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DS Question # 1 - X^2 + Y^2?

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Source: — Data Sufficiency |

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by Rahul@gurome » Sun Dec 19, 2010 10:32 am
Statement 1: 2xy < 100
Implies xy < 50
Only from this we cannot determine whether (x² + y²) is greater than 100 or not.

Not sufficient

Statement 2: (x + y)² > 200
Implies (x + y)² - 200 > 0
=> (x + y + √200)(x + y - √200) > 0
This implies either of the following
  • 1. Both (x + y + √200) and (x + y - √200) are greater than zero. Combining the two conditions (x + y) > √200 => y > (-x + √200) => y² > (-x + √200)² = (x - √200)²

    2. Both (x + y + √200) and (x + y - √200) are less than zero. Combining these two conditions (x + y) < -√200 => => y < (-x - √200) => y² > (-x - √200)² = (x + √200)²
Again we can combine both of the above conditions as y² > (x ± √200)²
Therefore, (x² + y²) > (x ± √200)² + x²

(x ± √200)² + x² = (x² ± (2√200)x + 200) + x² = (2x² ± (2√200)x + 200) = 2(x² ± (√200)x + 100) = 2(x² ± (2√50)x + 50 + 50) = 2((x ± √50)² + 50)

Minimum value of the above expression is 100. Thus (x ± √200)² + x² ≥ 100
Thus, (x² + y²) > 100

Sufficient

The correct answer is B.
Last edited by Rahul@gurome on Mon Dec 20, 2010 4:43 am, edited 2 times in total.
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by goyalsau » Sun Dec 19, 2010 11:35 pm
Rahul@gurome wrote:Statement 1: 2xy < 100
Implies xy < 50
Only from this we cannot determine whether (x² + y²) is greater than 100 or not.

Not sufficient

Statement 2: (x + y)² > 200
Implies (x² + 2xy + y²) > 200
=> (x² + y²) > (200 - 2xy)
=> (x² + y²) > 2*(100 - xy)

Now whether (x² + y²) is greater than 100 or not depends upon the value of xy.
For example,
  • 1. If xy < 50 => 2*(100 - xy) > 2*(100 - 50) = 100 => (x² + y²) > 100
    2. If xy > 50 => 2*(100 - xy) < 2*(100 - 50) = 100 => (x² + y²) may or not be greater than 100
Not sufficient


1 & 2 Together: xy < 50
Thus (x² + y²) > 100

Sufficient

The correct answer is C.

As you have already proved that statement 2 is insufficient

Can you please provide some example values for X and Y which proves that statement 2 is insufficient.
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by prachich1987 » Sun Dec 19, 2010 11:53 pm
Rahul@gurome wrote:Statement 1: 2xy < 100
Implies xy < 50
Only from this we cannot determine whether (x² + y²) is greater than 100 or not.

Not sufficient

Statement 2: (x + y)² > 200
Implies (x² + 2xy + y²) > 200
=> (x² + y²) > (200 - 2xy)
=> (x² + y²) > 2*(100 - xy)

Now whether (x² + y²) is greater than 100 or not depends upon the value of xy.
For example,
  • 1. If xy < 50 => 2*(100 - xy) > 2*(100 - 50) = 100 => (x² + y²) > 100
    2. If xy > 50 => 2*(100 - xy) < 2*(100 - 50) = 100 => (x² + y²) may or not be greater than 100
Not sufficient

1 & 2 Together: xy < 50
Thus (x² + y²) > 100

Sufficient

The correct answer is C.
I think option B should be sufficient
I tried with many values but couldn't find a single x,y pair where (x+y)^2>200 but x^2+y^2<100.
if (x+y)^2>200
then x+y>14 or x+y<-14
Take any x,y values & you will find that if (x+y)^2>200 then x^2+y^2 has to be greater than100

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by sumit.sinha » Mon Dec 20, 2010 2:44 am
I think option B should be sufficient
I tried with many values but couldn't find a single x,y pair where (x+y)^2>200 but x^2+y^2<100.
if (x+y)^2>200
Attaboy..that is exactly what happened with me! :)[/quote]
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by Rahul@gurome » Mon Dec 20, 2010 4:41 am
Sorry guys.
The analysis I posted earlier was wrong.
Statement 2 is sufficient indeed. I've edited the reply. But I must agree that it is complicated and time consuming.

