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# If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?

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If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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31 Jul 2014, 06:03
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If (x − y)^2 = x^2 − y^2, what is the value of nonzero integer xy?

(1) x = 5
(2) x − y = 0

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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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31 Jul 2014, 06:07
4
OA:

A. The left side of the equation expands to x^2−2xy+y^2=x^2−y^2. From here, you can subtract x^2 from both sides, simplifying to:

y^2−2xy=−y^2
This allows you to move all terms to the left:

2y^2−2xy=0
And then you can divide both sides by 2:

y^2−xy=0
And then because the stimulus guarantees that y is not zero, you can divide both sides by y to see that:

y−x=0 QUERY: But, y can also equal zero (y=0). If we take the value y=0 then xy=0 -- and not 25

Because statement 1 gives you the value of x, it is sufficient as x = y, so xy = (5)(5) = 25. Note that statement 2 is just a restatement of what you can prove from the stimulus, so it adds no new value and is not sufficient. The correct answer is A.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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31 Jul 2014, 06:10
2
pratikshr wrote:
If (x − y)^2 = x^2 − y^2, what is the value of nonzero integer xy?

(1) x = 5
(2) x − y = 0

OA:

A. The left side of the equation expands to x^2−2xy+y^2=x^2−y^2. From here, you can subtract x^2 from both sides, simplifying to:

y^2−2xy=−y^2
This allows you to move all terms to the left:

2y^2−2xy=0
And then you can divide both sides by 2:

y^2−xy=0
And then because the stimulus guarantees that y is not zero, you can divide both sides by y to see that:

y−x=0 QUERY: But, y can also equal zero (y=0). If we take the value y=0 then xy=0 -- and not 25

Because statement 1 gives you the value of x, it is sufficient as x = y, so xy = (5)(5) = 25. Note that statement 2 is just a restatement of what you can prove from the stimulus, so it adds no new value and is not sufficient. The correct answer is A.

The stem explicitly states that xy is nonzero integer, which means that neither x nor y could be 0.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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31 Jul 2014, 06:13
Thanks Bunuel. That was super quick.

Kudos.
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If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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28 Sep 2015, 10:43
1
2
If (x − y)^2 = x^2 − y^2, what is the value of nonzero integer xy?

(1) x = 5
(2) x − y = 0

x^2+y^2-2xy=x^2-y^2
2y^2=2xy
y=x (Now we understand from the given equation, non zero solution of the equation x=y, so as long as we know either one of those it will be sufficient to solve the problem)

Statement 1 is sufficient, since it gives the value of x, and from that we can calculate xy as 5*5=25
Statement 2 is not sufficient, since it does not give us any value for x or y.

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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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12 Oct 2015, 14:16
Hi, i tried solving it as shown below. Can someone guide what mistake i made here?
(X-y)(x-y) = (x-y)(x+y)
S, (x-y)=(x+y)
0 = 2y
Y=0
St 1-
X=5. So, x(tens digit)y(units digit)= 50
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If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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12 Oct 2015, 14:26
1
Meetup wrote:
Hi, i tried solving it as shown below. Can someone guide what mistake i made here?
(X-y)(x-y) = (x-y)(x+y)
S, (x-y)=(x+y)
0 = 2y
Y=0
St 1-
X=5. So, x(tens digit)y(units digit)= 50

Be very careful with cancelling variables in equations or inequalities.

After you get, (x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0 ---> either y=0 or xy=0 but as xy is a NON ZERO integer, y can not be =0. Thus x=y. So for knowing the value of xy, either knowing the value of x or y will be sufficient.

Per statement 1, x=5, exactly what you need. Sufficient.

Per statement 2, x=y , this is the same as the given information and will thus make it NOT sufficient.

You can not cancel (x-y) from both sides as cancelling this expression means that you are assuming that x $$\neq$$ y

Additionally, xy doesnt mean that x is the 10s digit and y is the unit digit. xy will give you a non zero integer value.

Hope this helps.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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12 Oct 2015, 16:45
Thanks a lot for the quick revert,

didnt get how did u get this
(x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0
Moreover, for the below equation,
(x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0
I see it as below,
(x-y) [x-y-x-y]=0 --- (x-y) [ -2y] = 0 and not +2y as mentioned in your post... Can you see it pls? Thanks.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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12 Oct 2015, 17:02
Meetup wrote:
Thanks a lot for the quick revert,

didnt get how did u get this
(x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0
Moreover, for the below equation,
(x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0
I see it as below,
(x-y) [x-y-x-y]=0 --- (x-y) [ -2y] = 0 and not +2y as mentioned in your post... Can you see it pls? Thanks.

Sure look below.

