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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
 28 Oct 2012, 09:44
Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
28 Oct 2012, 21:56
rajathpanta wrote: Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply! I found this solution from Bunuel. Statement 2. x^2 + y^2 + 2xy > 200Since (x-y)^2 is cannot be lesser than 0, (square of a number is always positive or 0), the minimum value that 2xy can take is equal to x^2 + y^2. So statement 2 can be changed to be, x^2 + y^2 + x^2 + y^2 > 200. So, x^2 + y^2 > 100. That is a convincing method to show that B is infact the right answer. Kudos Please... If my post helped.
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
29 Oct 2012, 04:41
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Re: Is x^2 + y^2 > 100? [#permalink]
31 Oct 2012, 15:49
It's B for me.
(x + y)^2 = x^2 + 2xy + y^2
The largest possible value 2xy can reach is (x + y)^2/2, that only occurs when x = y.
When x = y, it turns out that x^2 + 2xy + y^2 > 200 is x^2 + 2xx + x^2 > 200, which can be rearranged in 2x^2 + 2x^2 > 200, meaning that x^2 + y^2 is at least half of the stated value, while 2xy can be at most the other half.
Any value above 200 will require that x^2 + y^2 > 100, making 2) sufficient.
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Is x² + y² > 100 ?
1) 2xy < 100
2) (x + y)² > 200
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Re: Is x^2 + y^2 > 100? [#permalink]
18 Feb 2013, 19:58
Bunuel wrote: (2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} Hi Bunuel, could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?
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Re: Is x^2 + y^2 > 100? [#permalink]
19 Feb 2013, 05:43
LinaNY wrote: Bunuel wrote: (2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} Hi Bunuel, could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive? We have (x + y)^2 > 200 which is the same as x^2+2xy+y^2>200. Now, we need to find the relationship between x^2+y^2 and 2xy. Next, we know that (x-y)^2\geq{0} --> x^2-2xy+y^2\geq{0} --> x^2+y^2\geq{2xy}. So, we can safely substitute 2xy with x^2+y^2 in x^2+2xy+y^2>200 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient. Hope it's clear.
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Re: Is x^2 + y^2 > 100? [#permalink]
19 Feb 2013, 21:14
Thanks, it's clear now!
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Re: Is x^2 + y^2 > 100? [#permalink]
30 Mar 2013, 06:32
Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 (1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES. (2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient. Answer: B. Are you sure the OA is C? Brunel I have one confusion, As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice) So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200
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Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 08:46
anujtsingh wrote: Bunuel wrote: arvindg wrote: Problem source: Veritas Practice Test Is x^2 + y^2 > 100? (1) 2xy < 100 (2) (x + y)^2 > 200 (1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES. (2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient. Answer: B. Are you sure the OA is C? Brunel I have one confusion, As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice) So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200 No. If x^2 +y^2 = 2xy, then we can simply substitute 2xy to get the same: x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200. How did you get 2(x^2+y^2) <200?
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Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 12:34
If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.
Pls correct me if I am wrong
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Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 12:44
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Re: Is x^2 + y^2 > 100?
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31 Mar 2013, 12:44
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