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Is x^2  y^2 > 0?
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21 Mar 2017, 18:52
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37% (02:28) correct 63% (02:25) wrong based on 101 sessions
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Is \(x^2y^2>0\) ? (1) \(x^2y^2>(x  y)\) (2) \(x^2y^2<(x + y)\)
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Re: Is x^2  y^2 > 0?
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22 Mar 2017, 02:01
ziyuen wrote: Is \(x^2y^2>0\) ?
(1) \(x^2y^2>(x  y)\) (2) \(x^2y^2<(x + y)\) Hi In other words "Is \(x^2 > y^2\) ? (1) \(x^2y^2>(x  y)\) \(x(x  1) > y(y  1)\) We can have \(x>y>0\) > \(3>2>0\) \(3*2>2*1\) > \(6^2 > 2^2\) Yes. Or \(0<x<y\) > \(0<\frac{1}{3}<\frac{1}{2}\) \(\frac{1}{3}*(\frac{2}{3}) "?" \frac{1}{2} * (\frac{1}{2})\) > \(\frac{2}{9} > \frac{1}{4}\) > \((\frac{1}{3})^2 < (\frac{1}{2})^2\) No. Insufficient. (2) \(x^2y^2<(x + y)\) \(x(x  1) < y(y + 1)\) \(x=2, y=4\) > \(2*1 < 4*5\) > \(2^2 < 4^2\) No. \(x=\frac{1}{2}, y=\frac{1}{3}\) > \(\frac{1}{2}*(\frac{1}{2}) "?" \frac{1}{3}*\frac{4}{3}\) > \(\frac{1}{4} < \frac{4}{9}\) > \((\frac{1}{2})^2 > (\frac{1}{4})^2\) Yes. Insufficient. (1)&(2) \(x^2  y^2 > x  y\) & \(x^2  y^2 < x + y\) or \(y^2  x^2 < y  x\) & \(x^2  y^2 < x + y\) Adding we'll get: 0 < 2y > y>0. But we already have different answers when y is positive (x can be >0, <0, integer or a fraction). Insufficient. Answer E. What's the source of the question?



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Re: Is x^2  y^2 > 0?
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09 Jan 2019, 11:30
Here's how I solved this  please critique! Is \(x^{2}\)\(y^{2}\)>0? Rewriting the stem: Is (xy)(x+y)>0? Statement 1) \(x^{2}\)\(y^{2}\)>(xy) Rewriting: (xy)(x+y)>(xy) If (xy)>0Dividing both sides by (xy) we are left with (x+y)>1. If (xy)<0 Dividing both sides by (xy) we are left with (x+y)<1. Because we don't know if (xy) is positive, negative or zero, Statement 1 is insufficient. Statement 2) \(x^{2}\)\(y^{2}\)<(x+y) Rewriting: (xy)(x+y)<(x+y) If (x+y)>0Dividing both sides by (x+y) we are left with (xy)<1. If (x+y)<0 Dividing both sides by (x+y) we are left with (xy)>1. Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient. Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient. E is the answer.
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Re: Is x^2  y^2 > 0?
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09 Jan 2019, 13:05
hazelnut wrote: Is \(x^2y^2>0\) ?
(1) \(x^2y^2>(x  y)\) (2) \(x^2y^2<(x + y)\) Notice that x^2y^2 factors to (x+y)(xy) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(xy) then the question becomes ab>0 Yes/NO? (1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=1/2 a=1/2, then ab<0 NO. Since we get a Yes and a No. (1) is NS (2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=1 b=2, then ab<0 No Since we get a Yes and a No NS (1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E.