Plugging number is always a safest technique for PS problems. But for DS (for this particular type of problems), you can never be sure. That's why I tend to avoid that technique for DS.
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by ronaldramlan » Mon Dec 27, 2010 5:04 am
I truly hope that someone from Veritas can jump in and clarify this issue as the OA is C, and the explanation offered does not form a strong logical basis to claim that statement 2 is INSUFFICIENT.
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by anshumishra » Mon Dec 27, 2010 8:01 am
Why is 2 sufficient ? Here is how easy it is :

For any two distinct numbers x and y : (x-y)^2 >= 0
=> x^2 + y^2 >= 2xy

So, statement 2 :

(x+y)^2 > 200
=> x^2+y^2+2xy > 200
=> 2(x^2+y^2) > 200
=> x^2 + y^2 > 100

So it is sufficient .

Hope that makes it clear.
Last edited by anshumishra on Mon Dec 27, 2010 12:24 pm, edited 1 time in total.
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by goyalsau » Mon Dec 27, 2010 10:30 am
anshumishra wrote:Why is 2 sufficient ? Here is how easy it is :

For any two distinct numbers x and y : (x-y)^2 > 0
=> x^2 + y^2 > 2xy

So, statement 2 :

(x+y)^2 > 200
=> x^2+y^2+2xy > 200 ( Where you 2xy gone in the nest step)

=> 2(x^2+y^2) > 200
=> x^2 + y^2 > 100

So it is sufficient .

Hope that makes it clear.
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by anshumishra » Mon Dec 27, 2010 11:14 am
goyalsau wrote:
anshumishra wrote:Why is 2 sufficient ? Here is how easy it is :

For any two distinct numbers x and y : (x-y)^2 > 0
=> x^2 + y^2 > 2xy

So, statement 2 :

(x+y)^2 > 200
=> x^2+y^2+2xy > 200 ( Where you 2xy gone in the nest step)

=> 2(x^2+y^2) > 200
=> x^2 + y^2 > 100

So it is sufficient .

Hope that makes it clear.
If (x^2+y^2)+2xy > 200 and since x^2+y^2 > 2xy

Then (x^2+y^2)+(x^2+y^2) > 200 {You are adding something which greater than 2xy, So the sum should be definitely greater, right ?}.

Does that answer your question ?
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by goyalsau » Mon Dec 27, 2010 11:51 am
anshumishra wrote:
For any two distinct numbers x and y : (x-y)^2 > 0
=> x^2 + y^2 > 2xy

So, statement 2 :

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

=> 2(x^2+y^2) > 200
=> x^2 + y^2 > 100

So it is sufficient .

Hope that makes it clear.

If (x^2+y^2)+2xy > 200 and since x^2+y^2 > 2xy

Then (x^2+y^2)+(x^2+y^2) > 200 {You are adding something which greater than 2xy, So the sum should be definitely greater, right ?}.

Does that answer your question ?
You used the concept that the square of the difference of any two distant real numbers should always be greater than zero.
But they can be same as well.
As it is not given that x and y are different.........
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by anshumishra » Mon Dec 27, 2010 12:09 pm
goyalsau wrote:
anshumishra wrote:
For any two distinct numbers x and y : (x-y)^2 > 0
=> x^2 + y^2 > 2xy

So, statement 2 :

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

=> 2(x^2+y^2) > 200
=> x^2 + y^2 > 100

So it is sufficient .

Hope that makes it clear.

If (x^2+y^2)+2xy > 200 and since x^2+y^2 > 2xy

Then (x^2+y^2)+(x^2+y^2) > 200 {You are adding something which greater than 2xy, So the sum should be definitely greater, right ?}.

Does that answer your question ?
You used the concept that the square of the difference of any two distant real numbers should always be greater than zero.
But they can be same as well.
As it is not given that x and y are different.........
In any case X^2+Y^2 >= 2xy ("=" when x =y) Also, I have edited that in my first post
and You can substitute [x^2+Y^2 in place of 2xy, in the inequality : (x^2+y^2)+2xy > 200 ], to get the same result.[/i]
Thanks
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