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# M03 Q33

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M03 Q33 [#permalink]

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02 Nov 2008, 19:03
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Are integers $$x$$, $$y$$, and $$z$$ consecutive?

1. $$y$$ equals the arithmetic mean of $$x$$ and $$z$$
2. $$x = -z$$

[Reveal] Spoiler: OA
E

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I don't understand the answer...
shouldn't 1. be enough?

if y = x+z / 2
if they were consecutive, y should be (x+1).
z would then be x+2.
if x + (x+2) / 2, you do get y = (X+1)

so why isn't 1. enough?
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Re: M03 Q33 [#permalink]

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05 Oct 2010, 06:08
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(1) y = average of x and z:
x(1), y(2), z(3) --> y is avrg of x & z and x,y,z are consecutives
x(2), y(2), z(2) --> y is avrg of x & z but x,y,z are not consecutives
INSUFFICIENT

(1) x= -z says nothing about y...
INSUFFICIENT

(1) and (2):
x(-1), y(0), z(1)... y is avrg of x and z and x,y,z are consecutives
it might be tempting to pick C without trying another set of nos.
x(-2), y(0), z(2)... y is avrg of x and z but x,y,z are not consecutives
INSUFFICIENT

So, E.
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Re: M03 Q33 [#permalink]

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02 Nov 2009, 01:34
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2, 4, 6 would be consecutive even integers; 4, 8, 12 would be consecutive multiples of 4. Consecutive integers are 1, 2, 3, 4 and so on unless otherwise stated.
hogann wrote:
Is this the correct definition of consecutive integers?

To me consecutive integers could be 1,2,3 but also 2,4,6 or 4,8,12
Is this called something else? Consecutive series?

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Re: M03 Q33 [#permalink]

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05 Oct 2010, 17:58
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Statement 1: y = (x+z)/2. y is the middle term but we do not know whether the difference between x, y and/or y,z is 1.
Insufficient.

Statement 2: x = -z. This tells us that y = 0, however we do not know whether about the difference between x,y and y,z.

Insufficient.

Combining both: we know that y is the middle term and the mean of x and z but still we cannot determine that the difference between x,y and y,z is 1 (condition for x, y, z to be consecutive.)

Answer E.
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Re: M03 Q33 [#permalink]

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03 Nov 2009, 05:51
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Nice one. Thanks for clarifying the term “consecutive XXXX”
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Re: M03 Q33 [#permalink]

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03 Nov 2009, 21:13
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x, y and z are integers. Are they consecutive?

1. y equals to the arithmetic mean of x and z
2. x = -z

So is the correct answer E?

Taking statement 1, y = (x+z)/2 Not sufficient.

Taking statement 2, if x is 1, then the x, y, z could be -1, 0, 1. However, they could also be -2, 0, 2. Not sufficient.

Combining the statements and substituting, y = (-z+z)/2, which simplifies to y = 0/2 or 0. If y is 0 and x = -z, we still do not have enough information to answer the question.
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Re: M03 Q33 [#permalink]

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07 Oct 2011, 07:04
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Its E.
St1: Every AP has the same property b = (a+c)/2 . So insufficient.
St2: There are infinite series with a = -c eg -5,0,5 ; -2,0,2 So insufficient.

Both statements together doesn't narrow it down. Hence E.
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CR notes
http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142
http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133
http://gmatclub.com/forum/gmat-prep-critical-reasoning-collection-106783.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html
http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html?hilit=chineseburned

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Re: M03 Q33 [#permalink]

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09 Oct 2012, 04:07
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REVISED VERSION OF THIS QUESTION IS BELOW:

Are integers $$x$$, $$y$$, and $$z$$ consecutive?

(1) The average (arithmetic mean) of $$x$$ and $$z$$ equals to $$y$$ --> $$y=\frac{x+z}{2}$$: if $$x=y=z=0$$ then the answer is NO but if $$x=-1$$, $$y=0$$ and $$z=1$$ then the answer is YES. Not sufficient.

(2) $$x= -z$$. Not sufficient since no info about $$y$$.

(1)+(2) Examples discussed for statement (1) are still valid, so we can have a NO ($$x=y=z=0$$) as well as an YES ($$x=-1$$, $$y=0$$ and $$z=1$$) answers. Not sufficient.

