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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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New post 15 Jan 2016, 02:06
VeritasPrepKarishma wrote:
Bunuel wrote:
VeritasPrepKarishma wrote:

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?




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.



Thanks Bunuel and Karishma for detailed explanation... one confusion though...
from point 2) if 2xy is less than x^2 + y^2 .... lets say it is - (x^2 +y^2)... than the equation becomes 0 > 200 ? Am I missing some point
( SInce we need to know the minimum value of left side ... we should be able to substitute (x^2 + y^2 ) by 2xy..... but not 2xy by (x^2 + y^2 )

Thanks for any inputs...

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Re: Is x^2 + y^2 > 100? [#permalink]

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 30 May 2017, 16:21
MacFauz wrote:
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.

Kudos Please... If my post helped.


Bunuel this is the same reasoning I used- is this actually valid?

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 30 May 2017, 21:21
Nunuboy1994 wrote:
MacFauz wrote:
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.

Kudos Please... If my post helped.


Bunuel this is the same reasoning I used- is this actually valid?


Algebraic approach is more precise. The fact that you cannot find some set of values does not necessarily mean that there is no such set of values.
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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 24 Aug 2017, 11:56
VeritasPrepKarishma wrote:
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.


VeritasPrepKarishma
hi mam

for statement 1, we cannot set x to equal to y, as any maximum value can untrue the statement, but we can duly set x equals to y for statement 2, to find out the minimum value that complies to the information provided by statement 2...

is that the logic behind this operation, or anything else ....? please say to me, mam...

thanks in advance

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 27 Aug 2017, 11:46
Hi Bunuel / VeritasPrepKarishma,

In the post, link provided (https://gmatclub.com/forum/advanced-num ... 94102.html),

Quote:
"Question: If x and y are positive, is x^2 + y^2 > 100?

Statement 1: 2xy < 100

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

Solution:

We need to find whether x^2 + y^2 must be greater than 100.

Statement 1: 2xy < 100

Plug in some easy values to see that this is not sufficient alone.

If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100

If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100"



Unquoted: In the question, we are given that x and y are positive then why you considered x=0, y=0 in statement 1 because 0 is neither positive nor negative.

Please clarify for the doubt.

Thanks.

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 28 Aug 2017, 02:56
Bunuel wrote:
arvindg wrote:
Problem source: Veritas Practice Test

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.


OMG!!
Does this kind of question can be found in the real test?
I would have spent 3-5 min on this and got it wrong....

The quoted part of Brunnel solution is just... ingenious

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Is x^2 + y^2 > 100? [#permalink]

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New post 05 Sep 2017, 11:05
Although there is no substitute for Bunuel's solution, as it is very neat, and requires clever ideas like (x−y)^2≥0 (square of any number is more than or equal to zero) and requires crystal clear knowledge of algebra/equations. Since I could not think of that, I solved it in the following manner and reached the answer B:

1 is insufficient

2 says (x+y)^2>200 --> (x+y)^2>(14.something)^2
Let's say either x or y is zero then either of x or y has to be greater than 14, and inequality will hold for both the -ve and +ve values both and any combination of values of x and y. Hence, the value of x^2+y^2 will be more than 100

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 22 Sep 2017, 14:58
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?



Hi Bunuel,
I understand the algebraic solution now but I had a different reasoning when I did this problem.

Let's say (x + y)^2 = 200. In this case x+y would be 10\sqrt{2}. I considered the minimum possibility where x=y which makes each of them 5\sqrt{2}. So x^2+y^2=100.

Since it our question says (x + y)^2 > 200, based on above deduction, I assumed that their (x + y) should definitely be more than 10 \sqrt{2} implying that x^2+y^2 should also be greater than 100.

Is this a reasonable deduction? I honestly didn't even think of any other numbers to prove my case incorrect after this and I know that doesn't stand well with other questions maybe but I was wondering if it was a good method for this question.

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Re: Is x^2 + y^2 > 100? [#permalink]

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New post 25 Sep 2017, 02:54
rajudantuluri 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?



Hi Bunuel,
I understand the algebraic solution now but I had a different reasoning when I did this problem.

Let's say (x + y)^2 = 200. In this case x+y would be 10\sqrt{2}. I considered the minimum possibility where x=y which makes each of them 5\sqrt{2}. So x^2+y^2=100.

Since it our question says (x + y)^2 > 200, based on above deduction, I assumed that their (x + y) should definitely be more than 10 \sqrt{2} implying that x^2+y^2 should also be greater than 100.

Is this a reasonable deduction? I honestly didn't even think of any other numbers to prove my case incorrect after this and I know that doesn't stand well with other questions maybe but I was wondering if it was a good method for this question.


rajudantuluri
hi

I have a similar question to Bunuel the great
Question stem:

Is "x^2 + y^2 > 100"
Now taking the minimum values for x and y

x^ + x^2 > 100
or, 2x^2 > 100

Now the question becomes:
Is 2x^2 > 100

statement 1 says:

twice the value, the product of x and y, is less than 100
If maximized, it becomes, 2x^2 < 100
but, we don't know whether the maximum value is less than 100
not sufficient

statement 2 says:

(x + y)^2 > 200
taking minimum values

(x + x)^2 > 200
or (2x)^ 2 > 200
or 4x^2 > 200
reducing by 1/2

2x^2 > 100

sufficient ...
B

But, rajudantuluri, here I have a question, why it has to be assumed that the minimum value of (x + y)^2 will exceed 200 ...?

thanks in advance

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Re: Is x^2 + y^2 > 100?   [#permalink] 25 Sep 2017, 02:54

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