Is x^2 + y^2 100? (1) 2xy 200

Question

Is x^2 + y^2 > 100?

(1) 2xy < 100

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

Bunuel · Accepted Answer

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?

tarunjagtap · Answer

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

arvindg · Answer

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

veritas_30.JPG

Bunuel · Answer

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?

arvindg · Answer

Here's the explanation! Veritas_30_expln.jpg

tarunjagtap · Answer

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

Bunuel · Answer

Well I think that solution provided by Veritas is just wrong.

144144 · Answer

Damn Bunuel - nice trick. very nice!

KarishmaB · Answer

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.

tarunjagtap · Answer

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.. :?:

Bunuel · Answer

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.

abhicoolmax · Answer

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.

KarishmaB · Answer

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.

LinaNY · Answer

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?

Bunuel · Answer

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.

anujtsingh · Answer

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

Bunuel · Answer

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?

GMATGuruNY · Answer

Statement 1: 2xy < 100 Thus, xy < 50. If x=1 and y=1, then x²+y² < 100. If x=2 and y=10, then x²+y² > 100. INSUFFICIENT. Statement 2: (x+y)² > 200 Since the square of a value cannot be negative, (x-y)² ≥ 0. Adding together (x+y)² > 200 and (x-y)² ≥ 0, we get: (x+y)² + (x-y)² > 200+0 (x² + 2xy + y²) + (x² - 2xy + y²) > 200 2x² + 2y² > 200 x² + y² > 100 SUFFICIENT. B

EllaWoodMN · Answer

Hi, Please help verify my approach. it also ends up with B.

So rephrase the question to |x|+|y| > 10?

From statement 2, |x+y| > 10\sqrt{2}

Since |x+y| is always less than or equal to |x|+|y| then if |x+y| is more than 10\sqrt{2} which is more than 10, then |x| + |y| must be more 10.

Bunuel  @KarismaB

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

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