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DS: xy=1? 1.xyx=x 2.yxy=y

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DS: xy=1? 1.xyx=x 2.yxy=y [#permalink] New post 06 Jul 2004, 23:02
00:00
A
B
C
D
E

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(N/A)

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DS: xy=1?

1.xyx=x
2.yxy=y
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 [#permalink] New post 07 Jul 2004, 00:03
I want to warn people, D is wrong
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 [#permalink] New post 07 Jul 2004, 01:15
will go with E

1 x(xy-1)=0 either x=0 or xy=1
2 y(xy-1)=0 either y=0 or xy=1
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 [#permalink] New post 07 Jul 2004, 01:25
My try

a)X Y X = X, in this case, if X is -1 then Y is 1, and if X is 1 then y is 1. Hence, xy can have 2 values .. xy =-1 or xy =1

b) Y X Y=Y, the same as above. Y can be -1, or 1, and x is 1, therefore XY can be -1, or 1.

If x is other -1 and y is 1, then only one statement is satisfied. (Statement - A). And if y is -1 and x is 1, then it satisfies statement B.

Combining both, there is only one value, that is if X is 1, and y is 1 only then both satifies both the statments. If either of them is -1 or if both are -1, we do not satisfy both statements.


Answer is C.
Correct me, if wrong.
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Last edited by carsen on 07 Jul 2004, 04:30, edited 1 time in total.
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 [#permalink] New post 07 Jul 2004, 03:40
Will agree with Srijay here, (E) should be it.

(Would like to see how one would refute a common solution to two simultaneous equation, in our case common solution to two equations is xy=1)
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 [#permalink] New post 07 Jul 2004, 16:11
[quote="carsen"]My try

a)X Y X = X, in this case, if X is -1 then Y is 1, and if X is 1 then y is 1. Hence, xy can have 2 values .. xy =-1 or xy =1

b) Y X Y=Y, the same as above. Y can be -1, or 1, and x is 1, therefore XY can be -1, or 1.

Answer is C.
Correct me, if wrong.[/quote]

carsen : in part A if x =-1 then y = -1 and not 1
similarly in part B if y =-1 then x = -1 and not 1
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 [#permalink] New post 07 Jul 2004, 16:15
MBA,
the only place the common solution (xy = 1 ) failing is when x= y =0
so the answer I think is E
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 [#permalink] New post 07 Jul 2004, 17:42
Hi smandalika

I feel that your calculation is wrong.

You have mentioned

in part A if x =-1 then y = -1 and not 1
similarly in part B if y =-1 then x = -1 and not 1


1.xyx=x
2.yxy=y

If we substitute the values as mentioned by you...

If X is -1, then Y=-1, then we get the anser for statement A as -1
Statement 1 = -1 x -1 x -1 = -1 (this violate the original equation)

The same case for Statement 2.
If y=-1, and X=-1, then the final result will be -1 (-1 x -1 x -1 =-1). This again will violate the original statement 2.

Hence, if x is -1, then y has to 1, to satisfy statment 1, and similarly, for statement b, if y is -1, then x has to be 1 to satisfy the original statement 2.

Combing both, only one value can satisfy both equations, that is when, x is 1, and y is 1. Hence the answer is C.

I hope, i have explained better in here. If not, let me know, or perhaps, correct me, if my focus is wrong.
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 [#permalink] New post 07 Jul 2004, 18:40
Dude,
You are confusing me even more ....

consider the first eq
XYX = X
if X= -1

(-1)Y(-1) = -1
then Y = -1 and not 1

similarly

consider the second eq
YXY = Y
if Y= -1

(-1)X(-1) = -1
then X = -1 and not 1

what am I missing here????
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 [#permalink] New post 07 Jul 2004, 21:33
Hi Smandalika

Got ya point.

You know, I got my mind fixed to this equation instead ..

1.xyx=x => my mind got fixed to this equation as xyx=1
2.yxy=y => my mind got fixed this equation as yxy=1

My mistake. I did crap on the above explaination, with the misinterptn. Sorry about girl. I need to re-read the whole thing again. Thanks for correcting me.


The answer is E. (Man, i need to focus, and I have just 2 weeks to go).
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relax [#permalink] New post 08 Jul 2004, 05:08
that is OK Carsen.Dont worry too much about the test. Take a deep breath. Relax . you will do just fine.

Good luck with the test . Let us all know how it went.
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 [#permalink] New post 08 Aug 2004, 13:37
I think it is E. But OA was not posted..

Can some expert please comment?
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 [#permalink] New post 08 Aug 2004, 14:33
DS: xy=1?

1.xyx=x
2.yxy=y

My answer is C.

(1)
xyx = x
xyx - x = 0
x (xy-1) = 0
x = 0 OR xy = 1
Cannot say. Insufficient.

(2)
yxy = y
yxy - y = 0
y (xy-1) = 0
y = 0 OR xy = 1
Cannot say. Insufficient.


(Together)
xy = 1 is the solution present in both statements. So, I can give a certain answer ("Yes") with C.

My asnwer = C.
(Did I miss anything?)
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 [#permalink] New post 08 Aug 2004, 23:34
Answer is E for sure.

(1) xyx = x --> xy = 1 or x = 0 --> insufficient
(2) yxy = y --> xy = 1 or y = 0 --> insufficient

(1)+(2) --> xy = 1 or x=y=0 --> insufficient.

hardworker_indian: In your approach, you should take the combination of the 2 statements rather than the interception.
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 [#permalink] New post 14 Aug 2004, 22:20
Is it possible that A and B. tells use that all the signs are the same?

Ex. 5 * 1/5 * 5 = 5 or -5 * - 1/5 * -5 = -5 (XYX=X)

Ex. 1/5 * 5 * 1/5 = 1/5 or -1/5 * -5 * -1/5 = -5 (YXY=Y)

Can we say for sure that x = y or x = 1/y (or the inverse of this), either way X * Y would = 1

I might be wrong, this is my first post so go easy on me.

But this would rule out E ?
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 [#permalink] New post 17 Aug 2004, 13:53
Answer is E.

From 1) x = 0 OR xy = 1. Not sufficient. Rule out A and D.
From 2) y = 0 OR xy = 1. Again Not sufficient. Rule out B.
Combining the 2. xy = 1 OR xy = 0 (since x and y can both be 0). Again Not sufficient. Rule out C.

A, B, C, D are ruled out. We are left with E.
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 [#permalink] New post 17 Aug 2004, 19:46
bigtooth81 wrote:
Answer is E for sure.

(1) xyx = x --> xy = 1 or x = 0 --> insufficient
(2) yxy = y --> xy = 1 or y = 0 --> insufficient

(1)+(2) --> xy = 1 or x=y=0 --> insufficient.

hardworker_indian: In your approach, you should take the combination of the 2 statements rather than the interception.


I have a fundamental difficulty accepting E here. I could've sworn that Kaplan teaches you to look for interception -- hence my thought-flow was same as hardworking indian's.

What am I missing here?
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 [#permalink] New post 17 Aug 2004, 19:51
alright, I finally realized why answer is E here
x and y could both be zeros in both equations (thus eliminating the possibility that xy=1), and both equations would still equal 0.

the lesson learned here is that interception method does not apply when alternative solutions exist
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 [#permalink] New post 18 Aug 2004, 05:58
YES!!!!!!!!!!!!!! X & Y could be zero!!

Thanks guys
  [#permalink] 18 Aug 2004, 05:58
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