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Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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11 Feb 2012, 00:54
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Is xy > y^2? (1) x > y (2) y > 0 I rephrased the question as xy>y (since y^= y. On solving this I rephrased as x>1?
basis this rephrased version. the answer id D. however OA is C..
Have I solved the equation wrongly?
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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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11 Feb 2012, 01:10
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devinawilliam83 wrote: is xy>y^2? 1.x>y 2.y>0
I rephrased the question as xy>y (since y^= y. On solving this I rephrased as x>1?
basis this rephrased version. the answer id D. however OA is C..
Have I solved the equation wrongly? Yes, your rephrasing is wrong. You cannot substitute \(y^2\) with \(y\), because generally they are not equal: \(\sqrt{y^2}=y\). Next, even if it were "is \(xy>y\)?" you still cannot reduce it by \(y\) and write "is \(x>1\)", as \(y\) can be zero and you cannot reduce/divide by zero. \(xy>y\) can be rephrased as \(y(x1)>0\). Is xy>y^2?(1) x>y > if \(x>y>0\) then obviously \(xy>y^2\) but if \(0>x>y\) then \(xy<0<y^2\). Not sufficient. (2) y>0 > \(y=y\). No info about x. Not sufficient. (1)+(2) Since \(x>y>0\) then \(xy>y^2\). Sufficient. Answer: C. Hope it's clear.
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Re: Xy > y^2 [#permalink]
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26 Sep 2012, 19:56
Asked : is xy > y^2
Simplifying, is xy > y*y i.e is x > y
i) x > y. But this doesn't mean x > y unless y is positive ii)y > 0 ie. y is positive
Hence, both statements together are sufficient. Hence C.



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Re: Xy > y^2 [#permalink]
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26 Sep 2012, 23:23
piyatiwari wrote: Asked : is xy > y^2
Simplifying, is xy > y*y i.e is x > y
i) x > y. But this doesn't mean x > y unless y is positive ii)y > 0 ie. y is positive
Hence, both statements together are sufficient. Hence C. Simplifying, is xy > y*y i.e is x > yYou can simplify, or more precisely, divide through by y if you know that y is nonzero. The question "Is xy > y^2?" is equivalent to "Is y nonzero and x > y?"(1) y can be 0. Also, x can be greater than y. Not sufficient. (2) Not sufficient, as nothing is known about x. (1) and (2) together: y > 0 and x > y = y. Sufficient. Answer C.
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Re: Xy > y^2 [#permalink]
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Re: Xy > y^2 [#permalink]
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carcass wrote: I 'd like to know if my approach is correct x  y  > y^2 > x <  y or x > y 1) x > y but we do not know if x is less of  y insuff 2) y> 0 we know only that y is positive but nothing about x 1) + 2) y is positive and x must be greater than y. suff. is correct ?? x  y  > y^2 > x <  y or x > y NO First of all, y must be nonzero. Then dividing through by y, we obtain x > y > 0. From the given inequality, \(xy > y^2>0\) we deduce that x > 0, x cannot be negative. And you cannot automatically assume that y is positive.
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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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07 Jun 2013, 07:45
Can we take
\(Y^2\) = \([y][y]\) and cancel [y] from both sides if y is not equal to zero.



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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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07 Jun 2013, 07:59



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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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30 Jun 2013, 13:33
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Is xy>y^2?
(1) x>y
Though we know that x>y, we don't know anything about the signs of x and y. For example:
x=2, y=1 xy>y^2 21>1^2 2>1 Valid
x=2, y=3 xy>y^2 23>3^2 6>9 Invalid
INSUFFICIENT
(2) y>0 This tells us nothing about x. For example:
x=10, y=1 xy>y^2 101>1^2 10>1 Valid
x=1, y=10 xy>y^2 110>10^2 10>100 Invalid
INSUFFICIENT
1+2) x>y, y>0 ===> x>y>0 If y>0 then x is greater than zero (and y) x=10, y=9 xy>y^2 109>9^2 90>81
Think of it like this: The RHS is the smaller number times the smaller number. The LHS is the smaller number times a number larger than the smaller number. SUFFICIENT
(C)



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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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22 Jul 2013, 09:16
Is xy>y^2? (1) x>y x may be greater than y but we still cannot be sure if xy>y^2. For example:
x>y 2>1 21 > 1^2 2>1 Valid
x>y 1>2 12 > 2^2 2 > 4 Invalid INSUFFICIENT
(2) y>0 This tells us nothing about x. INSUFFICIENT
1+2) x>y and y>0. Therefore, x>y>0. We saw in #1 that when y is greater than zero, the inequality holds true. When it is less than zero, the inequality does not hold true. Just to be sure: x>y>0 2>1>0 xy>y^2 21>1^2 2>1
Also, take note that xy is y*a number larger than y. This will always be greater than y*y (y^2) so long as x is positive.
(C)



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Re: Is x*y > y^2? [#permalink]
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17 May 2014, 19:31
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russ9 wrote: Is x·y > y2?
(1) x > y (2) y > 0
I'll paste my question in the second post. Good Luck! We have xy> y2 Now, to solve this, we need to understand, what do we actually want to know. y2 = y2 xy>y2 xy  y2 > 0 or (xy)y>0 Now, we know that y is always greater than 0, Thus, our inequality yield xy>0 So Is xy>y2 translates to Is x> y Now, lets come to the statements, Statement 1 says, x>y; which is not sufficient. In case y is negative, we can't say anything about x being greater than absolute value of y Statement 2 say, y>0, which in itself is insufficient as we don't know anything about x here. But when we combine these two statements, we would get the answer. As we know when y > 0; y = y and statement 1 says x > y, and if y > 0 we can write x > y Hence, the answer is C. Hope it helps!!! Kudos if it helped!!!



