MathRevolution wrote:
If xy^2 = 1, is x > y?
(1) -1 < y < 1
(2) xy < 0
Let's do it with
case testing Question stem: xy^2 = 1. That's what you know so far. What you
want to know, is whether x is bigger than y.
Statement 1: This limits the values that y could have. Try out some values for y, using both positive and negative numbers.
y = -0.5
xy^2 = 1
x(-0.5)^2 = 1
0.25x = 1
x = 4
x IS bigger than y, so the answer is "yes"
y = 0
Oops, we can't actually test this case, because it goes against the info in the question stem. The question stem says that xy^2 = 1, but that can't be true if y = 0. So, y must not equal 0, and we can ignore this case.
y = 0.5
xy^2 = 1
x(0.5)^2 = 1
0.25x = 1
x = 4
x IS bigger than y, so the answer to the question is "yes"
It seems like we're always getting "yes" answers. Just to be sure, test some extreme cases. What if y = 0.9999999 or -0.99999999?
y = 0.999999999 (very close to 1):
x(0.999999999)^2 = 1
x(something a tiny bit smaller than 1) = 1
x = 1/(something a tiny bit smaller than 1) = a little bigger than 1
So, x is greater than y again.
y = -0.9999999 (very close to -1):
x(-0.9999999999)^2 = 1
x(something a tiny bit smaller than 1) = 1
x = 1/(something a tiny bit smaller than 1) = a little bigger than 1
x is once again greater than y.
It seems like x is always greater than y, and this statement is sufficient.
Statement 2:
xy is negative. So, either x is negative and y is positive, or x is positive and y is negative.
Try to find some cases where x is negative and y is positive, and xy^2 = 1:
-2(0.5^2) = -2(0.25) = -0.5 : that doesn't work...
-1(1^2) = -1(1) = -1: that doesn't work....
In fact, there ARE no such cases, because xy^2 will only be positive if x itself is positive.
So, we're stuck with the cases where x is positive, and y is negative. But no matter what, a positive number will always be bigger than a negative one! So, x is bigger than y, and this statement is sufficient.
Since both statements are sufficient individually, the answer is D.
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