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if x is NOT = 0, is (x^2 +1)/x > Y? (1) x = y (2) y>0

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if x is NOT = 0, is (x^2 +1)/x > Y? (1) x = y (2) y>0 [#permalink] New post 06 Oct 2011, 13:06
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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!?
[Reveal] Spoiler: OA

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Re: tough inequality [#permalink] New post 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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Re: tough inequality [#permalink] New post 06 Oct 2011, 13:30
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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

so answer = C
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Re: tough inequality [#permalink] New post 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

Start with statement 2, to make the solution simpler.
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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Re: tough inequality [#permalink] New post 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.
Re: tough inequality   [#permalink] 07 Oct 2011, 07:10
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