logitech wrote:1) x = y + 1/2 OKAY WE KNOW WHO IS THE DADDY! X>YII wrote:Please illustrate your logic behind the answer.
Thanks.
2) X/Y > 1 HERE IS THE TRAP!
(Y+1/2) / Y > 1
OR 1 + Y/2 > 1 hmmm...so, Y/2>0 ---> Y>0
(small mistake here: 1+1/2y > 1, or 1/2y>0, or y>0)
Remember X > Y
So these guys are both POSITIVE
Thing to remember:
X/Y > 1
X/Y - 1 > 0 Learn this!
(X-Y)/Y > 0
Y < 0 and X-Y < 0
OR
Y > 0 and X-Y> 0
Equatons: Are x and y both positive ?
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deepakteja1988
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Even I intially chose E as the choice. But we need to be careful. If we are substituting values, Please see that we are not changing the choice. For example,
x/y > 1
Please dont change the choice to x > y.
keep it as it is.
Everyone by now must be clear why A is not sufficient.
x - y = 1/2
(i) 3/2 - 1 = 1/2 both +ve (3/2, 1)
(ii) -3/2 - (-2) = 1/2 both -ve (-3/2, -2)
So unsufficient.
x/y > 1
(3/2)/1 > 1 for sure.
(-3/2)/(-2) < 1 for sure. --> so when we take two negative values to satisfy the first condition x-y=1/2. They can never satisfy the second condition x/y > 1.
The trap is changing the given condition to x > y and substituting the negative values in it.
x/y > 1
Please dont change the choice to x > y.
keep it as it is.
Everyone by now must be clear why A is not sufficient.
x - y = 1/2
(i) 3/2 - 1 = 1/2 both +ve (3/2, 1)
(ii) -3/2 - (-2) = 1/2 both -ve (-3/2, -2)
So unsufficient.
x/y > 1
(3/2)/1 > 1 for sure.
(-3/2)/(-2) < 1 for sure. --> so when we take two negative values to satisfy the first condition x-y=1/2. They can never satisfy the second condition x/y > 1.
The trap is changing the given condition to x > y and substituting the negative values in it.
