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M32-07

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If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



(1) \(|xy|\) is NOT a square of an integer

(2) Point \((x, y)\) is above x-axis
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Re M32-07 [#permalink]

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Official Solution:


If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



This is a hard question. You should pay attention to every detail and read the solution very carefully



First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.



Next, \(-x \leq x\) implies that \(x \geq 0\).



And finally, before moving to the statements, let's rephrase the question:

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).

Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?



(1) \(|xy|\) is NOT a square of an integer.



If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.

If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).



Therefore, since we are told that \(|xy|\) is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.



(2) Point \((x, y)\) is above x-axis



If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.

If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.



Therefore, since we are told that point \((x, y)\) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D
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Re: M32-07 [#permalink]

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New post 17 Jul 2017, 05:23
Bunuel wrote:
Official Solution:


If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



This is a hard question. You should pay attention to every detail and read the solution very carefully



First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.



Next, \(-x \leq x\) implies that \(x \geq 0\).



And finally, before moving to the statements, let's rephrase the question:

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).

Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?



(1) \(|xy|\) is NOT a square of an integer.



If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.

If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).



Therefore, since we are told that \(|xy|\) is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.



(2) Point \((x, y)\) is above x-axis



If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.

If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.



Therefore, since we are told that point \((x, y)\) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D



Hi Bunuel,

Can we say that x = -y is the same as -x = y in this case? And if that is the case, will that affect the fact that statement (2) is sufficient?

Thanks
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Re: M32-07 [#permalink]

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duahsolo wrote:
Bunuel wrote:
Official Solution:


If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



This is a hard question. You should pay attention to every detail and read the solution very carefully



First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.



Next, \(-x \leq x\) implies that \(x \geq 0\).



And finally, before moving to the statements, let's rephrase the question:

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).

Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?



(1) \(|xy|\) is NOT a square of an integer.



If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.

If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).



Therefore, since we are told that \(|xy|\) is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.



(2) Point \((x, y)\) is above x-axis



If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.

If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.



Therefore, since we are told that point \((x, y)\) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D



Hi Bunuel,

Can we say that x = -y is the same as -x = y in this case? And if that is the case, will that affect the fact that statement (2) is sufficient?

Thanks


x = -y is ALWAYS the same as -x = y. But this won't affect the answer for (2) because we know from the stem that \(x \geq 0\). So, \((x, -x)\), will be (non-negative, non-positive), which would mean that it's either on x-axis or below it.
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Re: M32-07 [#permalink]

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New post 17 Jul 2017, 19:13
Hi,
question :sqrt(x 2 −y 2) = x+y?
squaring on both sides, we get x2-y2 = (x+y)^2
(x+y)(x-y) = (x+y)^2 => (x-y) = (x+y) => y= -y?

1.|xy| is not square of an integer. => I don't know how this information is useful to get to stem.
2. point(x,y) is above x - axis, means we have a value of y, so, answer to problem statement is No, = > S

Can someone help me out here.. atleast is my statement y= -y? is correct?
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Re: M32-07 [#permalink]

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New post 17 Jul 2017, 20:39
sasidharrs wrote:
Hi,
question :sqrt(x 2 −y 2) = x+y?
squaring on both sides, we get x2-y2 = (x+y)^2
(x+y)(x-y) = (x+y)^2 => (x-y) = (x+y) => y= -y?

1.|xy| is not square of an integer. => I don't know how this information is useful to get to stem.
2. point(x,y) is above x - axis, means we have a value of y, so, answer to problem statement is No, = > S

Can someone help me out here.. atleast is my statement y= -y? is correct?


Unfortunately nothing is correct.

You cannot reduce (x + y)(x - y) = (x + y)^2 by x + y, because x + y can be 0, and we cannot divide by 0. By doing so you are loosing a root, namely x + y = 0, or which is the same x = -y.

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.

It seems that you did not read the solution above. Again, this is a hard question. You should pay attention to every detail and read the solution very carefully.
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Re: M32-07 [#permalink]

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New post 18 Jul 2017, 07:49
buenel , can we expect questions of this difficulty in real GMAT ?? just wanted to know.
btw very tough.
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Re: M32-07 [#permalink]

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New post 18 Jul 2017, 07:59
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Re: M32-07 [#permalink]

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New post 21 Jul 2017, 16:35
For statement 2)

\((x,y)=(-1,1)\), which satisfies the question stem, leads to \(1(1-1)=0\)

\((x,y)=(1,1)\), which satisfies the question stem, leads to \(1(1+1)=2\)

I've read the solution several times, but I still see this a proving insufficiency. Any clarification is appreciated!
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Re: M32-07 [#permalink]

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New post 22 Jul 2017, 01:10
spence11 wrote:
If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?



This is a hard question. You should pay attention to every detail and read the solution very carefully



First of all, \(-x \leq y \leq x\) ensures two things:

1. \(x^2-y^2\geq 0\), so the square root of this number will be defined.

2. \(x+y\geq 0\), so the square root won't be equal to negative number.



Next, \(-x \leq x\) implies that \(x \geq 0\).


And finally, before moving to the statements, let's rephrase the question:

Does \(\sqrt{x^2 - y^2} = x + y\)?

Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?

Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).

Does \(y(x+y)=0\)?

Does \(y=0\) or \(x=-y\)?



(1) \(|xy|\) is NOT a square of an integer.



If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.

If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).



Therefore, since we are told that \(|xy|\) is NOT a square of an integer, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.



(2) Point \((x, y)\) is above x-axis



If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.

If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so (non-negative, non-positive), which would mean that it's either on x-axis or below it.



Therefore, since we are told that point \((x, y)\) is above x-axis, then neither \(y=0\) nor \(x=-y\) is true. Sufficient.


Answer: D

For statement 2)

(x,y)=(-1,1), which satisfies the question stem, leads to \(1(1-1)=0\)

\((x,y)=(1,1)\), which satisfies the question stem, leads to \(1(1+1)=2\)

I've read the solution several times, but I still see this a proving insufficiency. Any clarification is appreciated!


Please re-read again. This time paying attention to the highlighted part.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M32-07 [#permalink]

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New post 22 Jul 2017, 06:18
Bunuel,

Thanks - I see it now. I'm a little bit weak at simplifying compound inequalities I suppose. Back to the books!

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Re: M32-07 [#permalink]

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New post 03 Aug 2017, 01:09
hi bb

can we re write the question stem as does |x| + |y| = x^2 - y^2 ?
Re: M32-07   [#permalink] 03 Aug 2017, 01:09
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