Veritas Confusing Question

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Veritas Confusing Question

by anandhimanshu85 » Sat Feb 11, 2012 4:54 am
IS X^ 2 + Y^2>100

1.2XY<100

2.(X + Y)^2>200

IMO OA [spoiler]= B as in order to have second statement true a+b should be atleast 15 and any sum for x and y to 15 have their sum of the square greater than 100 but OA was C.Please suggest [/spoiler]

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by Brent@GMATPrepNow » Sat Feb 11, 2012 7:32 am
anandhimanshu85 wrote:IS X^ 2 + Y^2>100

1.2XY<100

2.(X + Y)^2>200
Target Question: Is x^2 + y^2 > 100?

Statement 1: 2xy < 100
The only way to show that this statement is not sufficient is by counter-example.
There are several pairs of values that satisfy the condition that 2xy < 100
case a: x=0 y=0, in which case x^2 + y^2 is less than 100
case b: x=0 y=100, in which case x^2 + y^2 is greater than 100
So, statement 1 is NOT SUFFICIENT

Statement 2: (x+y)^2 > 200
Expand to get: x^2 + 2xy + y^2 > 200
Rearrange to get: x^2 + y^2 + 2xy > 200

Aside: (this is the tricky part)
Notice that for any values of x and y, it must be true that (x-y)^2 > 0
Expand to get x^2 - 2xy + y^2 > 0
Rearrange to get x^2 + y^2 > 2xy
Now add x^2 + y^2 to both sides to get: x^2 + y^2 + x^2 + y^2 > x^2 + y^2 + 2xy
Simplify to get: 2(x^2 + y^2) > x^2 + y^2 + 2xy

Now notice that the right-hand side of this inequality matches the left-hand side of the inequality we derived from statement 2.
So, we can write: 2(x^2 + y^2) > x^2 + y^2 + 2xy > 200
Now ignore the middle part to get: 2(x^2 + y^2) > 200
Divide both sides by 2 to get x^2 + y^2 > 100
Since we can answer the target question with certainty, statement 2 is SUFFICIENT and the answer is B

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by anandhimanshu85 » Sat Feb 11, 2012 11:07 am
Thanks Brent, I also opted for a B but I was surprised to see a C as the Answer, so thought to verify here and you have confirmed it I was right...:)
Brent@GMATPrepNow wrote:
anandhimanshu85 wrote:IS X^ 2 + Y^2>100

1.2XY<100

2.(X + Y)^2>200
Target Question: Is x^2 + y^2 > 100?

Statement 1: 2xy < 100
The only way to show that this statement is not sufficient is by counter-example.
There are several pairs of values that satisfy the condition that 2xy < 100
case a: x=0 y=0, in which case x^2 + y^2 is less than 100
case b: x=0 y=100, in which case x^2 + y^2 is greater than 100
So, statement 1 is NOT SUFFICIENT

Statement 2: (x+y)^2 > 200
Expand to get: x^2 + 2xy + y^2 > 200
Rearrange to get: x^2 + y^2 + 2xy > 200

Aside: (this is the tricky part)
Notice that for any values of x and y, it must be true that (x-y)^2 > 0
Expand to get x^2 - 2xy + y^2 > 0
Rearrange to get x^2 + y^2 > 2xy
Now add x^2 + y^2 to both sides to get: x^2 + y^2 + x^2 + y^2 > x^2 + y^2 + 2xy
Simplify to get: 2(x^2 + y^2) > x^2 + y^2 + 2xy

Now notice that the right-hand side of this inequality matches the left-hand side of the inequality we derived from statement 2.
So, we can write: 2(x^2 + y^2) > x^2 + y^2 + 2xy > 200
Now ignore the middle part to get: 2(x^2 + y^2) > 200
Divide both sides by 2 to get x^2 + y^2 > 100
Since we can answer the target question with certainty, statement 2 is SUFFICIENT and the answer is B

Cheers,
Brent

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by GMATGuruNY » Sat Feb 11, 2012 1:42 pm
I offered a solution (similar to Brent's) in my two posts here:

https://www.beatthegmat.com/x2-y2-t92678.html
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For more information, please email me (Mitch Hunt) at [email protected].
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by [email protected] » Wed Feb 15, 2012 11:39 pm
Hi Brent and Mitch!!! I solved the question in a different way... I hope its right...