You are given, $$(x-y)^2=x^2-y^2$$---> $$(x-y)(x-y)=(x+y)(x-y)$$ ...(applying the formulae, $$a^2=a*a$$ and $$a^2-b^2 = (a+b)(a-b)$$)

--->$$(x-y)(x-y)-(x+y)(x-y) = 0$$ ---> $$(x-y)*[(x-y)-(x+y)] = 0$$ ---> $$(x-y)*[x-y-x-y] = 0$$ etc

$$(x-y) [x-y-x-y]=0$$--->$$(x-y) (-2y)=0$$---> multiply both sides by (-1/2) ---> $$(x-y)(y)=0$$ ----> either $$y=0$$ or $$x-y=0$$ (--> $$x=y$$)

Hope this helps.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 12:37
Engr2012 wrote:
Meetup wrote:
Thanks a lot for the quick revert,

didnt get how did u get this
(x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0
Moreover, for the below equation,
(x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0
I see it as below,
(x-y) [x-y-x-y]=0 --- (x-y) [ -2y] = 0 and not +2y as mentioned in your post... Can you see it pls? Thanks.

Sure look below.

You are given, $$(x-y)^2=x^2-y^2$$---> $$(x-y)(x-y)=(x+y)(x-y)$$ ...(applying the formulae, $$a^2=a*a$$ and $$a^2-b^2 = (a+b)(a-b)$$)

--->$$(x-y)(x-y)-(x+y)(x-y) = 0$$ ---> $$(x-y)*[(x-y)-(x+y)] = 0$$ ---> $$(x-y)*[x-y-x-y] = 0$$ etc

$$(x-y) [x-y-x-y]=0$$--->$$(x-y) (-2y)=0$$---> multiply both sides by (-1/2) ---> $$(x-y)(y)=0$$ ----> either $$y=0$$ or $$x-y=0$$ (--> $$x=y$$)

Hope this helps.

Hi, Can you please explain why we can not cancel out (x-y) from both sides? This is not inequality so sign should not matter, right?
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 12:51
neeraj609 wrote:
Engr2012 wrote:
Meetup wrote:
Thanks a lot for the quick revert,

didnt get how did u get this
(x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0
Moreover, for the below equation,
(x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0
I see it as below,
(x-y) [x-y-x-y]=0 --- (x-y) [ -2y] = 0 and not +2y as mentioned in your post... Can you see it pls? Thanks.

Sure look below.

You are given, $$(x-y)^2=x^2-y^2$$---> $$(x-y)(x-y)=(x+y)(x-y)$$ ...(applying the formulae, $$a^2=a*a$$ and $$a^2-b^2 = (a+b)(a-b)$$)

--->$$(x-y)(x-y)-(x+y)(x-y) = 0$$ ---> $$(x-y)*[(x-y)-(x+y)] = 0$$ ---> $$(x-y)*[x-y-x-y] = 0$$ etc

$$(x-y) [x-y-x-y]=0$$--->$$(x-y) (-2y)=0$$---> multiply both sides by (-1/2) ---> $$(x-y)(y)=0$$ ----> either $$y=0$$ or $$x-y=0$$ (--> $$x=y$$)

Hope this helps.

Hi, Can you please explain why we can not cancel out (x-y) from both sides? This is not inequality so sign should not matter, right?

x - y could be 0 and we cannot divide by 0.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 12:52
1
neeraj609 wrote:
Engr2012 wrote:
Meetup wrote:
Thanks a lot for the quick revert,

didnt get how did u get this
(x-y)(x-y) = (x-y)(x+y) ---> (x-y) [x-y-x-y]=0
Moreover, for the below equation,
(x-y) [x-y-x-y]=0 ---> (x-y)(2y)=0
I see it as below,
(x-y) [x-y-x-y]=0 --- (x-y) [ -2y] = 0 and not +2y as mentioned in your post... Can you see it pls? Thanks.

Sure look below.

You are given, $$(x-y)^2=x^2-y^2$$---> $$(x-y)(x-y)=(x+y)(x-y)$$ ...(applying the formulae, $$a^2=a*a$$ and $$a^2-b^2 = (a+b)(a-b)$$)

--->$$(x-y)(x-y)-(x+y)(x-y) = 0$$ ---> $$(x-y)*[(x-y)-(x+y)] = 0$$ ---> $$(x-y)*[x-y-x-y] = 0$$ etc

$$(x-y) [x-y-x-y]=0$$--->$$(x-y) (-2y)=0$$---> multiply both sides by (-1/2) ---> $$(x-y)(y)=0$$ ----> either $$y=0$$ or $$x-y=0$$ (--> $$x=y$$)

Hope this helps.

Hi, Can you please explain why we can not cancel out (x-y) from both sides? This is not inequality so sign should not matter, right?

It does not matter whether you are working with inequalities or equations. The concept remains the same. You can not cancel variables out until you know for sure about the signs or their relative values.

In this case, if you cancel x-y from both sides, you are assuming the x can not be equal to y. This is a very dangerous assumption as you can clearly see from the correct solution. You were not given in the main question stem that 'x can not be equal to y'.

If you do cancel out such variables , you will end up with a case that might be a "maybe true" in a DS question. A "maybe true" is not sufficient in DS but after you cancel out the variabes, you will end up assuming that the case is 'always true', leading to wrong selection of option.

Hope this helps.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 13:00
Sure look below.

