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.
Answer is A
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cramya
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Take easier of the the two
Stmt II
No info about n .INSUFF
Stmt I
Lets assume that:
The order can be mn(m<n) or nm((m>n)
If its mn then m-1 and n+1 can never be consecutive. Therefore it has to be nm where m-1 (i.e. n) and n+1 (i.e m) are again conseutive satisfying Stmt I
SUFF
Hence A
Stmt II
No info about n .INSUFF
Stmt I
Lets assume that:
The order can be mn(m<n) or nm((m>n)
If its mn then m-1 and n+1 can never be consecutive. Therefore it has to be nm where m-1 (i.e. n) and n+1 (i.e m) are again conseutive satisfying Stmt I
SUFF
Hence A
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Bidisha800
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m-1 and n+1 are consecutive ,
therefore, either (m-1) + 1= (n+1) OR m-1 = (n+1) +1
if (m-1) + 1= (n+1)
m= n+1
therefore, m>n
if m-1 = (n+1) +1
m = n +3
m>n
(A) suff
(B) is insuff
therefore, either (m-1) + 1= (n+1) OR m-1 = (n+1) +1
if (m-1) + 1= (n+1)
m= n+1
therefore, m>n
if m-1 = (n+1) +1
m = n +3
m>n
(A) suff
(B) is insuff
Drill baby drill !
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