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I need a strategy for this one.

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I need a strategy for this one. [#permalink] New post 09 Sep 2009, 09:11
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Question Stats:

53% (02:20) correct 47% (01:12) wrong based on 32 sessions
if |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

a) x-y> 0
b) x-y< 0
c) x+y> 0
d) xy>0
e) xy<0


this one is from manhattan, the answer is
[Reveal] Spoiler:
E


I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you
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Re: I need a strategy for this one. [#permalink] New post 09 Sep 2009, 09:44
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|x| - |y| = |x+y|
=> (|x| - |y|)^2 = |x+y|^2 = (x+y)^2
=> x^2 - 2|x||y| + y^2 = x^2 + 2xy + y^2
=> |xy| = -xy
because xy != 0 so xy <0
Ans: E


Pedros wrote:
if |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

a) x-y> 0
b) x-y< 0
c) x+y> 0
d) xy>0
e) xy<0


this one is from manhattan, the answer is
[Reveal] Spoiler:
E


I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you
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Re: I need a strategy for this one. [#permalink] New post 09 Sep 2009, 10:35
there can be 4 instances:-
i)x +,y -....|x| - |y| = |x+y| => x-y=x-y
ii) both +... x-y=x+y.. not possible as xy not equal zero
iii) both -....x-y=-(x+y) .. not possible as xy not equal zero
iv) x-,y+...x-y=y-x......x=y..
therefore x and y are of different signs, when multiplied ,it should be -ive
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Re: I need a strategy for this one. [#permalink] New post 14 Jun 2011, 02:39
solution exists only for x,y<0 which is x+y = 0
thus xy < 0
E
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Re: I need a strategy for this one. [#permalink] New post 14 Jun 2011, 20:05
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Expert's post
Pedros wrote:
if |x| - |y| = |x+y| and xy not equal zero , which of the following must be true ?

a) x-y> 0
b) x-y< 0
c) x+y> 0
d) xy>0
e) xy<0


this one is from manhattan, the answer is
[Reveal] Spoiler:
E


I want to know wether there is a rule involved that i am missing, or an effective strategy to tackle that kind of questions

Thank you


mnpqxyzt has given a great solution above. I would like to add here that it is possible that it doesn't occur to you that you should square both sides. If you do get stuck with such a question, notice that it says 'which of the following MUST be true'. So as a back up, you can rely on plugging in numbers. If you get even one set of values for which the condition does not hold, the condition is not your answer.

|x| - |y| = |x+y|
First set of non-zero values that come to mind is x = 1, y = -1
This set satisfies only options (A) and (E).
Now, the set x = -1, y = 1 will also satisfy the given equation.
But this set will not satisfy option (A).
Hence answer (E).
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Re: I need a strategy for this one. [#permalink] New post 18 Jun 2011, 20:49
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.
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Re: I need a strategy for this one. [#permalink] New post 19 Jun 2011, 03:57
Expert's post
AnkitK wrote:
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.


Ok.
Given: |x| - |y| = |x+y|
There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1
a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True
b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate
c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate
d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate
e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y.
We need to eliminate one of them.
They are
a) x-y> 0
e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but
a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate
e) xy<0................. (-1)(1) < 0; True
Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y|
when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y|
when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)
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Re: I need a strategy for this one. [#permalink] New post 19 Jun 2011, 07:05
Thnkx for the wonderfull explanation karishma.
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Re: I need a strategy for this one. [#permalink] New post 21 Jun 2011, 09:25
VeritasPrepKarishma wrote:
AnkitK wrote:
Krishma can you pls explain how to eliminate answer choices precisely.How can xy<0 not be true when one is negative and the other positive.


Ok.
Given: |x| - |y| = |x+y|
There are infinite set of values for x and y that satisfy this equation. Let us try one of them. Say x = 1, y = -1
a) x-y> 0 .......... 1 - (-1) > 0; 2 > 0; True
b) x-y< 0........... 1 - (-1) < 0; 2 < 0; Eliminate
c) x+y> 0........... 1 + (-1) > 0; 0 > 0; Eliminate
d) xy>0.............. 1 *(-1) > 0; -1 > 0; Eliminate
e) xy<0.............. 1 *(-1) < 0; -1 < 0; True

So I have two options that satisfy the assumed values of x and y.
We need to eliminate one of them.
They are
a) x-y> 0
e) xy<0

We see that x = 1, y = -1 satisfies both these inequalities. But option (a) is not symmetric i.e. if you interchange the values of x and y, it will not hold. That is, if x = -1 and y = 1, our original equation |x| - |y| = |x+y| is still satisfied but
a) x-y> 0 .............. (-1) - (1) > 0; -2>0; False. Eliminate
e) xy<0................. (-1)(1) < 0; True
Since option (e) still holds, it is the answer.

xy<0 is certainly true when one of them is negative and the other is positive.

Takeaways:

|x| + |y| = |x+y|
when x and y have the same signs - either both are positive or both are negative (or one or both of them are 0)

|x| - |y| = |x+y|
when x and y have opposite signs - one is positive, the other negative (or y is 0 or both are 0)


Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y|
if and only if
1) Both have opposite signs and
2) |x| >= |y|

because
x, y, |x|-|y| , |x+y|
5 ,-6, -1,1
5 ,-5 , 0 , 0
5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur
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Re: I need a strategy for this one. [#permalink] New post 21 Jun 2011, 10:42
Expert's post
krishp84 wrote:

Karishma, I think the 2nd takeaway has some extra condition missed out :

|x|-|y|=|x+y|
if and only if
1) Both have opposite signs and
2) |x| >= |y|

because
x, y, |x|-|y| , |x+y|
5 ,-6, -1,1
5 ,-5 , 0 , 0
5, -4, 1 , 1

so, |5| cannot be greater than |-6| for the condition to occur


The takeaway is that if x and y satisfy the condition |x|-|y|=|x+y|, then they must have opposite signs (or y is 0 or both are 0).

But, x and y having opposite signs is not sufficient to satisfy the condition |x|-|y|=|x+y|. As you said, in that case we will also need to check for their absolute values. (Good thinking, btw)
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Re: I need a strategy for this one. [#permalink] New post 21 Jun 2011, 11:40
Great explanation!
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Re: I need a strategy for this one. [#permalink] New post 08 Jul 2011, 20:23
I plugged in numbers:

X = -6, Y =3 and X = 6, Y = -3. The answer is E.
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Re: I need a strategy for this one.   [#permalink] 08 Jul 2011, 20:23
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