Answer: E.
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Re: M03 Q33 [#permalink]

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03 Nov 2008, 02:23
Take the example of y = x+5 and z = x+10. In this case also, y = (x+z)/2, but x,y and z are not consecutive.
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Re: M03 Q33 [#permalink]

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29 Oct 2009, 11:17
Is this the correct definition of consecutive integers?

To me consecutive integers could be 1,2,3 but also 2,4,6 or 4,8,12
Is this called something else? Consecutive series?
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Re: M03 Q33 [#permalink]

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04 Nov 2009, 01:05
You're absolutely right. The OA is E.

phillypointgod21 wrote:
x, y and z are integers. Are they consecutive?

1. y equals to the arithmetic mean of x and z
2. x = -z

So is the correct answer E?

Taking statement 1, y = (x+z)/2 Not sufficient.

Taking statement 2, if x is 1, then the x, y, z could be -1, 0, 1. However, they could also be -2, 0, 2. Not sufficient.

Combining the statements and substituting, y = (-z+z)/2, which simplifies to y = 0/2 or 0. If y is 0 and x = -z, we still do not have enough information to answer the question.

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Re: M03 Q33 [#permalink]

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07 Oct 2010, 02:51
Agree with E for sure.

State 1: y = A.M of X & Y

We get inconclusive answers as if X=3 , Y=4, Z=5
Then Y=(3+5)/2=4 = A.M is true ans consecutive no. but say if X=2, Y=5, Z=8
then also Y=A.M =5 but the number are no consecutive so insufiecient

Answers could be B, C or e

Stat II: X=-Z gives us nothing so insuffiecient .

Taking them together:

If z=2 then X=-2 then y=0 not consecutive
Watch out here as all values except 1 the no will come out to be not consecutive.
Except 1 If z=1 then x=-1 Y=0 consecutive no. so we have conflicting answers so correct Answer is [E]

Hope this helps
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Re: M03 Q33 [#permalink]

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14 Oct 2010, 07:10
gmatbull wrote:
(1) y = average of x and z:
x(1), y(2), z(3) --> y is avrg of x & z and x,y,z are consecutives
x(2), y(2), z(2) --> y is avrg of x & z but x,y,z are not consecutives
INSUFFICIENT

(1) x= -z says nothing about y...
INSUFFICIENT

(1) and (2):
x(-1), y(0), z(1)... y is avrg of x and z and x,y,z are consecutives
it might be tempting to pick C without trying another set of nos.
x(-2), y(0), z(2)... y is avrg of x and z but x,y,z are not consecutives
INSUFFICIENT

So, E.

Thanks. To me, you did the best job of explaining why the answer isn't "A". Just because all consecutive integers will work for X+Z/2=Y doesn't mean that ONLY consecutive integers will do that.
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Re: M03 Q33 [#permalink]

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07 Oct 2011, 12:33
Great question. Made me make a mistake. I simply took Z=1 and X=-1 according to stmt 2. Then according to stmt 1, y must be 0 here. So the three numbers comprise a consecutive integer -1,0,1. But haha what about -2, 0 , 2 which is supported by the statements? It's no a consecutive integer series. So E
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Re: M03 Q33 [#permalink]

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08 Oct 2011, 16:17
I picked E

Stm.1- y=(x+z)/2
2y=x+z
3y=x+y+z
y=(x+y+z)/3 ...this statement only means that the mean=median... consecutive and nonconsecutive integers satisfy this condition...therefore not Sufficient.

Stm.2- No data about y...therefore insufficient.

Stm1 and stm2 together (x,y,z)=(-3,0,3) or (-1,0,1) therefore not sufficient.
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Re: M03 Q33 [#permalink]

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09 Oct 2012, 07:16
Are integers $$x$$, $$y$$, and $$z$$ consecutive?

1. $$y$$ equals the arithmetic mean of $$x$$ and $$z$$
2. $$x = -z$$

Nice workout !

(i) take 5,6,7
6 is the AM of 5&7 yes Consecutive
take 10,12,14
12 is the AM of 10&14 Not consecutive

insufficient

(ii)take -1,0,1 as x,y & z- possible because x=(-z)
but if we take z=-5, x will be 5 , then it will not be consecutive-not possible

insufficient

Answer-E
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Re: M03 Q33   [#permalink] 09 Oct 2012, 07:16
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