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If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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Updated on: 11 Nov 2017, 09:17
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Originally posted by Baten80 on 27 Jan 2012, 13:44.
Last edited by Bunuel on 11 Nov 2017, 09:17, edited 2 times in total.
Edited the OA.



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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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27 Jan 2012, 14:12



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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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27 Jan 2012, 14:14
Baten80 wrote: If X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2Y<1
I did B. But it is not OA. In statement 2, the inequality is well written? Is this the original expresion? (XY/Y^2) Y<1 or? XY/(Y^2Y)<1 If it is the first scenario, the answer is E. Please confirm.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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27 Jan 2012, 14:24
metallicafan wrote: Baten80 wrote: If X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2Y<1
I did B. But it is not OA. In statement 2, the inequality is well written? Is this the original expresion? (XY/Y^2) Y<1 or? XY/(Y^2Y)<1 If it is the first scenario, the answer is E. Please confirm. If (2) is \(\frac{xy}{y^2}y<1\), then the answer is indeed E. If x>1 and y>1, is x<y?(1) \(\frac{x^2}{xy+x}<1\) > reduce by \(x\): \(\frac{x}{y+1}<1\) > cross multiply, notice that we can safely do that since \(y+1>0\): \(x<y+1\) > if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient. (2) \(\frac{xy}{y^2}y<1\) > reduce by \(y\): \(\frac{x}{y}y<1\) > \(x<y^2+y\) > if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient. (1)+(2) If \(x=2\) and \(y=3\), then the answer is YES but If \(x=2\) and \(y=2\) (notice that this set of numbers satisfy both statements), then the answer is NO. Not Sufficient. Answer: E.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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05 Jul 2013, 02:43
Bunuel wrote: Baten80 wrote: If X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2Y<1
I did B. But it is not OA. OA is wrong. Answer is B. If x>1 and y>1, is x<y?(1) \(\frac{x^2}{xy+x}<1\) > reduce by \(x\): \(\frac{x}{y+1}<1\) > cross multiply, notice that we can safely do that since \(y+1>0\): \(x<y+1\) > if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient. (2) \(\frac{xy}{y^2y}<1\) > reduce by \(y\): \(\frac{x}{y1}<1\) > cross multiply, notice that we can safely do that since \(y1>0\): \(x<y1\) > \(x+1<y\) (\(y\) is more than \(x\) plus 1) > \(y>x\). Sufficient. Answer: B. Hi Bunuel, Can you explain statement 2 using numbers ? Rohan



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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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05 Jul 2013, 03:01
RohanKhera wrote: Bunuel wrote: Baten80 wrote: If X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2Y<1
I did B. But it is not OA. OA is wrong. Answer is B. If x>1 and y>1, is x<y?(1) \(\frac{x^2}{xy+x}<1\) > reduce by \(x\): \(\frac{x}{y+1}<1\) > cross multiply, notice that we can safely do that since \(y+1>0\): \(x<y+1\) > if \(x=2\) and \(y=2\), then the answer is NO but if \(x=2\) and \(y=3\), then the answer is YES. Not sufficient. (2) \(\frac{xy}{y^2y}<1\) > reduce by \(y\): \(\frac{x}{y1}<1\) > cross multiply, notice that we can safely do that since \(y1>0\): \(x<y1\) > \(x+1<y\) (\(y\) is more than \(x\) plus 1) > \(y>x\). Sufficient. Answer: B. Hi Bunuel, Can you explain statement 2 using numbers ? Rohan Number plugging is not the best approach to prove that a statement IS sufficient. On DS questions when plugging numbers, goal is to prove that the statement is NOT sufficient.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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05 Jul 2013, 03:40
Hi bunuel In first statement if x<y+1 can't we say that x<y because even though we add 1 to y x is less than y Can u plz explain Thanks Posted from GMAT ToolKit



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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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05 Jul 2013, 03:45
[quote="sravanigayatri"]Hi bunuel In first statement if xthan y Can u plz
They both could be equal.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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05 Jul 2013, 03:48



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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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23 Jan 2016, 10:48
The question basically asks if xy<0 Statement 1. (x^2)/(x(y+1))<1 Since x and y both>1 then divide by x and multuply by (y+1) to get x<y+1 ==>xy<1 Hence xy can be 0.5 for example or can be 2. Not sufficient Statement 2. (xy)/(y(y1)) Again since both x and y are positive we can divide by y and multiply by (y1) to get x<y1 Hence xy<1 and thus <0 Sufficient
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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23 Jan 2016, 23:42
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 In the original condition, there are 2 variables(x,y) and 1 equation(x>1 and y>1), which should match with the number of equations. so you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer. In 1), from x^2/(xy+x)<1, x/(y+1)<1, x<y+1 is possible but you can't figure out x<y, which is not sufficient. In 2), from xy/(y^2y)<1, x/(y1)<1, x<y1 is possible. Also, in x<y1<y, it is always x<y, which is yes and sufficient. Therefore, the answer is B. > For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1 [#permalink]
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2y)<1
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