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x, y, and z are consecutive integers, and x < y < z.

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Manager
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Kudos [?]: 88 [0], given: 2

x, y, and z are consecutive integers, and x < y < z. [#permalink] New post 25 Jan 2010, 19:38
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D
E

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50% (01:46) correct 50% (00:00) wrong based on 4 sessions
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, 13:54, 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] New post 26 Jan 2010, 00:34
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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] New post 26 Jan 2010, 06:22
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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Re: x, y, and z are consecutive integers, and x < y < z.   [#permalink] 26 Jan 2010, 06:22
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x, y, and z are consecutive integers, and x < y < z.

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