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Is x^2 + y^2 > 100?

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Is x^2 + y^2 > 100? [#permalink]

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Is x^2 + y^2 > 100?

(1) 2xy < 100

(2) (x + y)^2 > 200
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Oct 2012, 03:38, edited 1 time in total.
OA edited.

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Re: Is x^2 + y^2 > 100? [#permalink]

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

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Hey arvindg,
Thanks for pointing that out. The explanation is indeed incorrect. Sometimes, errors just creep up unwittingly. We will fix it soon. Your strategy of using numbers was spot on and that's exactly what I was thinking while reading the question too... Though the switch from <100 to >100 takes place at decimals so I was a little unhappy about that. The algebraic solution given by Bunuel is neat.
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Re: Is x^2 + y^2 > 100? [#permalink]

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tarunjagtap wrote:
Bunuel's explanation is clear to me..
I think what bunuel said could be correct..

bunuel can you please explain why they say 2xy value cannnot be found hence combining both the equation..

kudos to bunuel, for thinking (x-y)^2 >= 0


Well I think that solution provided by Veritas is just wrong.
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Re: Is x^2 + y^2 > 100? [#permalink]

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tarunjagtap wrote:
well from (x-y)^2 >=0 to x^2 + y^2 >= 2xy case
there could be a situation.. wen x^2 + y^2 is almost equal to 4xy or 6xy(since greater than 2xy is also applicable) then in that case we can not decide x^2 + y^2 > 100

can we vouch for this case as x^2 + y^2 equal to 4xy.
??? confused.. :?:


Not sure understood what you meant by that but anyway: 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 \(2xy\)): \(x^2+(x^2+y^2)+y^2>200\) --> \(x^2+y^2>100\).

Hope it's clear.
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first option 1:

2xy < 100 ------ 1

cant say from option 1.

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

substitute 2xy in above equation so x^2 + y^2 > 200 - 2xy
2xy is less than 100 from equation 1.
implies x^2 + y^2 > 100

so using both options.. Ans C

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Re: Is x^2 + y^2 > 100? [#permalink]

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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?


I am feeling proud about myself :). I got just the ONE question wrong in my Veritas CAT test today, and this was the one. I marked B as I proved it during the test - took 3.42 mins though. I should have got 100% correct otherwise. Good to know Veritas guys were wrong! I was a bit baffled by their explanation.

My proof (lengthy BUT conceptual proof):
(x+y)^2 > 200
=> |x + y| > 10*Sqrt(2)
=> For x^2 + y^2 to MINIMUM x=y : Why? Because squaring a number SPREADS the value exponentially. For given {x,y} such that x+y is known, x^2 + y^2 will ONLY spread MORE as we spread x and y away from their mean which is (x+y)/2 - regardless of the signs of x and y; in-fact opposite signs will spread the sum of squares even further. Hence x,y both MUST be > 5*sqrt(2). Hence MIN(x^2 + y^2) MUST be > 25*2 + 25*2 = 100. Hence, x^2 + y^2 > 100.

If there is any Veritas Instructor here, please let me know if I got it wrong in some freakish way.

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

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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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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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New post 27 Jan 2011, 07:56
Thanks Tarun. I get that (1) alone doesn't work, but how did you rule out (2)?
I chose B by trying different numbers and not finding any numbers such that (2) was true but the stem was not true.

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 27 Jan 2011, 08:15
Thanks Bunuel!

Yes, the OA was C, I've attached a screen capture. Hope I don't run into something like this on the actual GMAT :)

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New post 27 Jan 2011, 08:28
arvindg wrote:
Thanks Bunuel!

Yes, the OA was C, I've attached a screen capture. Hope I don't run into something like this on the actual GMAT :)

Attachment:
veritas_30.JPG


Yes OA is indeed given as C. So I think Veritas is wrong with this one. By the way what was their reasoning while eliminating the second statement?
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New post 27 Jan 2011, 19:28
Here's the explanation!

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New post 27 Jan 2011, 21:44
Bunuel's explanation is clear to me..
I think what bunuel said could be correct..

bunuel can you please explain why they say 2xy value cannnot be found hence combining both the equation..

kudos to bunuel, for thinking (x-y)^2 >= 0

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 29 Jan 2011, 09:42
Damn Bunuel - nice trick. very nice!
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New post 29 Jan 2011, 17:39
well from (x-y)^2 >=0 to x^2 + y^2 >= 2xy case
there could be a situation.. wen x^2 + y^2 is almost equal to 4xy or 6xy(since greater than 2xy is also applicable) then in that case we can not decide x^2 + y^2 > 100

can we vouch for this case as x^2 + y^2 equal to 4xy. ??? confused.. :?:

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New post 30 Jan 2011, 22:09
x*2+y*2 > 100, Bunuel proved it algebraically . the explanation is clear , so the answer should be B, not C , it is so much easier to fall for the trap under time pressure.
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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 02 Feb 2011, 04:41
Problem source: Veritas Practice Test

I was not able to rule out B as "Not sufficient" at first. After Bunuel's explanation, I tried to solve this with numbers and here's what I got.

