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St1: m-1 and n+1 are consecutive positive integers.

Case 1:
m is even, n is odd and m < n
Let m = 2, n = 3
m-1 = 1, n+1 = 4 -> not consecutive positive integers.

Case 2:
m is even, n is odd and m > n
Let m = 2, n = 1
m-1 = 1, n+1 = 2 -> consecutive positive integers.

Case 3:
m is odd, n is even and m < n
Let m = 3, n = 4
m-1 = 2, n+1 = 5 -> not consecutive positive integers.

Case 4:
m is odd, n is even and m > n
Let m = 3, n = 2
m-1 = 2, n+1 = 3 -> consecutive positive integers.

Not sufficient. Dow to B, C or E.

St2: m is an even integer. Not sufficient. If m = 2 and n = 1 -> m > n but if m = 2 and n = 3 -> m < n.

St 1 + St2: Not sufficient as illustrated by case 1 and case 2.

Answer (E).
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St1: m-1 and n+1 are consecutive positive integers.

Case 1:
m is even, n is odd and m < n
Let m = 2, n = 3
m-1 = 1, n+1 = 4 -> not consecutive positive integers.

Case 2:
m is even, n is odd and m > n
Let m = 2, n = 1
m-1 = 1, n+1 = 2 -> consecutive positive integers.

Case 3:
m is odd, n is even and m < n
Let m = 3, n = 4
m-1 = 2, n+1 = 5 -> not consecutive positive integers.

Case 4:
m is odd, n is even and m > n
Let m = 3, n = 2
m-1 = 2, n+1 = 3 -> consecutive positive integers.

Not sufficient. Dow to B, C or E.

St2: m is an even integer. Not sufficient. If m = 2 and n = 1 -> m > n but if m = 2 and n = 3 -> m < n.

St 1 + St2: Not sufficient as illustrated by case 1 and case 2.

Answer (E).

I think you are making a mistake somewhere !! Well from the different cases you have illustrated it is clear that "A" is sufficient !! Because both cases where m-1,n+1 were consecutive you had m>n therefore statement "1" is clearly sufficient. Hope that helped ;)
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The question states that \(m\) &\(n\) are consecutive positive integers. The question can then be boiled down to whether the order is \(m,n\) or \(n,m\).

(1) m-1 and n+1 are consecutive positive integers.

We know that \(m-1\) and \(n+1\)are consecutive positive integers. If the order is \(m,n\) then this statement clearly fails. If the order is \(n,m\) then this statement is true, hence \(A\) is sufficient.

(2) m is an even integer.

If \(m\) is an even integer, \(n\) must be odd. \(n\) can still fall to the left or right of \(m\), and therefore statement (2) is insufficient.

Answer: A
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Bahadur
If m and n are consecutive positive integers, is m greater than n ?

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




Arithmetic Properties of numbers
For two integers x and y to be consecutive, it is
both necessary and sufficient that I x - y I = 1.
(1) Given that m-1 and n + 1 are consecutive
integers, it follows that ~ m - 1)- ( n + 1 ~ = 1,
or lm - n - ~ = 1. Therefore, m - n - 2 = 1 or
m - n - 2 = -1. The former equation implies
that m- n = 3, which contradicts the fact
that m and n are consecutive integers.
Therefore, m - n - 2 = -1, or m = n + 1,
and hence m > n; SUFFICIENT.
(2) If m = 2 and n = 1, then m is greater than n.
However, if m = 2 and n = 3, then m is not
greater than n; NOT sufficient

I know all of the topics of the question. But my confusion is that question describes m and n are consecutive positive integers, so why we need
I x - y I = 1. Why we use absolute for positive integers. Would any one like to explain me please.

Hi,
we take ABSOLUTE MOD because we are not aware whether x is big or y is big..
if x is bigger, x-y=1..
if y is bigger, x-y=-1..
so |x-y|=1..
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I have one small confusion. The question says that "if m and n are consecutive positive integers..." Doesn't this imply automatically that the order is m and then n? How are we unsure that its not m and n rather it could be n and m? Please help me.

Thanks!
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csaluja
I have one small confusion. The question says that "if m and n are consecutive positive integers..." Doesn't this imply automatically that the order is m and then n? How are we unsure that its not m and n rather it could be n and m? Please help me.

Thanks!

If it were the way you are saying, we won't have this question. So, x and y are consecutive integers does NOT necessarily mean that y > x.

P.S. Also, note that this is an official question.
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VeritasKarishma chetan2u MentorTutoring nick1816

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

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

I find the highlighted part and St 1 to be conflicting (usually I take the approach if I could SOLVE q stem by using the statements)
If m and n are two consecutive no, are not m-1 and n-1 ALSO consecutive by default?

