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If x!=0, is (x^2 + 1)/x > y? (1) x = y (2) y > 0 What

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If x!=0, is (x^2 + 1)/x > y? (1) x = y (2) y > 0 What [#permalink]  14 Jun 2011, 05:23
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If x!=0, is (x^2 + 1)/x > y?
(1) x = y
(2) y > 0

What will be the correct answer to this question? I am confused.

Approach 1:
I tried to simplify this question as follows
x^2 + 1 - xy > 0
x(y-x) < 1
But we can't be sure of the sign of the inequality as I have cross multiplied by x (whose sign I do now know).
So in essence the question can be rephrased as:
x(y-x) [greater than, equal to or less than] 1 --> This is what is actually being asked.

I want to know if i have made any fundamental mistake when doing this rephrasing. As I have already mentioned I know that the sign of the inequality should be reversed when multiplied by a negative number.

So going by this approach the answer is A

Approach 2:

I will consider both cases of x before cross multiplying.

Case 1: x is negative
x^2+1 < xy
x^2-xy + 1 < 0
x(x-y) + 1 <0

Case 2: x is positive
x^2+1 > xy
x^2-xy + 1 > 0
x(x-y) + 1 > 0

So based on these 2 equations we find that statement (1) alone is not sufficient because it gives different answers for Case 1 and 2 i.e. 1 < 0 and 1 > 0.

Statement 2 is also insufficient because it does not say anything about x.

Combining 1 and 2 -
y > 0
x = y
Therefore, x > 0
x(x-y) + 1 > 0

The second approach looks more methodical but I want to understand why the first approach is INVALID (if it is so).
[Reveal] Spoiler: OA
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 05:51
rampa wrote:
If x!=0, is (x^2 + 1)/x > y?
(1) x = y
(2) y > 0

1. x=y
$$\frac{x^2 + 1}{x} > y$$
$$\frac{x^2 + 1}{x} > x$$
$$\frac{x^2 + 1}{x}-x > 0$$
$$\frac{x^2 + 1 -x^2}{x} > 0$$
$$\frac{1}{x} > 0$$

Means; $$x>0$$

Thus, the expression would be true when x>0
But, we don't know whether x>0.

Not Sufficient.

2. y>0

y=1, x=10; LHS>RHS
x=1, y=10; LHS<RHS
Not Sufficient.

Combining both we know;
y>0 & x=y; means x>0; We got our condition we required in statement 1.
Sufficient.

Ans: "C"
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 06:20
Expert's post
rampa wrote:
...
So in essence the question can be rephrased as:
x(y-x) [greater than, equal to or less than] 1 --> This is what is actually being asked.

You can't rephrase it in that way. First of all you need to drop "equal" and then add conditions:
x(y-x) [greater than (for negative x) or less than (for positive x)] 1
It actually means your second approach.

I don't know what x!=0 means. There is no x for which x!=0
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 06:29
walker wrote:
I don't know what x!=0 means. There is no x for which x!=0

x != 0
x Not Equal 0
$$x \ne 0$$
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 06:56
fluke wrote:
walker wrote:
I don't know what x!=0 means. There is no x for which x!=0

x != 0
x Not Equal 0
$$x \ne 0$$

Fluke to be honest to dint get this?
! ( factorial) symbol mean #( not equal to )?
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 07:07
rampa wrote:
If x!=0, is (x^2 + 1)/x > y?
(1) x = y
(2) y > 0

What will be the correct answer to this question? I am confused.

Approach 1:
I tried to simplify this question as follows
x^2 + 1 - xy > 0
x(y-x) < 1
But we can't be sure of the sign of the inequality as I have cross multiplied by x (whose sign I do now know).
So in essence the question can be rephrased as:
x(y-x) [greater than, equal to or less than] 1 --> This is what is actually being asked.

I want to know if i have made any fundamental mistake when doing this rephrasing. As I have already mentioned I know that the sign of the inequality should be reversed when multiplied by a negative number.

So going by this approach the answer is A

There is no problem in re-phrasing . But how do u get A?
st.1 says X=Y
from rephrase
X(y-X)<1
X(X-X)<1
X*0<1
0<1........ not sufficient
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 07:19
sudhir18n wrote:
fluke wrote:
walker wrote:
I don't know what x!=0 means. There is no x for which x!=0

x != 0
x Not Equal 0
$$x \ne 0$$

Fluke to be honest to dint get this?
! ( factorial) symbol mean #( not equal to )?

I agree!! This symbol is more a C programer's symbol.

In C: != means NOT EQUAL
In Java: <> means NOT EQUAL

And so on...

It's better to use $$\ne$$ to avoid ambiguity.
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 07:39
Expert's post
Got it!
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Joined: 17 Nov 2007
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 07:41
Expert's post
By the way, GMAC "not equal" usually says by words, does it? (For example, OG12 DS 125)
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Re: DS - Inequalities - 2 rephrasings give 2 different answers! [#permalink]  14 Jun 2011, 07:56
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walker wrote:
By the way, GMAC "not equal" usually says by words, does it?

*******************
9. If $$xy \ne 0$$, is $$\frac{x}{y}<0$$ ?

(1) x = –y

(2) –x = – (– y)
**********************

This question is from GMAT paper test, Test Code 25, Section 5, exactly how it appears there.
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Re: DS - Inequalities - 2 rephrasings give 2 different answers!   [#permalink] 14 Jun 2011, 07:56
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