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Re M3207
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17 Jul 2017, 03:30
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. \(x^2y^2\geq 0\), 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 xaxis If \(y = 0\) were true, then point \((x, y)\) would be ON the xaxis. If \(x = y\) were true, then then point \((x, y)\) would be \((x, x)\), so (nonnegative, nonpositive), which would mean that it's either on xaxis or below it. Therefore, since we are told that point \((x, y)\) is above xaxis, then neither \(y=0\) nor \(x=y\) is true. Sufficient. Answer: D
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Re: M3207
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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. \(x^2y^2\geq 0\), 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 xaxis
If \(y = 0\) were true, then point \((x, y)\) would be ON the xaxis. If \(x = y\) were true, then then point \((x, y)\) would be \((x, x)\), so (nonnegative, nonpositive), which would mean that it's either on xaxis or below it.
Therefore, since we are told that point \((x, y)\) is above xaxis, 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: M3207
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17 Jul 2017, 05:35
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. \(x^2y^2\geq 0\), 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 xaxis
If \(y = 0\) were true, then point \((x, y)\) would be ON the xaxis. If \(x = y\) were true, then then point \((x, y)\) would be \((x, x)\), so (nonnegative, nonpositive), which would mean that it's either on xaxis or below it.
Therefore, since we are told that point \((x, y)\) is above xaxis, 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 (nonnegative, nonpositive), which would mean that it's either on xaxis or below it.
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Re: M3207
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17 Jul 2017, 19:13
Hi, question :sqrt(x 2 −y 2) = x+y? squaring on both sides, we get x2y2 = (x+y)^2 (x+y)(xy) = (x+y)^2 => (xy) = (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: M3207
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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 x2y2 = (x+y)^2 (x+y)(xy) = (x+y)^2 => (xy) = (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: M3207
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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: M3207
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21 Jul 2017, 16:35
For statement 2)
\((x,y)=(1,1)\), which satisfies the question stem, leads to \(1(11)=0\)
\((x,y)=(1,1)\), 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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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. \(x^2y^2\geq 0\), 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 xaxis
If \(y = 0\) were true, then point \((x, y)\) would be ON the xaxis. If \(x = y\) were true, then then point \((x, y)\) would be \((x, x)\), so (nonnegative, nonpositive), which would mean that it's either on xaxis or below it.
Therefore, since we are told that point \((x, y)\) is above xaxis, 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 \(1(11)=0\)
\((x,y)=(1,1)\), 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 reread 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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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: M3207
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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: M3207
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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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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 xaxis 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.