To avoid, confusion, can I safely use number picking as below:
Case 1: I take m-1 less than n+1
m-1 = 2
n+1 = 3
hence m=3, n=2 m>n

In next case, I purposely take m-1 greater than n+1
m-1= 4
n+1= 3
Hence m=6, n=2 Still m>n

Clearly St 1 is suff.

St 2 is clearly insuff. Ans: A
Can you share your two cents?
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adkikani
VeritasKarishma chetan2u MentorTutoring nick1816

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

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

I find the highlighted part and St 1 to be conflicting (usually I take the approach if I could SOLVE q stem by using the statements)
If m and n are two consecutive no, are not m-1 and n-1 ALSO consecutive by default?

To avoid, confusion, can I safely use number picking as below:
Case 1: I take m-1 less than n+1
m-1 = 2
n+1 = 3
hence m=3, n=2 m>n

In next case, I purposely take m-1 greater than n+1
m-1= 4
n+1= 3
Hence m=6, n=2 Still m>n

Clearly St 1 is suff.

St 2 is clearly insuff. Ans: A
Can you share your two cents?

Yes, if m and n are consecutive, m-1 and n-1 would surely be consecutive, but this is a peculiar case where m-1 and n+1 are also consecutive.
This now should tell you that n+1>m-1 or n+1=(m-1)+1...n+1=m...m>n
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Hello, adkikani. My in-line responses are below.

adkikani
VeritasKarishma chetan2u MentorTutoring nick1816

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

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

I find the highlighted part and St 1 to be conflicting (usually I take the approach if I could SOLVE q stem by using the statements)
If m and n are two consecutive no, are not m-1 and n-1 ALSO consecutive by default?
I do not understand the conflict. Statement (1) presents the information in a distorted way, making it such that the larger value has changed places. That is, it is clear that m is greater than n within the constraints of the question stem and this statement, but n + 1 yields the greater value than m - 1. This is the sort of confusion the GMAT™ likes to create, just enough to make test-takers doubt themselves.
adkikani
To avoid, confusion, can I safely use number picking as below:
Case 1: I take m-1 less than n+1
m-1 = 2
n+1 = 3
hence m=3, n=2 m>n

In next case, I purposely take m-1 greater than n+1
m-1= 4
n+1= 3
Hence m=6, n=2 Still m>n
This second case does not conform to the given constraints. If m and n are consecutive positive integers, then m - 1 cannot be greater than n + 1. Assume, for the sake of argument, that you had originally picked n to be the greater of the two integers. For instance, let n = 2 and m = 1.

m - 1 = 1 - 1 = 0
n + 1 = 2 + 1 = 3

These values do not conform to the information given in Statement (1), so you would know you had leaned in the wrong direction.
adkikani
Clearly St 1 is suff.

St 2 is clearly insuff. Ans: A
Can you share your two cents?
Yes, clearly Statement (1) is indeed SUFFICIENT, but be careful about testing, willy-nilly, whatever numbers you want, since that sort of method could lead to incorrect conclusions.

I hope that helps. Thank you for bringing my attention to the question.

- Andrew
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Question:- If m and n are consecutive positive integers, is m greater than n?
What I inferred> Question is basically asking, is M=N+1? (as M and N are consecutive, Difference between them is 1)

Statement 1: m-1 and n+1 are consecutive positive integers.

Assume; M=6 & N=5
> M-1 = N = 5
> N+1 = M = 6 ...... (Yes, M>N) .. Thus, Statement sufficient

Statement 2: m is an even integer.
>If m is an even integer, n must be odd. m can still fall to the left or right of n, and therefore statement (2) is insufficient.

Answer: A

Is my approach, Correct?
Bunuel VeritasKarishma chetan2u
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Arpit22
Question:- If m and n are consecutive positive integers, is m greater than n?
What I inferred> Question is basically asking, is M=N+1? (as M and N are consecutive, Difference between them is 1)

Statement 1: m-1 and n+1 are consecutive positive integers.

Assume; M=6 & N=5
> M-1 = N = 5
> N+1 = M = 6 ...... (Yes, M>N) .. Thus, Statement sufficient

Statement 2: m is an even integer.
>If m is an even integer, n must be odd. m can still fall to the left or right of n, and therefore statement (2) is insufficient.

Answer: A

Is my approach, Correct?
Bunuel VeritasKarishma chetan2u

Yes, though if it holds for one set of values it doesn't mean it will hold for all. Understand the logic I discussed above to know why it will hold for all.
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let's choose one random positive integer as m: exp: m=5 => n could be: 4 or 6

(1) (m-1) and (n+1) are positive consecutive integers
(m-1)= 5-1=4
#1: n=4 => n+1= 5
#2: n=6 => n+1= 6
Keep #1 (4 and 5 are consecutive positive integers) and eliminate #2 based on the (1) statement, we could conclude m>n : SUFFICIENT

(2) m is even => n must be odd. But cannot determine m is consecutively after n or not : INSUFICIENT

===> A
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