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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
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).
Re: If x ≠ 0, is (x^2 + 1)/x > y? [#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 _________________
Re: If x ≠ 0, is (x^2 + 1)/x > y? [#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
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 _________________
Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.
Re: tough inequality [#permalink]
07 Oct 2011, 06:10
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: If x ≠ 0, is (x^2 + 1)/x > y? [#permalink]
02 Oct 2015, 05:10
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