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Re: If X>1 and Y>1, is X<Y? [#permalink]
27 Jan 2012, 13:12

1

This post received KUDOS

Expert's post

Baten80 wrote:

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.

If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]
27 Jan 2012, 13:24

Expert's post

metallicafan wrote:

Baten80 wrote:

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.

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.

Re: If X>1 and Y>1, is X<Y? [#permalink]
05 Jul 2013, 01:43

Bunuel wrote:

Baten80 wrote:

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.

Re: If X>1 and Y>1, is X<Y? [#permalink]
05 Jul 2013, 02:01

Expert's post

RohanKhera wrote:

Bunuel wrote:

Baten80 wrote:

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

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. _________________

Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]
16 Sep 2014, 18:39

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