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Is x^2 + y^2 > 100?

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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink] New post 28 Oct 2012, 05:54
rajathpanta wrote:
Is x^2+y^2>100??

(1) 2xy<100
(2) (x+y)^2>200

To me its only B. because statement 2 boils down to x+y>\sqrt{200}

Can someone explain the OA


Its B for me as well.

2) (x+y)^2 > 200,
So., (x+y) > 10\sqrt{2} or (x+y) < -10\sqrt{2}

I cant find any two numbers that would satisfy statement 2 and give a value of x^2 + y^2 < 100

Confusing.

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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink] New post 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!
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Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink] New post 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.

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Re: Is x^2 + y^2 > 100? [#permalink] New post 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.
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Re: Is x^2 + y^2 > 100? [#permalink] New post 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?
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Re: Is x^2 + y^2 > 100? [#permalink] New post 19 Feb 2013, 04:43
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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] New post 19 Feb 2013, 20:14
Thanks, it's clear now!
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Re: Is x^2 + y^2 > 100? [#permalink] New post 30 Mar 2013, 05:32
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test

Is x^2 + y^2 > 100?

(1) 2xy < 100

(2) (x + y)^2 > 200


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?




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] New post 31 Mar 2013, 07:46
Expert's post
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


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?




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] New post 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.

Pls correct me if I am wrong
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Re: Is x^2 + y^2 > 100? [#permalink] New post 31 Mar 2013, 11:44
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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}.
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Re: Is x^2 + y^2 > 100? [#permalink] New post 05 Jul 2013, 01:44
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Re: Is x^2 + y^2 > 100? [#permalink] New post 14 Jan 2014, 06:45
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test

Is x^2 + y^2 > 100?

(1) 2xy < 100

(2) (x + y)^2 > 200


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 ...

:) :)
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Re: Is x^2 + y^2 > 100? [#permalink] New post 14 Jan 2014, 07:03
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joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test

Is x^2 + y^2 > 100?

(1) 2xy < 100

(2) (x + y)^2 > 200


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.

By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573
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Re: Is x^2 + y^2 > 100? [#permalink] New post 14 Jan 2014, 20:30
Expert's post
joe26219 wrote:
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test

Is x^2 + y^2 > 100?

(1) 2xy < 100

(2) (x + y)^2 > 200


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 ...

:) :)


To put it in words, think of it this way:

Is x^2 + y^2 > 100?
(2) (x + y)^2 > 200
which means: x^2 + y^2 + 2xy > 200

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.
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Re: Is x^2 + y^2 > 100? [#permalink] New post 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.

By the way the OA is B, not C. VeritasPrep corrected it: is-x-2-y-108343.html#p860573



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.
:-D :-D
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Re: Is x^2 + y^2 > 100? [#permalink] New post 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.
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Re: Is x^2 + y^2 > 100? [#permalink] New post 10 May 2014, 07:29
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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}.
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Re: Is x^2 + y^2 > 100?   [#permalink] 10 May 2014, 07:29
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