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M32-08

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M32-08  [#permalink]

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New post 17 Jul 2017, 03:41
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Question Stats:

35% (01:06) correct 65% (01:28) wrong based on 23 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 NOT on x-axis.

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Re M32-08  [#permalink]

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New post 17 Jul 2017, 03:41
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.

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then \(xy\) would be \(-x^2\), so negative, which cannot be a square of an integer. So, \(x = -y\) IS possible.

Of course, \(x\) can be not equal to \(-y\), and the statemented still could hold true. For example, \(x = 2\) and \(y =1\).



Thus, we have have two different answers to the question: Not sufficient.



(2) Point \((x, y)\) is NOT on x-axis.



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

However, if \(x = -y\) and \(x ≠ 0\) were true, for example if \(x = -y = 1\), then then point \((x, y)\) would be \((x, -x)\), so (positive, negative), which would mean that it's below x-axis. So, \(x = -y\) IS possible.

Of course, \(x\) can be not equal to \(-y\), and the statemented still could hold true. For example, \(x = 2\) and \(y =1\).



(1)+(2) We could have the same cases: if \(x = -y\) and \(x ≠ 0\), for example if \(x = -y = 1\), then the answer is YES but if say \(x = 2\) and \(y =1\), then the answer is NO. Not sufficient.


Answer: E
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Re: M32-08  [#permalink]

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New post 11 Sep 2017, 06:25
Can this question be solved by plugging in values?
the approach explained by bunuel is tough to comprehend does someone else has an easy method or trick for this question?
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Re: M32-08  [#permalink]

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New post 11 Sep 2017, 06:38
haardiksharma wrote:
Can this question be solved by plugging in values?
the approach explained by bunuel is tough to comprehend does someone else has an easy method or trick for this question?


Not every question has silver bullet solution. This is not an easy question. You can check alternative solutions here: https://gmatclub.com/forum/if-x-and-y-a ... 42602.html
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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

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M32-08  [#permalink]

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New post 27 Oct 2018, 05:56
By plugging in numbers -

Hey Bunuel amazing question.
I solved it a bit differently. Please correct me if i am wrong.

Reducing the equation.
\(\sqrt{x^2 - y^2}\) \(= x + y\)
\(x^2 - y^2 = x^2 + y^2 + 2xy\)
\(-2y^2 = 2xy\)
\(y^2 = -xy\) ?


Statement 1
- xy is NOT a square of an integer.
It means x is not equal to y.
If x = 3 and y = -3. Then equation condition satisfied.
But, if x = 3 and y = -2 then equation condition not satisfied.
So INSUFFICIENT

Statement 2 -
- Point (x,y) is NOT on x-axis.
From this we can infer that y is not equal to zero.
Still the above to examples are possible.
So INSUFFICIENT.

Both statements together -

We can still see that the above examples can happen and hence it is not Sufficient.

Answer - E
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M32-08 &nbs [#permalink] 27 Oct 2018, 05:56
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