IS X^ 2 + Y^2>100

1.2XY<100

2.(X + Y)^2>200



Statement 1: xy < 50

therefore when x = 5 and y = 6 yes it comes out to be 25 and 36.

But when when x = 9 and y = 5; xy = 45 but x^2 + y^2 = 81 + 25 which turns out to be greater than 100...

Hence statement 1 is insufficient...


STATEMENT 2:

(X + Y)^2 > 200

NOW lets us solve this equation first...

taking sq root on both sides,

(X + Y) > 10sqrt root of 2 or (X + Y) < -10 sqr root of 2

basically it comes out to be approximately + or - 15

Hence either (X + Y) > 15 or (X + Y) < -15

In both the cases take as many values as possible,

the answer comes out to be (X)^2 + Y^2 > 100

Hence statement 2 is sufficient...

Hope this really helped...... took a lot of time to figure out this question...
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by Brent@GMATPrepNow » Thu Feb 16, 2012 6:47 am
[email protected] wrote:Hi Brent and Mitch!!! I solved the question in a different way... I hope its right...


IS X^ 2 + Y^2>100

1.2XY<100

2.(X + Y)^2>200



Statement 1: xy < 50

therefore when x = 5 and y = 6 yes it comes out to be 25 and 36.

But when when x = 9 and y = 5; xy = 45 but x^2 + y^2 = 81 + 25 which turns out to be greater than 100...

Hence statement 1 is insufficient...


STATEMENT 2:

(X + Y)^2 > 200

NOW lets us solve this equation first...

taking sq root on both sides,

(X + Y) > 10sqrt root of 2 or (X + Y) < -10 sqr root of 2

basically it comes out to be approximately + or - 15

Hence either (X + Y) > 15 or (X + Y) < -15

In both the cases take as many values as possible,

the answer comes out to be (X)^2 + Y^2 > 100


Hence statement 2 is sufficient...

Hope this really helped...... took a lot of time to figure out this question...
Looking for contradictory sets of values (as you did with statement 1) is a great way to show insufficiency. Unfortunately, you really can't do it to prove sufficiency.

There's a problem with the part in blue above.

To be certain that statement 2 is sufficient, it's not enough to say "In both the cases take as many values as possible, the answer comes out to be X^2 + Y^2 > 100"

Perhaps it's possible that out of the infinite possibilities out there, there's one pair of values for x and y such that X^2 + Y^2 is not greater than 100. We won't know for sure unless we try every pair of values.

That said, this method is useful for getting a general feeling about whether a statement is sufficient. For example, if you plug in several pairs of numbers and, each time, X^2 + Y^2 > 100, then you might feel confident that the statement is sufficient. However, you can't be absolutely certain of your answer.

I cover this in greater detail at the end of this video: https://www.beatthegmat.com/mba/2010/10/ ... ble-method

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Brent
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by ArunangsuSahu » Fri Feb 17, 2012 12:18 pm
Statement 2:
Let's take x=y and
(x+y)^2=(10*2^1/2)^2

x=5*2^1/2=y that's the least

and x^2+y^2=100

So for all other cases for (x+y)^2>200
x^2+y^2>100

(B)..SUFFICIENT

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by pappueshwar » Wed Feb 29, 2012 7:20 am
thanks mith,
ur solution was easy to understand ...
GMATGuruNY wrote:I offered a solution (similar to Brent's) in my two posts here:

https://www.beatthegmat.com/x2-y2-t92678.html

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by shantanugangal » Sun Apr 29, 2012 8:13 am
I agree with the above posts and am surprised that Veritas got the answer wrong!