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

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If x and y are both positive, is x greater than 4? [#permalink]

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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.
[Reveal] Spoiler: OA

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

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

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New post 12 Sep 2015, 04:45
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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New post 12 Sep 2015, 09:05
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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]

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New post 14 Sep 2015, 06:33
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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.


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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: If x and y are both positive, is x greater than 4? [#permalink]

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New post 16 Sep 2015, 10:12
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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New post 18 Sep 2017, 01:51
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.
If x and y are both positive, is x greater than 4?   [#permalink] 18 Sep 2017, 01:51
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