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Re: Is x*y > y^2? [#permalink]
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18 May 2014, 00:27
mittalg wrote: russ9 wrote: Is x·y > y2?
(1) x > y (2) y > 0
I'll paste my question in the second post. Good Luck! We have xy> y2 Now, to solve this, we need to understand, what do we actually want to know. y2 = y2 xy>y2 xy  y2 > 0 or (xy)y>0 Now, we know that y is always greater than 0, Thus, our inequality yield xy>0 So Is xy>y2 translates to Is x> y Now, lets come to the statements, Statement 1 says, x>y; which is not sufficient. In case y is negative, we can't say anything about x being greater than absolute value of y Statement 2 say, y>0, which in itself is insufficient as we don't know anything about x here. But when we combine these two statements, we would get the answer. As we know when y > 0; y = y and statement 1 says x > y, and if y > 0 we can write x > y Hence, the answer is C. Hope it helps!!! Kudos if it helped!!! Everything is correct except that y is more than or equal to 0. So, x*y > y^2 after reducing by y can be translated: is \(x > y>{0}\)?
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Re: Is x*y > y^2? [#permalink]
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05 Oct 2014, 15:19
Hi Experts,
Is the following solution correct? Question: Is x·y > y^2? True when: Case 1: y>0 and x > y Case 2: y < 0 and x < y
(1) NS; we don't know whether y > 0 (2) NS; we don't know anything about x
(1) + (2) Case 1, so sufficient



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Re: Is x*y > y^2? [#permalink]
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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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06 Feb 2016, 10:21
devinawilliam83 wrote: Is xy>y^2?
(1) x>y (2) y>0
I rephrased the question as xy>y (since y^= y. On solving this I rephrased as x>1?
basis this rephrased version. the answer id D. however OA is C..
Have I solved the equation wrongly? Can we rephrase the question as IxI>IyI?? reason.. if we square both side xIyI>y^2 we get x^2Y^2>Y^4 X^2>Y^2 or IXI>IYI...????



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Re: Is xy > y^2? (1) x > y (2) y > 0 [#permalink]
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Re: Is x·y > y^2? [#permalink]
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21 May 2016, 05:47
nishatfarhat87 wrote: Is x·y > y^2? (1) x > y (2) y > 0 Hi Bunuel, Can you please help me with a quick solution for this one. Hi, Bunuel too will be able to give a approach, but i will give you one \(x·y > y^2.....\) what does this mean... \(y^2xy<0.............\) since y^2 and y will be positive, we have to answer if x>y......OR x>0 and x>y...lets see the statements \((1) x > y\) say y is ive and x is also ive...... ans will be NO if x is +ive, ans is YES Insuff \((2) y > 0\) nothing about relation between x and y.. Insuff... Combined x>y and y>0, so x>0.. ans is YES Suff C
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Re: Is x·y > y^2? [#permalink]
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21 May 2016, 05:59
chetan2u wrote: nishatfarhat87 wrote: Is x·y > y^2? (1) x > y (2) y > 0 Hi Bunuel, Can you please help me with a quick solution for this one. Hi, Bunuel too will be able to give a approach, but i will give you one \(x·y > y^2.....\) what does this mean... \(y^2xy<0.............\) since y^2 and y will be positive, we have to answer if x>y......OR x>0 and x>y...lets see the statements \((1) x > y\) say y is ive and x is also ive...... ans will be NO if x is +ive, ans is YES Insuff \((2) y > 0\) nothing about relation between x and y.. Insuff... Combined x>y and y>0, so x>0.. ans is YES Suff C Hi Chetan, Thanks for your response. How did you get this: since y^2 and y will be positive, we have to answer if x>y......OR x>0 and x>y..



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Re: Is x·y > y^2? [#permalink]
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22 May 2016, 00:28
nishatfarhat87 wrote: Hi Chetan,
Thanks for your response. How did you get this: since y^2 and y will be positive, we have to answer if x>y......OR x>0 and x>y.. Hi, Quote: \(x·y > y^2.....\) what does this mean... \(y^2xy<0.............\) since y^2 and y will be positive, we have to answer if x>y......OR x>0 and x>y... Now when will x·y > y^2.... RHS= Y^2, which is always 0 or +ive.... LHS has two terms x and y, y is again 0 or +.. so if x is ive , xy will be ive and it will not be > + value... so x has to be +ive .. and the numeric value has to be >y for x*y to be > than y*y
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