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# If x ≠ 0, is (x^2 + 1)/x > y?

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If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 06:23
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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.

[Reveal] Spoiler:
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

Last edited by Bunuel on 02 Oct 2015, 09:02, edited 2 times in total.
Formatted the question.

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Re: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 06:51
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 07:20
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 07:29
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 07:56
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 08:07
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 08:19
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 08:39
Got it!
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Re: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 08:41
By the way, GMAC "not equal" usually says by words, does it? (For example, OG12 DS 125)
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Re: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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14 Jun 2011, 08:56
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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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if x is NOT = 0, is (x^2 +1)/x > Y? (1) x = y (2) y>0 [#permalink]

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06 Oct 2011, 13:06
if x is NOT = 0, is (x^2 +1)/x > Y?

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

I don't understand how (1) is not sufficient.

To multiply by variable, we need to take the positive and the negative:
(x^2 +1)/x > Y
x^2 +1 > xy ?

And stmt 1 says y=y hence the question becomes, x^2+1>x^2? In which case, YES! regardless of whether x is negative or even a fraction.

now if we take the negative of x:

(-x)(x^2 +1)/x > Y(-x)
-x^2 -1 > -xy
x^2 + 1 < xy which equals x^2 + 1 < x^2

Even here, statement (1) is sufficient to answer the question! how is it insufficient!?
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06 Oct 2011, 13:22
if x is NOT = 0, is (x^2 +1)/x > Y?

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

from question x can be negative or positive

Statement 1
X = Y

(X^2 + 1 )/X >Y
=>(X^2)/X+1/X>Y
=>x + 1/X > Y (since y = X)
=>X + 1/X > X -------(1)

If X is positive

(1) will become

X + 1/X > X

If x is negative then

(1) ==> -X - 1/x < -X

SO we are getting two solutions

not sufficient

Statement 2

Y>0

so Y is positive

Not sufficient

Both combined

X is positive because Y is positive and X = Y

(x^2 +1)/x > Y is true

Ans C
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06 Oct 2011, 13:30
1
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or you can solve in below way

Statement 1
X = Y

(X^2 + 1 )/X >Y
=>(X^2)/X+1/X>Y
=>x + 1/X > Y (since y = X)
=>X + 1/X > X -------(1)

If x is positive say x = 2

then
(1) will become

2+ 1/2 > 2

2.5 > 2

which is correct

if x is negative say x = -2

-2 - 1/2 > -2
-2.5 > -2

Which is not correct (-2.5 < -2)

so insufficient

Statement 2 says Y is positive which alone is not sufficient but combining with statement 1 it is sufficient

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06 Oct 2011, 18:27
386390 wrote:
if x is NOT = 0, is (x^2 +1)/x > Y?

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

I don't understand how (1) is not sufficient.

To multiply by variable, we need to take the positive and the negative:
(x^2 +1)/x > Y
x^2 +1 > xy ?

And stmt 1 says y=y hence the question becomes, x^2+1>x^2? In which case, YES! regardless of whether x is negative or even a fraction.

now if we take the negative of x:

(-x)(x^2 +1)/x > Y(-x)
-x^2 -1 > -xy
x^2 + 1 < xy which equals x^2 + 1 < x^2

Even here, statement (1) is sufficient to answer the question! how is it insufficient!?

You can't multiply both sides of an inequality by a variable unless the variable is positive. The inequality may be reversed
e.g.
1>-4
Multiplying both sides by -1 we get-
-1<4 & not -1>4

Another way to solve this is to just simplify the equation
(x^2+1)/x>Y
or x+1/x>y

2. y>0
we have no idea what the value of x is. x can be positive but lesser/greater than y or negative which would make the expression negative.
Eliminate B & D

1. x=y
We don't know what are x & y so it's hard to precisely determine whether the expression will be greater or lesser than y

1+2
y>0 & x=y
Positive+Positive is Positive.
x+1/x=y+1/y
i.e. positive number + some fraction
so y+1/y>y

Hence C
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07 Oct 2011, 07:10
Statement 1... If x = 2, 2^2+1 =5. 5/2 = 2.5, given x=y=2
Then 2.5>2
If x = -2, (-2^2+1) = 5, 5/-2 = -2.5.
Then -2.5 > -2 ...not true... 1 insufficient.

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Re: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]

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02 Oct 2015, 06:10
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Re: If x ≠ 0, is (x^2 + 1)/x > y?   [#permalink] 02 Oct 2015, 06:10
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