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If a, b, and c are consecutive integers such that a < b

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If a, b, and c are consecutive integers such that a < b [#permalink] New post 24 Apr 2010, 07:19
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A
B
C
D
E

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10% (02:15) correct 90% (00:48) wrong based on 10 sessions
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2
II. b = a + 1 = c – 1
III. abc is an even integer


A I only
B II only
C III only
D I and III only
E I, II, and III

Please explain with reasons?
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Re: Seems easy, but is it really??? [#permalink] New post 24 Apr 2010, 07:29
hardnstrong wrote:
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2 -> a, b, c are consecutive integers and c > a. So in any case this is true.
II. b = a + 1 = c – 1 -> same as above. a, b, c are consecutive Integers and b is between a and c.
III. abc is an even integer -> 0 is an even integer. Also, multiplication of odd and even is even. So in any case this would be true.


A I only
B II only
C III only
D I and III only
E I, II, and III

Please explain with reasons?


IMO E. Please feel free to correct if my explainations are wrong.
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Re: Seems easy, but is it really??? [#permalink] New post 24 Apr 2010, 07:31
hardnstrong wrote:
If a, b, and c are consecutive integers such that a < b < c, then which of the following must be true?
I. c – a = 2
II. b = a + 1 = c – 1
III. abc is an even integer

A I only
B II only
C III only
D I and III only
E I, II, and III
Please explain with reasons?

given
b= a+1 & c = a+2 & c=b+1
I is true
II is true
III: atleast one of a,b,c is even and odd * even is always even hence it is also true.
so its E.
Whats the OA
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Re: Seems easy, but is it really??? [#permalink] New post 24 Apr 2010, 07:35
Yes E only.

Take a = n-1 . b = n and c = n+1

substitute in all the equations, you will see its satisfying all of them.
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Re: Seems easy, but is it really??? [#permalink] New post 24 Apr 2010, 21:59
thanks guys ...........i got confused with the third option
Got it now
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Re: Seems easy, but is it really??? [#permalink] New post 15 Jun 2010, 15:51
what if one of is zero, then it can not satisfy the third option.
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Re: Seems easy, but is it really??? [#permalink] New post 15 Jun 2010, 16:08
what if one of is zero, then it can not satisfy the third option.


Zero is considered to be an even integer.
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Re: Seems easy, but is it really??? [#permalink] New post 15 Jul 2010, 04:56
nsp007 wrote:
what if one of is zero, then it can not satisfy the third option.


Zero is considered to be an even integer.


really? I didn't know that... thanks for the info.
Re: Seems easy, but is it really???   [#permalink] 15 Jul 2010, 04:56
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