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Source: — Data Sufficiency |
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by shankar.ashwin » Mon Oct 10, 2011 11:34 am
From (2) we have (x+y)^2 > 200

We try to prove a case where x^2 + y^2 would be less than 100.
Sum of the squares would be minimum when numbers are equal.

Here, sub x=y and try;

(2x)^2 > 200
x > Sqrt(200)/2
x>10/Sqrt(2)

When I sub x=y=10/Sqrt(2), I get (x+y)^2 = 200.

When from (2) we know x>10/Sqrt(2), Hence the sum of squares would be greater than 200 IMO. Sufficient B
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by Whitney Garner » Mon Oct 10, 2011 12:22 pm
shankar.ashwin wrote:From (2) we have (x+y)^2 > 200
We try to prove a case where x^2 + y^2 would be less than 100.
Sum of the squares would be minimum when numbers are equal.
Be careful Shankar.ashwin -

The sum of squares would only be at a minimum if x=y=0. For example:

x=y=2; (x+y)^2 = (2+2)^2 = 4^2
x=y=9; (x+y)^2 = (9+9)^2 = 18^2
...
None of these are the minimum value for (x+y)^2
...
x=y=0; (x+y)^2 = (0+0)^2 = 0

...because the ^2 makes any result positive, then 0 is the least result we can find. You might be thinking of the difference of squares [because (x-y)^2 would be at its minimum value when x=y].
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by shankar.ashwin » Mon Oct 10, 2011 12:47 pm
I think my wording in the previous reply was very poor.

If (x+y)^2 > 200

We could have x=14 and y =2
which would satisfy 14^2+2^2 = 200.

Here x+y=16.

and (x+y)^2 would be 256.(which is > 200)

We need to find the minimum value of x+y which would satisfy (x+y)^2 > 200.

I think that would be when x=y as solved before. Am I wrong here?

Whitney or any experts, could you please help?
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by n@resh » Mon Oct 10, 2011 12:59 pm
shankar.ashwin wrote:I think my wording in the previous reply was very poor.

If (x+y)^2 > 200

We could have x=14 and y =2
which would satisfy 14^2+2^2 = 200.

Here x+y=16.

and (x+y)^2 would be 256.(which is > 200)

We need to find the minimum value of x+y which would satisfy (x+y)^2 > 200.

I think that would be when x=y as solved before. Am I wrong here?

Whitney or any experts, could you please help?
As you have ( x + y )^2 > 200, then the nearest possible sum of x and y which satisfy will be: 15.Based on this lets say x = 8 and y = 7.. or say 13 or 2 or 11 or 4...etc, make sure sum must be 15, so here x^2 + y^2 >100 ..that's gonna be true always whenever ( x + y )^2 > 200!
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by shankar.ashwin » Mon Oct 10, 2011 1:05 pm
n@resh wrote:
As you have ( x + y )^2 > 200, then the nearest possible sum of x and y which satisfy will be: 15.Based on this lets say x = 8 and y = 7.. or say 13 or 2 or 11 or 4...etc, make sure sum must be 15, so here x^2 + y^2 >100 ..that's gonna be true always whenever ( x + y )^2 > 200!
You are assuming numbers are integers here. But yeah I guess its >200 anyways.
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by Whitney Garner » Mon Oct 10, 2011 2:26 pm
shankar.ashwin wrote:I think my wording in the previous reply was very poor.

If (x+y)^2 > 200

We could have x=14 and y =2
which would satisfy 14^2+2^2 = 200.

Here x+y=16.

and (x+y)^2 would be 256.(which is > 200)

We need to find the minimum value of x+y which would satisfy (x+y)^2 > 200.

I think that would be when x=y as solved before. Am I wrong here?

Whitney or any experts, could you please help?
AH - Now I see what you were trying to explain and I would definitely agree with you. To minimize the effect of squaring each piece in (x)^2 + (y)^2, we would want to make them both as small as possible. But their sum must still exceed the square root of 200. That means that we would want to make them equal and ever so slightly more than root(200). If we set them equal to sqrt(200)/2 to establish the boundary, we get the following:

(sqrt(200)/2)^2 + (sqrt(200)/2)^2
= 200/4 + 200/4
= 50 + 50
= 100.

