If x does not equal -y, is (x-y)/(x+y) > 1?
1. x > 0
2. y < 0
got the right answer here, but took me too long. i plugged in answers to get the right one. Wondering if there was a faster way to do this one. thanks.
Fast way to do this?
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IMO the answer is E.. both together not sufficient.fangtray wrote:If x does not equal -y, is (x-y)/(x+y) > 1?
1. x > 0
2. y < 0
got the right answer here, but took me too long. i plugged in answers to get the right one. Wondering if there was a faster way to do this one. thanks.
Here is the explanation-
is (x-y)/(x+y)>1 ?
=> (x-y)/(x+y) - 1 > 0 ?
=> -2y/(x+y) > 0 ?
=> y/(x+y) < 0 ?
Now, if you analyze the last equation.. y & (x+y) must be of different signs.
[A] If x>0 it tells nothing abt signs of y and (x+y)
If y< 0 .. sign of (x+y) is still unknown
[B} & [A] together.. if y is -ve and x is +ve still for (x+y) to be positive |x| > |y|.. this info is not given
Hence both together not sufficient.
Hope it helps!!
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This problem seems to be about POSITIVE VS. NEGATIVE.fangtray wrote:If x does not equal -y, is (x-y)/(x+y) > 1?
1. x > 0
2. y < 0
got the right answer here, but took me too long. i plugged in answers to get the right one. Wondering if there was a faster way to do this one. thanks.
Try numbers that are at varying DISTANCES FROM 0.
If x=2 and y=-1, then (x-y)/(x+y) = 3, which is greater than 1.
If x=1 and y=-2, then (x-y)/(x+y) = -3, which is not greater than 1.
Since each combination satisfies both statements, and in the first case (x-y)/(x+y)>1, and in the second case (x-y)/(x+y)<1, the two statements combined are INSUFFICIENT.
The correct answer is E.
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You can use your knowledge of concepts to solve rather than plugging in numbers (although picking numbers is also a great way to go).fangtray wrote:If x does not equal -y, is (x-y)/(x+y) > 1?
1. x > 0
2. y < 0
got the right answer here, but took me too long. i plugged in answers to get the right one. Wondering if there was a faster way to do this one. thanks.
First, let's ask ourselves when we'll get a YES answer to the question.
First, we need the sign of the top and bottom to be the same (if the signs were different, the result would be negative, giving us an automatic "NO").
Second, we need the absolute value of the numerator to be greater than the absolute value of the denominator (otherwise we'll get a fraction, giving us an automatic "NO").
To the statements!
Each of (1) and (2) is clearly insufficient alone, since each only gives info about one of the two variables, so let's jump right to combination.
If x > 0 and y < 0, then x > 0 > y.
So, we know that:
x - y = (+) - (-) = +; and
x + y = (+) + (-) = it depends on which has a greater magnitude (if x=10 and y=-1, then it's positive; if x=1 and y=-10, then it's negative).
So, we could have:
+/+ which gives a potential YES answer; or
+/- which gives a definite NO answer.
Maybe YES, maybe NO - insufficient, choose E.
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