive integers, y and z, is 12.5. How can any of them be negative?
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x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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x, y , z are consecutive integers and \(x<y<z\)

average of x, y , z ?

Statement 1

x=11

As \(x<y<z\) and x, y, z are consecutive integers
therefore y=12 and z=13
average = \(\frac{11+12+13}{3} = 12\)


Sufficient

Statement 2

average of y and z = 12.5

therefore y =12 and z =13 (as \(y<z\))

therefore x=11 (as \(x<y<z\) )

average = \(\frac{11+12+13}{3} = 12\)


Sufficient

Both A and B are alone sufficient to answer the question

OA = D

Hope it helps :)

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 09 Dec 2014, 07:48
x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11
(2) The average of y and z is 12.5.


x, y, and z are consecutive integers and x<y<z.
Therefore y is the average of x,y and z.
The statement is sufficient if value of y can be determined.

Statement (1)
x=11
Since y =x +1 or y =12 ......Sufficient......BCE
Statement (2)
Average of y and z is 12.5
y+z = 25 or 2y+3=25 or y= 12....Sufficient....A D

Answer: D

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera [#permalink]

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New post 07 Jun 2017, 23:13
glender wrote:
x, y, and z are consecutive integers, and x < y < z. What is the average of x, y, and z?

(1) x = 11
(2) The average of y and z is 12.5.

[Reveal] Spoiler:
The answer is apparently D (both statements alone are sufficient). But my concern is with statement 1. According to the solution:

(1) SUFFICIENT: This statement tells us that x is 11. This definitively answers the rephrased question “what is the value of x, y, or z?” To illustrate that this sufficiently answers the original question: since x, y and z are consecutive integers, and x is the smallest of the three, then x, y
and z must be 11, 12 and 13, respectively.


But couldn't x y and z be something like 11, 22, and 33?
Are those not consecutive integers as well? Aren't consecutive integers just integers with a fixed interval or exhibit a fixed pattern?


Thanks in advance for any help answering my question.


Consecutive in this context means monotonically increasing by 1 ( 11, 12, 13, 14 ,15) perhaps outside the context of the GMAT this argument could be made but strictly for the GMAT consecutive implies increasing by 1.

Statement 1

If x is 11 then y is 12 and z is 13

Sufficient
Statement 2

12.5 = (y +z)/2
25 = (y + z) - do not forget the restriction in the problem, because if y and z can only be consecutive integers then the only possibilities within the GMAT's definition of consecutive would be 12 and 13

Sufficient

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Re: x, y, and z are consecutive integers, and x < y < z. What is the avera   [#permalink] 07 Jun 2017, 23:13
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