Clearly C is the winner.
When we take both the options together, only then we can satisfy the condition given in the question
Hope this helps
DS
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raghavsarathy
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Is x-y+1 greater than x+y-1
1. x > 0
2. y < 0
In the 2 terms which are to be comapted , x has the same sign in both. Whereas the sign on y differs. Hence the answer surely depends on y.
Taking statement B and replcing y by -k we get
first eqn as x-(-k) +1 = x+k+1
Second Eqn is x+(-k) -1 = x-k-1
In these eqns K is positive. x can be positive or negative. It does not matter
We can also consider the case when x equals k
Examples
x= -2 k=2
eqn 1 has value 1
eqn 2 has value -5
eqn 1 > eqn 2
x= 2 , k =2
eqn 1 has value 5
eqn 2 has value -1
eqn 1 > eqn 2
The same holds good even for fractions (positive and negative)
Hence statement 2 is sufficient
1. x > 0
2. y < 0
In the 2 terms which are to be comapted , x has the same sign in both. Whereas the sign on y differs. Hence the answer surely depends on y.
Taking statement B and replcing y by -k we get
first eqn as x-(-k) +1 = x+k+1
Second Eqn is x+(-k) -1 = x-k-1
In these eqns K is positive. x can be positive or negative. It does not matter
We can also consider the case when x equals k
Examples
x= -2 k=2
eqn 1 has value 1
eqn 2 has value -5
eqn 1 > eqn 2
x= 2 , k =2
eqn 1 has value 5
eqn 2 has value -1
eqn 1 > eqn 2
The same holds good even for fractions (positive and negative)
Hence statement 2 is sufficient
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PussInBoots
- Master | Next Rank: 500 Posts
- Posts: 157
- Joined: Tue Oct 07, 2008 5:47 am
- Thanked: 3 times
To show that x - y + 1 > x + y - 1 is equivalent to (x - y + 1) - (x + y - 1) > 0 which is equivalent to showing that 2 - 2y > 0.
Answer is B and not C.
P.S.: what program is that screenshot from? I've asked this question like 5 times and non one bothers answering
Answer is B and not C.
P.S.: what program is that screenshot from? I've asked this question like 5 times and non one bothers answering
