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# If x < y, is x^2 < y^2 ?

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Re: If x < y, is x^2 < y^2 ? [#permalink]
Thanks Bunuel. Kept is simple while solving. In first case, I assumed Y = 1 in which case option 1 is ruled out as x^2 can be greater or lesser than y^2. in second case also assume x=1 in which case y would always be positive and data would be sufficient. To be on the safe side tried with x = 0.5 also in which case too option 2 was sufficient.
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Re: If x < y, is x^2 < y^2 ? [#permalink]
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Bunuel wrote:
If x < y, is x² < y² ?

(1) y > 0
(2) x > 0

Target question: Is x² < y² ?
STRATEGY: When I scan the target question and the two statements, I see that I can REPHRASE the target question as Is x² - y² < 0?
We can also factor the left side to get:
REPHRASED target question: Is (x + y)(x - y) < 0 ?

Given: x < y
If we subtract y from both sides, we get: x - y < 0
In other words, x - y is NEGATIVE
So, to answer the target question we need only determine whether x+y is positive or negative.

Statement 1: y > 0
There are several values of x and y that satisfy statement 1 (and the given information). Here are two:
Case a: x = 1 and y = 2. In this case, the answer to the REPHRASED target question is YES, (x + y)(x - y) is less than 0
Case b: x = -3 and y = 2. In this case, the answer to the REPHRASED target question is NO, (x + y)(x - y) is not less than 0
Since we can’t answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x > 0
So, x is positive.
Since we're told x < y, we know y is also positive, which means x+y is POSITIVE
Since we know x - y is NEGATIVE, we get: (x + y)(x - y) = (NEGATIVE)(POSITIVE) = NEGATIVE
In other words, the answer to the REPHRASED target question is YES, (x + y)(x - y) is less than 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

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Re: If x < y, is x^2 < y^2 ? [#permalink]
Quote:
If x < y, is x^2 < y^2 ?

(1) y > 0
(2) x > 0

Like Brent above, I thought to rephrase the question, but I took things in a different direction:

Is |x| < |y| ?

By removing the squares from consideration, it is clear that all we have to do is track which variable is more or less distant from 0.

(1) Conceptually, we can see that if y > 0, we still cannot say whether x is more negative than y is positive. However...

(2) If x > 0, then x is positive, and y must be more distant from 0, since x < y.

Hence, the answer must be (B). There is no need to test cases, even if I often resort to this strategy when I cannot find an algebraic or conceptual solution.

- Andrew
Re: If x < y, is x^2 < y^2 ? [#permalink]
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