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Re: If x and y are both positive, is x greater than 4? [#permalink]
Answer as per me is C. only first quadrant is to be considered. As x>y, area between x=0 and x=y is to be considered. thus, C.

Bunuel, is the approach right?
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Re: If x and y are both positive, is x greater than 4? [#permalink]
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Bunuel wrote:
If x and y are both positive, is x greater than 4?

(1) x > y
(2) In the coordinate plane, the point (x, y) lies outside a circle of radius 5 centered at the origin.

Kudos for a correct solution.


Given: x and y are both POSITIVE

Question : Is x>4?

Statement 1: x>y

Case 1: y = 1, x=3 i.e. x < 4
Case 2: y = 1, x=5 i.e. x > 4
NOT SUFFICIENT

Statement 2: In the coordinate plane, the point (x, y) lies outside a circle of radius 5 centered at the origin
i.e. \(x^2 + y^2 > 5^2\)
i.e. \(x^2 + y^2 > 25\)
Case 1: y = 4, x=3.5 i.e. x < 4
Case 2: y = 4, x=4.5 i.e. x > 4
NOT SUFFICIENT

Combining the two statements:
\(x^2 + y^2 > 25\) and \(x>y\)
Case 1: \(y = 5/\sqrt{2} = 3.54\), \(x=3.6\) i.e. x < 4
Case 2: y = 4, x=4.5 i.e. x > 4

Answer: Option E
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Re: If x and y are both positive, is x greater than 4? [#permalink]
Hi Experts,

I thought a lot on this but couldnt picture a values of x and Y when x>y -- Condition on combining both options.

Let us forget a equation x^2+Y^2 > 25 for a moment. Can you help me plotting the points on the graph which satisfies below conditions-

X Can be less then 4
X < Y
X,Y lie outside the circle of radius 5
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If x and y are both positive, is x greater than 4? [#permalink]
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Bunuel wrote:
If x and y are both positive, is x greater than 4?

(1) x > y
(2) In the coordinate plane, the point (x, y) lies outside a circle of radius 5 centered at the origin.

Kudos for a correct solution.


Given: x > 0, y > 0
x > 4?

Statement 1:
x > y
This statement is insufficient to prove whether x > 4 because there are innumerable pairs of positive numbers where x > y and x < 4 such as (2, 1) and also x > y and x > 4 such as (5, 3), etc.

Statement 2:
Given: the point (x, y) lies outside a circle of radius 5 centered at the origin.
The point (x, y) lies on a circle of radius \(\sqrt{x^2 + y^2}\) centered at the origin.
Since the point (x, y) lies outside the circle of radius 5, this implies that the radius of the circle with the point (x, y) is greater than 5.
So, \(\sqrt{x^2 + y^2} > 5\)
\(x^2 + y ^2 > 25\)
But, there are many pairs of numbers which satisfy this inequality without giving a definite answer on whether x > 4 such as (2, 5), (4, 4), (5, 1), etc.

Since the individual statements are insufficient individually, on combining them:
x > y and \(x^2 + y ^2 > 25\)
Since, x > y and x and y are positive
\(x^2 > y^2\)
\(x^2 + x^2 > y^2 + x^2 > 25\)
\(2x^2 > 25\)
x > \(5/\sqrt{2}\)
x > 3.5

But, combining the statements hasn't concluded if x > 4.
Hence, the statements are sufficient neither individually nor combined.

Therefore, the answer is E.
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If x and y are both positive, is x greater than 4? [#permalink]
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This is how I solved. Hope it helps.

Posted from my mobile device
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Originally posted by EncounterGMAT on 23 Oct 2018, 09:37.
Last edited by EncounterGMAT on 15 Sep 2019, 00:47, edited 1 time in total.
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Re: If x and y are both positive, is x greater than 4? [#permalink]
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Bunuel wrote:
If x and y are both positive, is x greater than 4?

(1) x > y
(2) In the coordinate plane, the point (x, y) lies outside a circle of radius 5 centered at the origin.

Kudos for a correct solution.



From the stem we know x > 0 and y > 0

(1) Clearly not sufficient. X = 2, y = 1 works and so does x = 5, y = 4 (although we are not limited to integers as in the choices I made)

(2) we now know x^2+y^2 > 25, and x > 0, y > 0 This is still not sufficient. Examples include x = 1, y = 100 and x = 6, y = 2

(+) now in the coordinate plane, we only care about points in quadrant one that are both below the line y=x in and outside the circle y^2+x^2 = 25

It is obvious that we can find values of x that satisfy this >4, so the question is are there any values of 4 or less. Well the minimum value has to be just greater than 2x^2 = 25 (subbing in y=x for y in the equation of the circle). That gives us the square root of 12.5 which is less than the square of 4 (16), so x could be less than or equal to 4 still.

Answer E
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Re: If x and y are both positive, is x greater than 4? [#permalink]
Hi,

here is why it's (C):

We find the point (5/root(2), 5/root(2))=(3.5,3.5) on the circle which is the only one in the positive quadrant in which both have equal values. Moving slightly to the right in order to get x out of the circle (I assume the border counts as inside), x increases while y decreases. For that point, x>y and x<4. Now pick (20,0). That point is outside, x>y, but x>20, so we can't answer
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Re: If x and y are both positive, is x greater than 4? [#permalink]
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