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can somebody explain this question

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can somebody explain this question

by davy420 » Wed Oct 31, 2007 11:19 pm
If xy > 0, does (x-1)(y-1) = 1

1. x+ y = xy
2. x = y

I thought the answer is E but it is A

I thought the only numbers that would work for solution 1 would be if you plug in 1 or 2 both x and y. ( x + y =xy) = 1+1=1 or 2+2=4, and 2 would work for the question but Since there are 2 choices either 1 or 2 and solution 2 seems like extra information. I thought the answer would be E. Can somebody explain this? Thanks
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Source: — Data Sufficiency |
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by samirpandeyit62 » Thu Nov 01, 2007 4:05 am
If xy > 0, does (x-1)(y-1) = 1

this means x & y are not 0
(x-1)(y-1) = 1
i.e xy -(x +y) +1 = 1

1. x+ y = xy

xy = x +y the expr xy -(x +y) +1 = 1

SUFF

2. x = y

so it is X^2 - 2x +1 = 1

here value may be 1 or not depending upon val of x

INSUFF
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Samir
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by gmatrant » Thu Nov 01, 2007 6:03 pm
samirpandeyit62 wrote:If xy > 0, does (x-1)(y-1) = 1

this means x & y are not 0
(x-1)(y-1) = 1
i.e xy -(x +y) +1 = 1

1. x+ y = xy

xy = x +y the expr xy -(x +y) +1 = 1

SUFF

2. x = y

so it is X^2 - 2x +1 = 1

here value may be 1 or not depending upon val of x

INSUFF
I solved the second part like this
2. x=y
Since x =y,
(x-1)^2 =1
Taking square root on both sides
so x-1 = -1 or x-1 =1
so x=0 or x =2, but x cannot be 0 as xy >0 . so x =2.
So this too is sufficient.

Samir, can you tell me where I am going wrong.
So thought D was the answer
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by samirpandeyit62 » Thu Nov 01, 2007 10:33 pm
Hi gmatrant,
There is a small trick in this question

the Q asks If xy > 0, does (x-1)(y-1) = 1

now here if x=y then (x-1)(y-1) = 1 will give us a quadratic eqn which will have two roots i.e 2,0 as u calculated

but here 2 is a root of the eqn which will satisfy X^2 - 2x +1 = 1 however if values of x are different viz 3,5..... etc then the eqn will not be satisfied i.e (x-1)(y-1) <> 1 so it INSUFF

we are not asked to find value of x which will satisfy X^2 - 2x +1 = 1

instead we are asked if x^2>0 i.e if x = any nos except 0 then Is X^2 - 2x +1 = 1

so this will be 1 for value =2 & a diff nos for value = any other nos

hence INSUFF

I hpe this makes sense.
Regards
Samir
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by gmatrant » Fri Nov 02, 2007 8:35 am
samirpandeyit62 wrote:Hi gmatrant,
There is a small trick in this question

the Q asks If xy > 0, does (x-1)(y-1) = 1

now here if x=y then (x-1)(y-1) = 1 will give us a quadratic eqn which will have two roots i.e 2,0 as u calculated

but here 2 is a root of the eqn which will satisfy X^2 - 2x +1 = 1 however if values of x are different viz 3,5..... etc then the eqn will not be satisfied i.e (x-1)(y-1) <> 1 so it INSUFF

we are not asked to find value of x which will satisfy X^2 - 2x +1 = 1

instead we are asked if x^2>0 i.e if x = any nos except 0 then Is X^2 - 2x +1 = 1

so this will be 1 for value =2 & a diff nos for value = any other nos

hence INSUFF

I hpe this makes sense.
hmm..yup..u rock dude..
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