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Is x²  y > 1/y² where y is nonzero?
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19 Oct 2018, 02:49
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Is x²  y > 1/y² where y is nonzero? (1) x < y (2) x² < y²
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Re: Is x²  y > 1/y² where y is nonzero?
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19 Oct 2018, 03:10
This is the solution that I found. I do not find this explanation satisfactory. Can any of the experts please explain the solution better? Thanks in advance Let's start by simplifying the original by multiplying both sides by y² (we can safely do so because we know that y² is positive): Is \(x^2y^3\)  \(y^3\) > 1? (1) x < y If x and y are big positive numbers, \(y^3\) will dominate x²y² and the left side will be negative; if x and y are big negative numbers, x²y² and (\(y^3\)) will both be positive and the left side will be a big positive number: insufficient. (2) x² < y² Tells us that x < y, but nothing about the signs of x and y. If we pick really big values with different signs we can generate both "yes" and "no" answers. Together: combining doesn't help us at all, since we have the same generalizations from both: choose E.
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Re: Is x²  y > 1/y² where y is nonzero?
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31 Jan 2019, 11:27
chetan2u please help with this question. I guessed the answer right but not sure about the approach.



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Re: Is x²  y > 1/y² where y is nonzero?
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31 Jan 2019, 12:21
PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² \(x^2\)  y > 1/\(y^2\) \(x^2\)  y  1/\(y^2\) > 0 Lets try 2 first (2) x² < y² or y^2 > x^2, 9 > 4, y can be 3 or 3 and x can be 2 or 2 1/4 > 1/16, y can be 1/2 or 1/2 and x can be 1/4 or 1/4 4 +3  1/9 > 0, This answers a yes to the question 1/16  1/2  1/4 < 0, This answers a no to the question from 1) x < y or y>x, here y & x can be again a fraction, integer or a number 1/2 > 1/4 or 3 > 2 1/16  1/2  1/4 < 0, This answers a no to the question 4 3  1/9 > 0, This answers a yes to the question The same test cases can be used again to show that after combining the 2, we wont get an unique answer E
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Is x²  y > 1/y² where y is nonzero?
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31 Jan 2019, 18:39
PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² Hi arpitkansal, What \(x^2<y^2\) means is \(x<y\).. X<y and x<y when combined mean y>x and y>0.. By looking at the equation, \(x^2y>\frac{1}{y^2}\), we should realise that \(\frac{1}{y^2}\) will be positive. We have to see if x^2y can be both negative and positive. 1) x=2 and y=3So \(2^23>\frac{1}{3^2}....43>\frac{1}{9}...1>\frac{1}{9}....\)yes 2) x=1 and y=4\(1^24>\frac{1}{4^2}.....14>\frac{1}{16}.....3>\frac{1}{16}...\).no Thus no definite answer.. E
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Re: Is x²  y > 1/y² where y is nonzero?
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31 Jan 2019, 21:08
PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² Take (x,y) = (0,1) & (0,1). We get E.
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Re: Is x²  y > 1/y² where y is nonzero?
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31 Jan 2019, 21:27
rahul16singh28 wrote: PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² Take (x,y) = (0,1) & (0,1). We get E. Hi Rahul, be careful with numbers you pick up.. (o, 1) does not fit into statement 1, which is x<y.. as 0>1, meaning x>y
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Is x²  y > 1/y² where y is nonzero?
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01 Feb 2019, 08:47
Here's how I solved, feel free to critique! Is \(x^{2}\)  y > \(\frac{1}{y^{2}}\)? If \(y\neq{0}\).I simplified the expression as follows: Because \(y^{2}\) is always positive I multiplied both sides by \(y^{2}\) to get \(x^{2}y^{2}\)  \(y^{3}\) > 1. From here I factored out \(y^{2}\) to get \(y^{2}\)(\(x^{2}\)  y) > 1. The question is now: Is \(y^{2}\)(\(x^{2}\)  y) > 1? If Y \(x\neq{0}\).Statement 1: x < yI picked x as 1 and y as 1. Plugging these values into the question the result is 0 > 1. I then picked x as 3 and y as 2. Plugging these values into the question the result is 44 > 1. Clearly insufficient. Statement 2: \(x^{3}\) < \(y^{3}\)I picked x as 1 and y as 1. Since they are raised to an odd power the signs remain the same for x and y (1 and 1, respectively). Plugging these values into the question the result is 0 > 1. I then picked x as 3 and y as 2. Since they are raised to an odd power the signs remain the same for x and y (3 and 2, respectively). Plugging these values into the question the result is 44 > 1. Clearly insufficient. Statement 1&2Clearly insufficient. Therefore, the answer is E.
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Re: Is x²  y > 1/y² where y is nonzero?
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01 Feb 2019, 10:00
PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² (1) x < y Let x = 3, y =2. Then we have (3)^2 2 > 1/(2)^2 > is 9 2 > 1/4? Yes Let x =1/2 let y =1/4 is (1/2)^2 1/4 > 1/(1/4)^2 ? is 1/4  1/4 >4 NO NS (2) x^2 <y^2 This is equivalent to x <y Let x =2 y =3 Is (2)^2 3 > 1/(3)^2 ? is 4+3 > 1/9? Yes let x = 3 y =4 is (3)^2 4 > 1/4^2? is 94 > 1/16 No NS (1) and (2) x < y and x<y so values like x = 4, y =3 won't work Let x=0 y =1 is 0^2 1 >1/1^2 No Let x =3, y=4 is 3^2 4 > 1/4^2 ? is 94 >1/4^2 Yes NS Answer is E.



