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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]

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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]

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

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29 Oct 2012, 03:41
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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

Merging similar topics. You are right, OA should be B, not C. See here: is-x-2-y-108343.html#p859197
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Re: Is x^2 + y^2 > 100? [#permalink]

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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]

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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]

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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]

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19 Feb 2013, 20:14
Thanks, it's clear now!
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Re: Is x^2 + y^2 > 100? [#permalink]

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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.

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]

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31 Mar 2013, 07:46
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

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.

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]

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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]

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31 Mar 2013, 11:44
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]

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05 Jul 2013, 01:44
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Re: Is x^2 + y^2 > 100? [#permalink]

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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.

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]

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14 Jan 2014, 07:03
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.

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]

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14 Jan 2014, 20:30
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.

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]

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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.

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

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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]

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10 May 2014, 07:29
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]

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27 May 2015, 11:36
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Re: Is x^2 + y^2 > 100? [#permalink]

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17 Jun 2015, 05:11
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Is X^2+Y^2>100 ..

----> is [(x+y)^2 + (x-y)^2]/2 >100

So we can change the question stem to.. is (x+y)^2+(x-y)^2>200?

Statement A can in invalidated easily. Use different values of x and y.

statement B says that (X+Y)^2>200

Substitute statement B in question stem -

is Something >200 [(X+Y)^2] + something >=0 [(X-Y)^2] >200 .... Ans is obviously yes.

So B is the correct Ans. I hope this makes sense!
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Re: Is x^2 + y^2 > 100?   [#permalink] 17 Jun 2015, 05:11

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