So we know that the minimum boundary (that actually cannot be reached because it is strictly >200) is that (x)^2 + (y)^2 > 100.

Would OP please note the source?

Thanks!
:)
Whit
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by eshwarjayanth » Tue Oct 11, 2011 12:58 am
Source: Veritas free CAT

OA given is C
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by GMATGuruNY » Tue Oct 11, 2011 5:02 am
eshwarjayanth wrote:Is x^2 + y^2 > 100

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

OA - C

Can someone pls explain why B is not sufficient
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.

The correct answer is B.

If the OA is C, then the OA is incorrect.
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by briology » Tue Oct 11, 2011 9:48 am
Where does (x-y)^2 come from? I'm confused about how/why we're adding this with (x+y)^2
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by GMATGuruNY » Tue Oct 11, 2011 10:16 am
briology wrote:Where does (x-y)^2 come from? I'm confused about how/why we're adding this with (x+y)^2
The question stem asks about the value of x² + y².
Statement 2 offers information about (x+y)².
(x+y)² = x² + 2xy + y².
We want to eliminate 2xy from the expression above in order to isolate x² + y².
(x-y)² = x² - 2xy + y².
When the two expressions are added, the 2xy term cancels out, allowing us to isolate x² + y²:
(x+y)² + (x-y)² = (x² + 2xy + y²) + (x² - 2xy + y²) = 2x² + 2y² = 2(x² + y²).
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by briology » Tue Oct 11, 2011 7:19 pm
Thanks. That makes a lot of sense. I didn't consider adding a foreign equation to the problem
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by GmatKiss » Sun Oct 16, 2011 7:20 am
Thanks Mitch!
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by Mo2men » Wed May 31, 2017 8:14 am
GMATGuruNY wrote:
eshwarjayanth wrote:Is x^2 + y^2 > 100

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

OA - C

Can someone pls explain why B is not sufficient
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.

The correct answer is B.

If the OA is C, then the OA is incorrect.
Dear Mitch,

I have another view for Fact 2 as follows:

(x+y)^2 > 200

x + y > 10√2 or x + y < - 10√2

We can test the threshold

x = 5√2 & y= 5√2 ... Apply in question stem x^2 + y^2 = 100..This implies that any raise in x or y will make x^2 + y^2 > 100.

The same can be done for x =- 5√2 & y= - 5√2..... This implies that any raise in x or y will make x^2 + y^2 > 100.

So sufficient.

is my reasoning above correct?

Do we have number like 6.1 √2? it is combines of 6.1 & √2?

Thanks
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by GMATGuruNY » Wed May 31, 2017 4:10 pm
Mo2men wrote:
eshwarjayanth wrote:Is x^2 + y^2 > 100

1) 2xy < 100
2) (x+y)^2 > 200
Dear Mitch,

I have another view for Fact 2 as follows:

(x+y)^2 > 200

x + y > 10√2 or x + y < - 10√2

We can test the threshold

x = 5√2 & y= 5√2 ... Apply in question stem x^2 + y^2 = 100..This implies that any raise in x or y will make x^2 + y^2 > 100.

The same can be done for x =- 5√2 & y= - 5√2..... This implies that any raise in x or y will make x^2 + y^2 > 100.

So sufficient.

is my reasoning above correct?

Do we have number like 6.1 √2? it is combines of 6.1 & √2?

Thanks
Nice approach.
It would be wise, however, to test cases in which x≠y.
Case 3: x=6√2 and y=4√2, with the result that (x+y)² = 200
In this case, x² + y² = (6√2)² + (4√2)² = 72 + 32 = 104.
Case 4: x=10√2 and y=0, with the result that (x+y)² = 200
In this case, x² + y² = (10√2)² + 0² = 200.
The values in red are all GREATER than the result yielded when x=y.
Implication:
If (x+y)² = 200, then the least possible value for x² + y² is yielded when x=y:
(5√2)² + (5√2)² = 100.
Since Statement 2 requires that (x+y)² > 200, it must be true that x² + y² is GREATER THAN the value in blue:
x² + y² > 100.
SUFFICIENT.
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