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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. Approach 1: I tried to simplify this question as follows x^2 + 1  xy > 0 x(yx) < 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(yx) [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^2xy + 1 < 0 x(xy) + 1 <0
Case 2: x is positive x^2+1 > xy x^2xy + 1 > 0 x(xy) + 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(xy) + 1 > 0
Therefore C is the answer.
The second approach looks more methodical but I want to understand why the first approach is INVALID (if it is so).
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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(yx) [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(yx) [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(yx) < 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(yx) [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 rephrasing . But how do u get A? st.1 says X=Y from rephrase X(yX)<1 X(XX)<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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Re: tough inequality [#permalink]
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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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Re: tough inequality [#permalink]
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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]
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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 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]
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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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