You are given, $$(x-y)^2=x^2-y^2$$---> $$(x-y)(x-y)=(x+y)(x-y)$$ ...(applying the formulae, $$a^2=a*a$$ and $$a^2-b^2 = (a+b)(a-b)$$)

--->$$(x-y)(x-y)-(x+y)(x-y) = 0$$ ---> $$(x-y)*[(x-y)-(x+y)] = 0$$ ---> $$(x-y)*[x-y-x-y] = 0$$ etc

$$(x-y) [x-y-x-y]=0$$--->$$(x-y) (-2y)=0$$---> multiply both sides by (-1/2) ---> $$(x-y)(y)=0$$ ----> either $$y=0$$ or $$x-y=0$$ (--> $$x=y$$)

Hope this helps.[/quote]

Hi, Can you please explain why we can not cancel out (x-y) from both sides? This is not inequality so sign should not matter, right?[/quote]

It does not matter whether you are working with inequalities or equations. The concept remains the same. You can not cancel variables out until you know for sure about the signs or their relative values.

In this case, if you cancel x-y from both sides, you are assuming the x can not be equal to y. This is a very dangerous assumption as you can clearly see from the correct solution. You were not given in the main question stem that 'x can not be equal to y'.

If you do cancel out such variables , you will end up with a case that might be a "maybe true" in a DS question. A "maybe true" is not sufficient in DS but after you cancel out the variabes, you will end up assuming that the case is 'always true', leading to wrong selection of option.

Hope this helps.[/quote]

Hi, thanks alot for the response!

I am sure I am missing something very basic here. Even if x=y then, (x-y) = 0 which satisfies the equation given right?
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 13:06
1
neeraj609 wrote:

Hi, thanks alot for the response!

I am sure I am missing something very basic here. Even if x=y then, (x-y) = 0 which satisfies the equation given right?

No. If you look carefully, the question stem mentions that xy = integer and NON ZERO value. Now if I give you the scenario x=y=0, then this goes against this given information and is hence not allowed.

Additionally, as Bunuel mentioned above, if x-y=0, then you will end up dividing the given expression $$(x-y)^2=x^2-y^2$$ by $$(x-y)$$ in order to 'eliminate' $$(x-y)$$. You can not divide by a number =0 as it becomes not defined function in maths.

Hope this helps.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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18 Oct 2015, 14:05
Engr2012 wrote:
neeraj609 wrote:

Hi, thanks alot for the response!

I am sure I am missing something very basic here. Even if x=y then, (x-y) = 0 which satisfies the equation given right?

No. If you look carefully, the question stem mentions that xy = integer and NON ZERO value. Now if I give you the scenario x=y=0, then this goes against this given information and is hence not allowed.

Additionally, as Bunuel mentioned above, if x-y=0, then you will end up dividing the given expression $$(x-y)^2=x^2-y^2$$ by $$(x-y)$$ in order to 'eliminate' $$(x-y)$$. You can not divide by a number =0 as it becomes not defined function in maths.

Hope this helps.

Perfect, thank you so much for explaining this!!!
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If (x−y)2=x2−y2, what is the value of nonzero integer xy? (1) x=5 (2)  [#permalink]

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07 Dec 2015, 03:40
If (x−y)^2=x^2−y^2, what is the value of nonzero integer xy?

(1) x=5
(2) x−y=0
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Re: If (x−y)2=x2−y2, what is the value of nonzero integer xy? (1) x=5 (2)  [#permalink]

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07 Dec 2015, 03:50
2
If (x−y)^2=x^2−y^2, what is the value of nonzero integer xy?

(1) x=5
(2) x−y=0

Hi,

after simplification, the equation tells you that either y=0 or y=x..
but it is given that both x and y are non zero integer, so x=y..
we are asked value of xy, so we require value of x or y..

1) x=5 suff..
2) x-y=0.. it does not give us any new info.. insuff

A
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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07 Dec 2015, 04:30
If (x−y)^2=x^2−y^2, what is the value of nonzero integer xy?

(1) x=5
(2) x−y=0

Merging similar topics. Please refer to the discussion above.
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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?  [#permalink]

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02 Oct 2018, 10:10
Statement 1.

Quote:
The left side of the equation expands to $$x^2−2xy+y^2=x^2−y^2$$
From here, we can subtract $$x^2$$ from both sides, simplifying to:

$$y^2−2xy=−y^2$$

This allows us to move all terms to the left:
$$2y^2−2xy=0$$

And then we can divide both sides by 2:

$$y^2−xy=0$$
And then because the stimulus guarantees that y is not zero, we can divide both sides by y to see that:

y−x=0

Because statement 1 gives us the value of x, it is sufficient as x=y,

so$$, xy=(5)(5)=25$$

Statement 2
Quote:
It is just a restatement of what we can prove from the stimulus, so it adds no new value and is not sufficient. The correct answer is A

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Re: If (x-y)^2=x^2-y^2, what is the value of nonzero integer xy?   [#permalink] 02 Oct 2018, 10:10
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