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M04#03

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M04#03 [#permalink]

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New post 22 Dec 2008, 19:19
If \(m\) and \(n\) are two consecutive positive integers, is \(m \gt n\) ?

1. \(m-1\) and \(n+1\) are consecutive positive integers
2. \(m\) is an even integer

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I am not able to understand S1. if m and n are consecutive +ve integers, how can m-1 and n+1 be consecutive +ve integers?

The explanation just says "S1 holds only when m=n+1 . It is sufficient." - couldnt make sense out of it.

can someone help?
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Re: M04#03 [#permalink]

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New post 22 Dec 2008, 21:30
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dadagreat wrote:
m and n are two consecutive positive integers. Is m > n ?
1. m-1 and n+1 are consecutive positive integers
2. m is an even integer

I am not able to understand S1. if m and n are consecutive +ve integers, how can m-1 and n+1 be consecutive +ve integers?

The explanation just says "S1 holds only when m=n+1 . It is sufficient." - couldnt make sense out of it.

can someone help?


Given : m and n are consecutive integers
Now there are two possibilities only. Obviously you cant make out because
m>n or n>m

Take any 2 numbers for first possibility where m=n+1. So no.s here we have are n and n+1
with S1=> m-1= n+1-1=n -----(1)
n+1= n+1---(2)
1 and 2 are consecutive from first possibility. This means if you know that m+1 and n-1 are consecutive, then you know that m and n were also consecutive, with m>n.

But say, if they are not, which will happen in case of another possibility, then this cant be deducted. If we go to another possibility and do same analysis we wont find this true. I have given that also below, however not needed.
With another possibility, m=n-1 also. we have n-1 and n.
Now S1=> m-1= n-1-1= n-2----(1)
n+1 n+1----(2)
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Re: M04#03 [#permalink]

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New post 09 Nov 2010, 07:28
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Another way, plugging numbers,
Case m < n: m = 3, n = 4
m - 1 = 2, n + 1 = 5, 2 and 5 are not consecutive.
Case m > n: m = 4, n = 3
m - 1 = 3, n + 1 = 4, 3 and 4 are consecutive.

hence, m > n. A is suff.

B gives no info about n. So, insuff.

HTH
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Re: M04#03 [#permalink]

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New post 09 Nov 2010, 11:23
This is simple. A is the answer.
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Re: M04#03 [#permalink]

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New post 13 Nov 2010, 00:23
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dadagreat wrote:
If \(m\) and \(n\) are two consecutive positive integers, is \(m \gt n\) ?

1. \(m-1\) and \(n+1\) are consecutive positive integers
2. \(m\) is an even integer

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I am not able to understand S1. if m and n are consecutive +ve integers, how can m-1 and n+1 be consecutive +ve integers?

The explanation just says "S1 holds only when m=n+1 . It is sufficient." - couldnt make sense out of it.

can someone help?


Absolute difference between two consecutive integres (positive or negetive) must always be equal to 1

Using 1:
|(m-1)-(n+1)|=1
|m-n-2|=1
m-n-2=1 OR m-n-2=-1

m-n-2=1 => m-n=3 => m=n+3 => Not possible since 'm' and 'n' are consecutive integers.
OR
m-n-2=-1 => m-n=1 => m=n+1 => m>n
SUFFICIENT

Using 2:
'm' is even so 'n' could either be 'm+1' or 'm-1'
NOT SUFFICIENT

A: 1 alone is sufficient.
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Vaibhav

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Re: M04#03 [#permalink]

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New post 11 Nov 2011, 08:17
A.

M= n +/- 1

Plug in either n+1 or n-1 in first statement

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Re: M04#03 [#permalink]

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New post 13 Nov 2011, 16:44
I plugged in an even positive number, an odd number and and a negative number for m and saw what options I can get for N with both being consecutive numbers. 2 was irrelevant. Answer is A.
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Re: M04#03 [#permalink]

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New post 13 Nov 2012, 06:39
Expert's post
dadagreat wrote:
If \(m\) and \(n\) are two consecutive positive integers, is \(m \gt n\) ?

1. \(m-1\) and \(n+1\) are consecutive positive integers
2. \(m\) is an even integer

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

I am not able to understand S1. if m and n are consecutive +ve integers, how can m-1 and n+1 be consecutive +ve integers?

The explanation just says "S1 holds only when m=n+1 . It is sufficient." - couldnt make sense out of it.

can someone help?


If m and n are consecutive positive integers, is m greater than n?

(1) m-1 and n+1 are consecutive positive integers --> if m were less than n than m-1 (integer less than m) and n+1 (integer more than n) wouldn't be consecutive, so m is greater than n. Sufficient.

Or look at this in another way: stem says that the distance between m and n is 1. Now, if m<n then the distance between m-1 and n+1 would be 3 and they couldn't be consecutive as (1) states. Thus it must be true that m>n.

(2) m is an even integer. Clearly insufficient.

Answer: A.
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Re: M04#03 [#permalink]

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New post 13 Nov 2012, 07:20
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Given m and n are consecutive integers. Therefore either m>n or n>m.
Statement 1) m-1 and n+1 are consecutive integers.
Two cases are possible:-
i)
If (n+1)>(m-1)
Then m-1+1=n+1
or m-1=n THAT is OK.
BUT
If (n+1)<(m-1)
Then m-1-1=n+1
m-2=n+1
m-n=3
Hence the difference between the two numbers is not equal to 1. It means the two numbers are not consecutive.
Therefore only first case is valid.
Statement 2) Insufficient
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Re: M04#03   [#permalink] 13 Nov 2012, 07:20
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