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# M32-07

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M32-07  [#permalink]

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17 Jul 2017, 03:29
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Question Stats:

30% (00:56) correct 70% (01:36) wrong based on 40 sessions

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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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17 Jul 2017, 03:30
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2
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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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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17 Jul 2017, 05:35
1
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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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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Posts: 51223
Re: M32-07  [#permalink]

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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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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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18 Jul 2017, 07:59
1
kumarparitosh123 wrote:
buenel , can we expect questions of this difficulty in real GMAT ?? just wanted to know.
btw very tough.

Probably only if you doing very well and are aiming to Q51.
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Re: M32-07  [#permalink]

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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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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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Re: M32-07  [#permalink]

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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!

SR

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Re: M32-07  [#permalink]

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03 Aug 2017, 01:09
hi bb

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

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24 Feb 2018, 23:18
Hi Bunuel,

Could you please explain what does this mean?

Next, −x≤x implies that x≥0

and if it is true in all cases?

Thank you.
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Re: M32-07  [#permalink]

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25 Feb 2018, 01:42
ra5867 wrote:
Hi Bunuel,

Could you please explain what does this mean?

Next, −x≤x implies that x≥0

and if it is true in all cases?

Thank you.

−x ≤ x;

Add x to both sides: 0 ≤ 2x;

Reduce by 2: 0 ≤ x.
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Re: M32-07  [#permalink]

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26 Feb 2018, 00:02
Bunuel wrote:
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

I did this question by plotting number which are as following:-
lets take 3 situation according to above condition
1. X=3,y=2 and -x=-3
2.X=3, y=3 and -x=-3
3.x=3, y=-3 and -x=-3
so if we plug these value in above question only statement 3 will prove above equation. Now come to given data.
only statement 1 is possible so above equation is not true statement 1 is sufficient.
2. from this we can say x can be positive or negative but y is always positive. So assumption 1 and 2nd is possible. still the answer is above statement is not true so b is also sufficient.
so final answer is option D.
Re: M32-07 &nbs [#permalink] 26 Feb 2018, 00:02
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# M32-07

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