If x > 1 and y > 1, is x < y ? (1) x^2/(xy + x) < 1 (2) xy/(y^2 - y)

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.

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.

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

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.

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.

The solution is as attached

Answer: Option B

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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