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Is x²  y > 1/y² where y is nonzero?
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Updated on: 09 Oct 2019, 07:20
PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² Hey VeritasKarishma Bunuel, could this be solved algebraically? \(x^2y>1/y^2…x^2y^2y^3>1?\) \([I.]…x=2,y=3:(4)(9)(27)>1?…3627>1?…answer=yes\) \([II.]…x=1,y=4:(1)(16)(64)>1?…1664>1?…answer=no\) (1) x < y: case [I.] and [II.] are possible; insufic. (2) x² < y²: case [I.] and [II.] are possible; insufic. (1&2): case [I.] and [II.] are possible; insufic. Answer (E)
Originally posted by exc4libur on 08 Oct 2019, 08:22.
Last edited by exc4libur on 09 Oct 2019, 07:20, edited 1 time in total.



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Re: Is x²  y > 1/y² where y is nonzero?
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09 Oct 2019, 07:07
exc4libur wrote: PriyankaPalit7 wrote: Is x²  y > 1/y² where y is nonzero?
(1) x < y (2) x² < y² Hey VeritasKarishma Bunuel, could this be solved algebraically? \(x^2y>1/y^2…x^2y^2y^3>1\) \([I.]…x=1,y=2:(1)(4)8>1?…48=4<1…answer=no\) \([II.]…x=3,y=4:(9)(16)(64)>1?…14464=80>1…answer=yes\) (1) x < y: case [I.] and [II.] are possible; insufic. (2) x² < y²: case [I.] and [II.] are possible; insufic. (1&2): case [I.] and [II.] are possible; insufic. Answer (E) You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = 3 and y = 4, x > y, not x < y x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
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Re: Is x²  y > 1/y² where y is nonzero?
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09 Oct 2019, 07:21
VeritasKarishma wrote: You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = 3 and y = 4, x > y, not x < y
x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
There I fixed it. Thanks But can you solve this algebraically? Or is it a waste of time?



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Re: Is x²  y > 1/y² where y is nonzero?
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10 Oct 2019, 03:34
exc4libur wrote: VeritasKarishma wrote: You are still using number plugging (which is pretty much the way to go in this question). Also, how come the two examples work for both statements? When x = 3 and y = 4, x > y, not x < y
x < y and x^2 < y^2 when x and y are both positive. Try x = 1, y = 2 x = 1/4, y = 1/2
There I fixed it. Thanks But can you solve this algebraically? Or is it a waste of time? I wouldn't even try algebra here. x < y and x^2 < y^2 should remind you of number properties. For what kind of numbers does this hold? For non negative numbers. We know that numbers behave differently 'between 0 and 1' and '1 onwards'. So I will try both type of numbers.
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Re: Is x²  y > 1/y² where y is nonzero?
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