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
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x2 + y2
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eshwarjayanth
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shankar.ashwin
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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
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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Whitney Garner
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Be careful Shankar.ashwin -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.
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].
Whitney Garner
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shankar.ashwin
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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?
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!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?
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shankar.ashwin
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You are assuming numbers are integers here. But yeah I guess its >200 anyways.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!
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Whitney Garner
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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: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?
(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
Whitney Garner
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Math is a lot like love - a simple idea that can easily get complicated
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eshwarjayanth
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Statement 1: 2xy < 100.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
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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The question stem asks about the value of x² + y².briology wrote:Where does (x-y)^2 come from? I'm confused about how/why we're adding this with (x+y)^2
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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Mo2men
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Dear Mitch,GMATGuruNY wrote:Statement 1: 2xy < 100.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
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.
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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Nice approach.Mo2men wrote:Dear Mitch,eshwarjayanth wrote:Is x^2 + y^2 > 100
1) 2xy < 100
2) (x+y)^2 > 200
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
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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