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Perplexing DS Question

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Perplexing DS Question

by edge » Mon Jul 25, 2011 5:09 pm
Hello all, I am new here.
If x ≠ 0, is (x² + 1)/x > y?
1. x = y
2. y > 0
[spoiler]According to me, the answer should be A, but the guide I am referring to says the answer is C. My rationale goes like this: If x = y, the sign of x or y should not matter. The sign of x² is the same as the sign of xy, which is the same as the sign of y² (all three are always positive).[/spoiler]
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by Nodr » Mon Jul 25, 2011 5:32 pm
ans is C.

your rationale is wrong. > & < signs change when multiplied or divided by a negative number.

Check this out

as x not 0,
simplified lhs.
x + 1/x > y

now x=y implies, 1/x > 0 -- the validity here depends on the sign of x(&y as x=y) .
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by edge » Mon Jul 25, 2011 5:47 pm
You're right. I tried to solve this mathematically using x = y = -2. LHS = -5/2 RHS = -2. Therefore, LHS < RHS. When I use x = y = 2, LHS = 5/2 and RHS = 2 giving LHS > RHS. Therefore, Statement 1 by itself is insufficient. I need to be more careful. Thanks Nodr.
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by naveen451 » Mon Jul 25, 2011 11:08 pm
the best method to do these kind of problems is substituting the equations with negative and positive numbers according to the options
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by SticklorForDetails » Tue Jul 26, 2011 8:19 am
IF you really want to do algebra on these kinds of problems (which you shouldn't; as has been suggested above, you should always plug in numbers), you have to remember there are always 2 cases for an Inequality with variables in denominators:

x^2 + 1 / x > y? -->
1. Is x^2 + 1 > xy, while x is positive? or
2. Is x^2 + 1 < xy, while x is negative?

Then, your initial approach works: if x=y, then case 1 always works, but case 2 always doesn't because x^2+1 > x^2. However, Statement (2) invalidates the "what if x is negative?" possibility when combined with Statement (1), so the answer is C.

The only reason I mention this is that your initial logical approach was very good and it's always really satisfying on test day to see through the problem to the theory. If you can handle the weird logic of two possible superposed cases, then you CAN, in fact, do algebra with inequalities.
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