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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) Updated on: 25 Jan 2022, 06:26
Originally posted by MHIKER on 27 Jan 2012, 13:44.
Last edited by Bunuel on 25 Jan 2022, 06:26, edited 3 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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If \(x > 1\) and \(y > 1\), is \(x < y\) ?


(1) \(\frac{x^2}{xy + x} < 1\)

(2) \(\frac{xy}{y^2 - y} < 1\)
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B
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 27 Jan 2012, 14:12
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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 \(y\): \(\frac{x}{y-1}<1\);

Cross multiply (notice that we can safely do that since \(y-1>0\)): \(x<y-1\);

Re-arrange: \(x+1<y\) (\(y\) is more than \(x\) plus 1), so \(y>x\). Sufficient.


Answer: B.
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 27 Jan 2012, 14:14
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Baten80
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.
Show more

In statement 2, the inequality is well written?
Is this the original expresion?
(XY/Y^2) -Y<1

or?
XY/(Y^2-Y)<1

If it is the first scenario, the answer is E.
Please confirm.
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 27 Jan 2012, 14:24
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Baten80
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<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^2-Y)<1

If it is the first scenario, the answer is E.
Please confirm.
Show more

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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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 05 Jul 2013, 02:43
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If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<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) \(\frac{xy}{y^2-y}<1\) --> reduce by \(y\): \(\frac{x}{y-1}<1\) --> cross multiply, notice that we can safely do that since \(y-1>0\): \(x<y-1\) --> \(x+1<y\) (\(y\) is more than \(x\) plus 1) --> \(y>x\). Sufficient.

Answer: B.
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Hi Bunuel,

Can you explain statement 2 using numbers ?

Rohan
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 05 Jul 2013, 03:01
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Bunuel
Baten80
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<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) \(\frac{xy}{y^2-y}<1\) --> reduce by \(y\): \(\frac{x}{y-1}<1\) --> cross multiply, notice that we can safely do that since \(y-1>0\): \(x<y-1\) --> \(x+1<y\) (\(y\) 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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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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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 05 Jul 2013, 03:40
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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

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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 05 Jul 2013, 03:48
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sravanigayatri
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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No, unfortunately that's not correct: x<y+1, does not necessarily means that x<y.

Consider x=2 and y=1.5 --> x<y+1 and x>y.
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 23 Jan 2016, 23:42
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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^2-y)<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^2-y)<1, x/(y-1)<1, x<y-1 is possible. Also, in x<y-1<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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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 27 Jan 2020, 23:43
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Baten80
If x>1 and y>1, is x<y?

(1) x^2/(xy+x)<1

(2) xy/(y^2-y)<1

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I did B. But it is not OA.
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The solution is as attached

Answer: Option B
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If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y) 25 Jul 2022, 01:01
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My 2 cents on this:

Won't add more than what has been mentioned, just two points:

--> We know the sign of the variables given that x>1 y>1 hence x,y are positive. So cross multiplication can happen in inequalities if sign is known.
--> Best to use numbers in this case. Will only take 30 seconds to rule out/find answer.

B
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