Question: Is x^2 + y^2 > 100?

(1)
x=1,y=1,2xy=2<100 but x^2+y^2=1+1=2<100. Answer to Question: NO
x=10,y=1,2xy=20<100 but x^2+y^2=100+1=101>100. Answer to Question: YES
Not sufficient.

(2)
\((x+y)^2>200\)
\(|x+y|>200\)
\(x+y>\sqrt{200}\)

i.e.
\((x+y) > \sqrt{200} or (x+y) < -\sqrt{200}\) --- Restriction

Now; if we can prove that at least one value of \(x^2 + y^2\) for the given condition is <=100; this statement becomes NOT SUFFICIENT. Let's find the least value for \(x^2 + y^2\) with above restriction;

Considering the above restriction, the least value pair for x and y so that \(x^2 + y^2\) should give us the least value should be greater than the following:

\(x=\sqrt{200}/2\) and \(y=\sqrt{200}/2\)
or
\(x=-\sqrt{200}/2\) and \(y=-\sqrt{200}/2\)

Solve for \(x^2 + y^2\)
\((200/4)+(200/4)\)
\(50+50=100\)

But; this result is for x+y=(\sqrt{200}/2)+(\sqrt{200}/2); we want it for x+y>(\sqrt{200}/2)+(\sqrt{200}/2); even a little fluctuation in x or y will push the result beyond 100.

Thus; there is no value pair for x and y that gives us a value for \(x^2 + y^2\) <=100. The second statement is SUFFICIENT.

Ans: B

However, I will remember the takeaway from Bunuel that \(x^2 + y^2 >= 2xy\) for solving similar types of questions later.
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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 29 Apr 2011, 23:37
I am weak (i think the weakest) in DS.

This problem tempted me today to give a try. I solved this and came up with the answer - Option C.

Here is why.

First I simplified the question stem in this was so that I can find both the statements' issues in the stem.
x^2 + y^2 > 100, or x^2 + y^2 + 2xy > 100 + 2xy, or (x + y)^2 > 100 + 2xy

Statement 1 is not sufficient. So, we eliminate option A and D. Then, statement 2 is insufficient as well. Thus, we can eliminate the option D. Because, both the statements don't give us any specific values and keep another set of unknown/variable in the inequality.

Now, if we consider Statements 1 and 2 together, we can find different values for (x + y)^2 > 100 + 2xy. Hence, C is the answer.

As I said before, I am really weak in DS; my explanation may be so hilarious and incorrect. I will appreciate if anyone could pinpoint where problem lies with my explanation though the answer is correct.

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 07 Sep 2011, 20:18
abhicoolmax 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?


I am feeling proud about myself :). I got just the ONE question wrong in my Veritas CAT test today, and this was the one. I marked B as I proved it during the test - took 3.42 mins though. I should have got 100% correct otherwise. Good to know Veritas guys were wrong! I was a bit baffled by their explanation.

My proof (lengthy BUT conceptual proof):
(x+y)^2 > 200
=> |x + y| > 10*Sqrt(2)
=> For x^2 + y^2 to MINIMUM x=y : Why? Because squaring a number SPREADS the value exponentially. For given {x,y} such that x+y is known, x^2 + y^2 will ONLY spread MORE as we spread x and y away from their mean which is (x+y)/2 - regardless of the signs of x and y; in-fact opposite signs will spread the sum of squares even further. Hence x,y both MUST be > 5*sqrt(2). Hence MIN(x^2 + y^2) MUST be > 25*2 + 25*2 = 100. Hence, x^2 + y^2 > 100.

If there is any Veritas Instructor here, please let me know if I got it wrong in some freakish way.


Your reasoning is fine. That's good thinking. Think of it in another way:
When 2 numbers are equal, their Arithmetic Mean = Geometric Mean
AM is least when it is equal to GM and GM is greatest when it is equal to AM.
So sum of the terms is least when the numbers are equal; product is maximum when they are equal.

For minimum value of \(x^2 + y^2\), we need \(x^2 = y^2\) or |x| = |y|

On the same line, if product is given to be constant, sum is minimum when numbers are equal.
If the sum is given to be constant, the product is maximum when the numbers are equal.
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Karishma
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Re: Is x^2 + y^2 > 100?   [#permalink] 07 Sep 2011, 20:18

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