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# If a and b are consecutive negative integers, is a less than

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Intern
Joined: 27 Sep 2009
Posts: 41
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Kudos [?]: 84 [0], given: 4

If a and b are consecutive negative integers, is a less than [#permalink]

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07 Jun 2010, 16:17
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Difficulty:

15% (low)

Question Stats:

87% (01:36) correct 13% (01:25) wrong based on 51 sessions

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If a and b are consecutive negative integers, is a less than b ?

(1) a + 1 and b - 1 are consecutive negative integers.
(2) a is an odd integer.
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 37654
Followers: 7409

Kudos [?]: 99758 [0], given: 11047

Re: Consecutive integers- 600 level question [#permalink]

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08 Jun 2010, 07:00
shekar123 wrote:
If a and b are consecutive negative integers, is a less than b ?

(1) a + 1 and b - 1 are consecutive negative integers.
(2) a is an odd integer.

$$a$$ and $$b$$ are consecutive integers means that either:
$$b=a+1$$ (for example $$a=-5$$ and $$b=-4$$), in this case $$a<b$$;
OR:
$$a=b+1$$ (for example $$a=-5$$ and $$b=-6$$), in this case $$a>b$$.

So the we should determine which case we have.

(1) $$a+1$$ and $$b-1$$are consecutive negative integers --> again either $$a+1=b-1+1$$ --> $$a+1=b$$ (first case so $$a<b$$) or $$b-1=a+1+1$$ --> $$b=a+3$$, which is not possible as $$a$$ and $$b$$ are consecutive integers (difference between two consecutive integers can not equal to 3). Sufficient.

(2) $$a$$ is an odd integer. Clearly insufficient.

Answer: A.
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Intern
Joined: 27 Sep 2009
Posts: 41
Followers: 0

Kudos [?]: 84 [0], given: 4

Re: Consecutive integers- 600 level question [#permalink]

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08 Jun 2010, 21:23
Could you please tell me as to how did u get this

$$b-1=a+1+1$$ --> $$b=a+3$$

Thanks

Bunuel wrote:
shekar123 wrote:
If a and b are consecutive negative integers, is a less than b ?

(1) a + 1 and b - 1 are consecutive negative integers.
(2) a is an odd integer.

$$a$$ and $$b$$ are consecutive integers means that either:
$$b=a+1$$ (for example $$a=-5$$ and $$b=-4$$), in this case $$a<b$$;
OR:
$$a=b+1$$ (for example $$a=-5$$ and $$b=-6$$), in this case $$a>b$$.

So the we should determine which case we have.

(1) $$a+1$$ and $$b-1$$are consecutive negative integers --> again either $$a+1=b-1+1$$ --> $$a+1=b$$ (first case so $$a<b$$) or $$b-1=a+1+1$$ --> $$b=a+3$$, which is not possible as $$a$$ and $$b$$ are consecutive integers (difference between two consecutive integers can not equal to 3). Sufficient.

(2) $$a$$ is an odd integer. Clearly insufficient.

Answer: A.
Math Expert
Joined: 02 Sep 2009
Posts: 37654
Followers: 7409

Kudos [?]: 99758 [0], given: 11047

Re: Consecutive integers- 600 level question [#permalink]

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09 Jun 2010, 04:51
shekar123 wrote:
Could you please tell me as to how did u get this

$$b-1=a+1+1$$ --> $$b=a+3$$

Thanks

Simple math:
$$b-1$$ and $$a+1$$ are consecutive integers --> $$(b-1)=(a+1)+1$$ --> $$b-1=a+2$$ --> add 1 to both sides --> $$b-1+1=a+2+1$$ --> $$b=a+3$$.
_________________
Re: Consecutive integers- 600 level question   [#permalink] 09 Jun 2010, 04:51
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# If a and b are consecutive negative integers, is a less than

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