St 1
(x + y)^2 > 0 just tells us that [spoiler](x+y) is not zero.[/spoiler]
The square of any number besides zero is > 0.
[spoiler]Another way of saying this is x NOT= -y[/spoiler]
St 2
[spoiler]This tells us nothing about y.
[/spoiler]
Combined:
All we know is that x is positive, and x NOT= -y. [spoiler]We don't have any way to know if x > y.[/spoiler]
Inequality
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smackmartine
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IMO E
Is x>y??? (Rephrase: what are the signs of the variable x & y)
Watch out for even squares. They are dangerous because they obscure the signs of variables.
1.
(x+y)^2>0
Any sq is positive , but we are more interested in the individual signs of x and y.From the statement 1, its not clear. So Insuff
2.
x >0
Whats the sign of y? We dnt know. So Insuff.
Combining st 1 and 2
we know x > 0
Rewrite st 1
case 1 (x+y) >0
x>-y
if x >0
y can either be a negative number ,but bigger in magnitude.
eg 4 > -(20)
x>y?? -------> No
or can be positive number but smaller in magnitude than x
eg 4 >-(2)
x>y?? -----> Yes
Because sign of y can be both positive and negative , combining sts is Insuff
Note: Case II can be (x+y)<0 & x>0, however we need not have to proceed further because in case 1 itself we have found both conditions in which y can be positive or negative.
Is x>y??? (Rephrase: what are the signs of the variable x & y)
Watch out for even squares. They are dangerous because they obscure the signs of variables.
1.
(x+y)^2>0
Any sq is positive , but we are more interested in the individual signs of x and y.From the statement 1, its not clear. So Insuff
2.
x >0
Whats the sign of y? We dnt know. So Insuff.
Combining st 1 and 2
we know x > 0
Rewrite st 1
case 1 (x+y) >0
x>-y
if x >0
y can either be a negative number ,but bigger in magnitude.
eg 4 > -(20)
x>y?? -------> No
or can be positive number but smaller in magnitude than x
eg 4 >-(2)
x>y?? -----> Yes
Because sign of y can be both positive and negative , combining sts is Insuff
Note: Case II can be (x+y)<0 & x>0, however we need not have to proceed further because in case 1 itself we have found both conditions in which y can be positive or negative.
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Anurag@Gurome
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Consider the following two cases,jayanti wrote:Is x > y?
(1) (x + y)² > 0
(2) x is positive
- 1. x = 2 and y = 1 --> x is positive and (x + y)² > 0 BUT x > y
2. x = 2 and y = 3 --> x is positive and (x + y)² > 0 BUT x < y
The correct answer is E.
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