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xy=? 1) y=x-1 2) y^2=-|x-1|

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xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 26 Jul 2017, 01:10
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Question Stats:

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\(xy=?\)

1) \(y=x-1\)
2) \(y^2=-|x-1|\)

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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 26 Jul 2017, 02:11
2
MathRevolution wrote:
\(xy=?\)

1) \(y=x-1\)
2) \(y^2=-|x-1|\)


(1) We don't know the value of x and y. Insufficient,

(2) Since \(y^2 \geq 0 \quad \forall y \in R\) and \(-|x-1| \leq 0 \quad \forall x \in R\), we have \(y^2=-|x-1|=0 \implies y=0\) and \(x=1 \implies xy=0\). Sufficient,

The answer is B.
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xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 26 Jul 2017, 03:49
nguyendinhtuong wrote:
MathRevolution wrote:
\(xy=?\)

1) \(y=x-1\)
2) \(y^2=-|x-1|\)


(1) We don't know the value of x and y. Insufficient,

(2) Since \(y^2 \geq 0 \quad \forall y \in R\) and \(-|x-1| \leq 0 \quad \forall x \in R\), we have \(y^2=-|x-1|=0 \implies y=0\) and \(x=1 \implies xy=0\). Sufficient,

The answer is B.


for 2) there could be other pairs of values of x and y which could satisfy the equation. For example the only possibility is the below equation
y^2=-|x-1|==> y^2= x-1==> which could mean different x and y pairs
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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 26 Jul 2017, 11:57
Supermaverick wrote:
nguyendinhtuong wrote:
MathRevolution wrote:
\(xy=?\)

1) \(y=x-1\)
2) \(y^2=-|x-1|\)


(1) We don't know the value of x and y. Insufficient,

(2) Since \(y^2 \geq 0 \quad \forall y \in R\) and \(-|x-1| \leq 0 \quad \forall x \in R\), we have \(y^2=-|x-1|=0 \implies y=0\) and \(x=1 \implies xy=0\). Sufficient,

The answer is B.


for 2) there could be other pairs of values of x and y which could satisfy the equation. For example the only possibility is the below equation
y^2=-|x-1|==> y^2= x-1==> which could mean different x and y pairs


Other values are not possible. y^2 is always greater than or equal to zero. The same is true for |x-1|. As there is the negative sign before |x-1|, the only possible option is when y^2=0 which means y is equal to 0. Also, |x-1|=0 which means x=0.
I hope it helps :)
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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 28 Jul 2017, 01:15
==> \(a^2+b^2=0\) or \(|a|+|b|=0\) is satisfied by \(a=b=0\) only. Thus, according to the same logic, \(a^2+|b|=0\) also must be \(a=b=0\). Then, you get 2) \(y^2=-|x-1|, y^2+|x-1|=0\), which is y=0 and x=1.

The answer is B.
Answer: B
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xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post Updated on: 30 Jul 2017, 05:51
3
MathRevolution wrote:
\(xy=?\)

1) \(y=x-1\)
2) \(y^2=-|x-1|\)



We need to find product of x and y



1)
y = x - 1

we do not know the value of either of x or y

Hence Insufficient
2)

y^2 = - | x-1 |


Case 1 : when | x - 1 | is negative

y^2 = -[-(x - 1)]
y^2 = x - 1 ....................equation 1

Case 2 : when | x - 1 | is positive

y^2 = -[(x - 1)]
y^2 = -x + 1 ....................equation 2

equation 1 = equation 2 (as both are equal to y^2 )

x-1 = -x + 1
x = 1

Now keeping value of x in any of the 2 equations

y^2 = -x + 1 (Now x = 1)
y^2 = -1 + 1
y^2 = 0 so y is also 0

It means x * y = 0

Hence B

Thanks :)

Originally posted by rocko911 on 29 Jul 2017, 01:47.
Last edited by rocko911 on 30 Jul 2017, 05:51, edited 1 time in total.
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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 29 Jul 2017, 04:59
Here we have to assume that Y^2 shouldn't be negative? In general in GMAT square of any number cant be negative?
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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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New post 29 Jul 2017, 05:08
ashokjha1986 wrote:
Here we have to assume that Y^2 shouldn't be negative? In general in GMAT square of any number cant be negative?



Not only in GMAT , in real world numbers and mathematics , SQUARE OF ANY NUMBER IS ALWAYS POSITIVE
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Re: xy=? 1) y=x-1 2) y^2=-|x-1|  [#permalink]

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Re: xy=? 1) y=x-1 2) y^2=-|x-1|   [#permalink] 26 Mar 2019, 07:21
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