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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
28 Oct 2012, 08:44
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! _________________
I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+
Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
28 Oct 2012, 20:56
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. \(x^2 + y^2 + 2xy > 200\)
Since \((x-y)^2\) 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. _________________
Did you find this post helpful?... Please let me know through the Kudos button.
Re: Is x^2 + y^2 > 100? [#permalink]
31 Oct 2012, 14:49
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.
Re: Is x^2 + y^2 > 100? [#permalink]
18 Feb 2013, 18:58
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?
Re: Is x^2 + y^2 > 100? [#permalink]
19 Feb 2013, 04:43
Expert's post
1
This post was BOOKMARKED
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.
(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
(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: \(x^2+(x^2+y^2)+y^2>200\) --> \(2(x^2+y^2)>200\).
Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 11:34
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.
Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 11:44
Expert's post
anujtsingh wrote:
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
Frankly I don't understand your point. I just wanted to show that we CAN substitute \(2xy\) with \(x^2+y^2\) if \(x^2+y^2\geq{2xy}\). _________________
(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?
Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...
(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?
Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: \(x^2+2xy+y^2>200\). Next, we know that \(x^2+y^2\geq{2xy}\) is true for any values of \(x\) and \(y\). So we can manipulate and substitute \(2xy\) with \(x^2+y^2\) in (2) (because \(x^2+y^2\) is at least as large as \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) --> \(x^2+y^2>100\).
(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?
Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...
Now you know that 2xy is less than or equal to x^2 + y^2. If 2xy is equal to \((x^2 + y^2)\), \((x^2 + y^2)\) will be greater than 100 since the total sum is greater than 200. If 2xy is less than \((x^2 + y^2)\), then anyway \((x^2 + y^2)\) will be greater than 100 (which is half of 200).
So statement 2 is sufficient alone. _________________
Re: Is x^2 + y^2 > 100? [#permalink]
16 Jan 2014, 03:04
Bunuel wrote:
joe26219 wrote:
Bunuel wrote:
Is x^2 + y^2 > 100?
(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?
Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...
There is no flaw in my reasoning. Statement (2) says: \(x^2+2xy+y^2>200\). Next, we know that \(x^2+y^2\geq{2xy}\) is true for any values of \(x\) and \(y\). So we can manipulate and substitute \(2xy\) with \(x^2+y^2\) in (2) (because \(x^2+y^2\) is at least as large as \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) --> \(x^2+y^2>100\).
Thank you Bunuel.Got it.Your explanation was really helpful (the highlighted line esp.) I guess any value of 2xy should be assumed to satisfy the given condition, as it is given.
Re: Is x^2 + y^2 > 100? [#permalink]
10 May 2014, 07:00
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
Re: Is x^2 + y^2 > 100? [#permalink]
10 May 2014, 07:29
Expert's post
mayank_sharma wrote:
Am still confused. Consider, x^2 + y^2 = 200 for the minimum value. That means, x + y is minimum 200 ^ 0.5 which means for a minimum sum of x + y, x has to be (200^0.5)/2 which gives x = 7.07. Sum up x^2 + y^2 at x = y = 7.07 we get x^2+ y^2 = approx. 98.30. Can someone please explain where I am going wrong.
Your question is not clear... For example, why are you considering x^2 + y^2 = 200?...
Anyway, x^2 + y^2 = 200 doe not mean that \(x+y=\sqrt{200}\). _________________
Re: Is x^2 + y^2 > 100? [#permalink]
27 May 2015, 11:36
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