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# DS: XY

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Manager
Joined: 01 Nov 2005
Posts: 126
Followers: 1

Kudos [?]: 24 [0], given: 0

DS: XY [#permalink]  07 Sep 2006, 09:10
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Attachments

25Q29.JPG [ 4.79 KiB | Viewed 689 times ]

SVP
Joined: 05 Jul 2006
Posts: 1519
Followers: 5

Kudos [?]: 114 [0], given: 39

B

FROM ONE

X^2/X(Y+1) < 1

X/(Y+1)<1 WE KNOW THAT Y+1 IS +VE

THERFORE

X<Y+1

WE CAN NEVER KNOW IF X<Y OR NOT

FROM TWO

XY/Y(Y-1) <1

X/(Y-1) < 1 AND SURE Y-1 IS +VE

THUS X<Y-1 THUS SURE X<Y

VP
Joined: 02 Jun 2006
Posts: 1267
Followers: 2

Kudos [?]: 42 [0], given: 0

Same calculation..

S1 is insufficient.

From S2: x < y-1 => x < y Sufficient.

yezz wrote:
B

FROM ONE

X^2/X(Y+1) < 1

X/(Y+1)<1 WE KNOW THAT Y+1 IS +VE

THERFORE

X<Y+1

WE CAN NEVER KNOW IF X<Y OR NOT

FROM TWO

XY/Y(Y-1) <1

X/(Y-1) < 1 AND SURE Y-1 IS +VE

THUS X<Y-1 THUS SURE X<Y

CEO
Joined: 20 Nov 2005
Posts: 2913
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
Followers: 18

Kudos [?]: 119 [0], given: 0

Either B or E. See the bold part below.

Question is
X-Y < 0 ???

St1: X^2/(XY+X) Since both X and Y are +ve both numerator and denominator are +ve and we can rewrite the equation as
X^2 < XY+X
X < Y+1
X-Y <1: INSUFF

St2: I am not sure if statement says XY/(Y^2-Y) OR (XY/Y^2)-Y

In case of XY/(Y^2-Y) <1
XY < Y^2 - Y
X < Y-1
X-Y < -1 : SUFF

In case of (XY/Y^2)-Y < 1
X/Y < Y
X < Y^2
X-Y^2 < 0 then X-Y could be +ve (For X = 5 and Y = 4) or -ve (for X = 5 and Y = 6): INSUFF
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
Joined: 11 May 2006
Posts: 260
Followers: 1

Kudos [?]: 6 [0], given: 0

i think its E

as 2) says (XY/Y^2)-Y < 1

i think there is no ambiguity because 1) specifically says

X^2/(XY+X) < 1

so the question writer knows to put the parantheses in the form a/(b). So we cannot assume 2 to be ambiguous.
Manager
Joined: 08 Jul 2006
Posts: 90
Followers: 1

Kudos [?]: 0 [0], given: 0

Statement 1:

x (x)
______ < 1

x (y + 1)

Therefore, x < y + 1

When x is 2, y could also be 2, 3, 4, 4.5 etc.
INSUFF

Statement 2 becomes,

xy < y^2 - y
then, xy + y < y^2
and finally, y(x+1) < y^2

x+1<y

In order for this statement to hold true, Y must always be greater than X. Right?

Ans. B
Intern
Joined: 04 May 2006
Posts: 40
Followers: 0

Kudos [?]: 0 [0], given: 0

I think it is E because 2) is written as XY/Y^2 - Y not XY/(Y^2-Y)
Director
Joined: 28 Dec 2005
Posts: 758
Followers: 1

Kudos [?]: 8 [0], given: 0

This is just a badly written question.....I'm pretty sure the questions on the GMAT should leave no ambiguity about what's in the denominator of an expression vs what's in another expression.

I went with B, but I see why the answer is E
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