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Re: Is x^2  y^2 > 0?
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09 Jan 2019, 20:13
emockus wrote: Here's how I solved this  please critique!
Is \(x^{2}\)\(y^{2}\)>0?
Rewriting the stem: Is (xy)(x+y)>0?
Statement 1) \(x^{2}\)\(y^{2}\)>(xy)
Rewriting: (xy)(x+y)>(xy)
If (xy)>0 Dividing both sides by (xy) we are left with (x+y)>1.
If (xy)<0 Dividing both sides by (xy) we are left with (x+y)<1.
Because we don't know if (xy) is positive, negative or zero, Statement 1 is insufficient.
Statement 2) \(x^{2}\)\(y^{2}\)<(x+y)
Rewriting: (xy)(x+y)<(x+y)
If (x+y)>0 Dividing both sides by (x+y) we are left with (xy)<1.
If (x+y)<0 Dividing both sides by (x+y) we are left with (xy)>1.
Because we don't know if (x+y) is positive, negative or zero, Statement 2 is insufficient.
Statement 1 & 2) We are left with a bunch of possibilities and no way to come to 1 final answer; as such, "C" is insufficient.
E is the answer. Thanks for the explanation I dont think we can divide by xy. since its no where mentioned that x is not equal to y similarly for the next given statement x can be equal to y.



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Is x^2  y^2 > 0?
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10 Jan 2019, 00:53
ocelot22 wrote: hazelnut wrote: Is \(x^2y^2>0\) ?
(1) \(x^2y^2>(x  y)\) (2) \(x^2y^2<(x + y)\) Notice that x^2y^2 factors to (x+y)(xy) for this reason, and all of the confusion that can occur from writing so much, missing a sign etc, I like to define a =(x+y), b=(xy) then the question becomes ab>0 Yes/NO? (1) ab>b if b>0, a>1. if b <0, a<1. Suppose b=1,a=2 then ab>0 Yes. Suppose b=1/2 a=1/2, then ab<0 NO. Since we get a Yes and a No. (1) is NS (2) ab<a if a>0,b <1. if a<0, b>1. Suppose a=1, b=1/2, then ab>0 Yes. Suppose a=1 b=2, then ab<0 No Since we get a Yes and a No NS (1) and (2) if b>0,a>1, if b<0, a<1, if b>0, a>1, if b<0, a <1, if a>0, b<1, if a<0, b>1. Suppose a=1/2, b=1 then ab<0 No. Suppose a=1/2, b=1/2, then ab>0 YES. Since we get a Yes and a NO (1) and (2) are NS E. Hi ocelot22Nice solution. However, I would take your concept in simplicity by the following: ab>b ab<a Subtract the inequalities as different signs ab ab > b a 0 > b a a > b....Form here it is easy to judge quickly If a = 10 & b= 4...Then answer is Yes If a = 10 & b = 4....Then answer No Answer: E hazelnutCan you please post Veritas solution



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Re: Is x^2  y^2 > 0?
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24 May 2019, 05:05
Bunuel, could you provide your insight on this question? I'm stuck on why E is the answer. Per Statement 1, when xy >0, x + y >1 (x=3, y=2) and x>y. When xy<0, x+y<1 (x=(2.5), y=3), and x<y. NS. Per Statement 2, when x+y>0, xy<1 (x=3, y=2.5) and x>y. When x+y<0, x1>1 (x=(2), y=(4), and x<y. NS. x=3 and y = 2, works for both statements. I can't find values that work for x<y fr both statements.
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Re: Is x^2  y^2 > 0?
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25 May 2019, 08:52
hazelnut wrote: Is \(x^2y^2>0\) ?
(1) \(x^2y^2>(x  y)\) (2) \(x^2y^2<(x + y)\) Hi, I answered E using the following approach. (1) \(x^2y^2>(x  y)\) \((x + y)(x  y)>(x  y)\) \((x + y)>1\) Insufficient. (2) \(x^2y^2<(x + y)\) \((x + y)(x  y)<(x + y)\) \((x  y)<1\) Insufficient. Together both statements are still insufficient as the possibility for (x  y) can range from positive fraction < 1 to unlimited negatives. Let me know if that makes sense.




Re: Is x^2  y^2 > 0?
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