If x and y are both positive, is x greater than 4?

Question

(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 .

Bunuel · Accepted Answer

MANHATTAN GMAT OFFICIAL SOLUTION:

We are told that both x and y are positive, and we are asked whether x is greater than 4. There’s not much rephrasing we can do to this question, so let’s move on to the statements.

Statement (1): INSUFFICIENT. All we know here is that x is larger than y. It is still possible for x to be any positive number (y just has to be a smaller positive number).

Statement (2): INSUFFICIENT. If the point (x, y) lies outside a circle of radius 5 centered at the origin, then that point lies at a distance of more than 5 units away from (0, 0). In the coordinate plane, distance can be computed with the Pythagorean Theorem. We can rephrase this statement to say that x^2 + y^2 > 25. However, x can still be as small or as large a positive number as we wish.

Statement (1) and (2) together: INSUFFICIENT. We can definitely pick a very large value for x to satisfy the statements and answer the question with a “Yes.” Just make y smaller to make the first statement true, and if x is bigger than 5, then we’ll definitely have statement (2) true as well.

The trick is that we can pick a value of x that is not greater than 4 and still satisfy all the conditions. Let’s try picking x equal to 4 (this would give us an answer of “No” to the question). So we need to see whether there are any values of y that satisfy the following 3 conditions:

1. y is positive (from the stem).
2. y is less than 4 (that is, less than x).
3. x^2 + y^2 > 25.

Let’s plug 4 in for x in the last inequality. We get

16 + y^2 > 25

y^2 > 9

y > 3

(since y must be positive, we don’t have to worry about the negative possibilities)

The conditions become these: y is greater than 3 and less than 4. Any number between 3 and 4 satisfies the conditions. Notice that y is not restricted to integer values; nothing in the problem indicates that it should be. Thus, we still cannot definitively answer the question of whether x is greater than 4. There are other values of x less than 4 that will work; one tricky part of this problem is that those values of x are greater than 3.

The correct answer is E.

Prajat · Answer

St 1 => no info about the value of y => Insufficient

St 2 => (x, y) must satisfy the inequality : x^2 + y^2 > 25
 Both points (1,6) & (6, 1) will satisfy this inequality
 i.e. x can be greater than or even less than 4 => hence, insufficient

Combining both statements => (x, y) such that x > y & x^2 + y^2 > 25
 We need to test only with points from Quadrant I since both x & y are positive
 Both points (6,1) & (3.9, 3.5) will satisfy both the given conditions
 We still do not get a definite YES/NO answer.

So, correct answer is E .......(Most Probably)

longfellow · Answer

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?

GMATinsight · Answer

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

maddy_595 · Answer

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

AC07 · Answer

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 .

EncounterGMAT · Answer

This is how I solved. Hope it helps.

Posted from my mobile device

kylebecker9 · Answer

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

hadimadi · Answer

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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If x and y are both positive, is x greater than 4?

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