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# In the xy-plane, region Q consists of all points (x,y) such that x^2 +

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Joined: 02 Sep 2009
Posts: 52906
In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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11 Nov 2014, 08:14
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:46) correct 20% (02:01) wrong based on 274 sessions

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Tough and Tricky questions: Coordinate Geometry.

In the $$xy$$-plane, region $$Q$$ consists of all points $$(x, y)$$ such that $$x^2 + y^2 \le 100$$. Is the point $$(a, b)$$ in region $$Q$$?

(1) $$a + b = 14$$

(2) $$a \gt b$$

Kudos for a correct solution.

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Joined: 02 Sep 2009
Posts: 52906
Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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12 Nov 2014, 03:47
1
5
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

In the $$xy$$-plane, region $$Q$$ consists of all points $$(x, y)$$ such that $$x^2 + y^2 \le 100$$. Is the point $$(a, b)$$ in region $$Q$$?

(1) $$a + b = 14$$

(2) $$a \gt b$$

Kudos for a correct solution.

Official Solution:

The first step is to figure out what region $$Q$$ represents. Let's consider the boundary of region $$Q$$ by ignoring the "less than" part: $$x^2 + y^2 = 100$$. This equation represents a circle in the $$xy$$-plane, centered on the origin, with a radius of 10 (the square root of 100). Thus, region $$Q$$ consists of all points on or inside this circle. We are asked whether point $$(a, b)$$ lies inside this region. We can rephrase the question by substituting $$a$$ for $$x$$ and $$b$$ for $$y$$: do the variables $$a$$ and $$b$$ always satisfy the inequality $$a^2 + b^2 \le 100$$?

Statement (1): INSUFFICIENT. Relatively quickly, we can find a point on the line $$a + b = 14$$ that does not satisfy the inequality. Choose $$a = 0$$. Then $$b = 14$$, and the sum of the squares is 196, which is greater than 100. Thus, in this case, $$(a, b)$$ would not fall within region $$Q$$.

However, can we find any point on or within the circle? If we make both $$a$$ and $$b$$ equal 7, then the sum of their squares is $$49 + 49 = 98$$, which is less than 100. We could also choose $$a = 8$$ and $$b = 6$$, which gives us the sum of squares $$64 + 36 = 100$$. Either case satisfies the inequality, and so $$(a, b)$$ in these cases would fall within region $$Q$$.

Statement (2): INSUFFICIENT. The condition that $$a$$ is greater than $$b$$ is not very restrictive. We can find points that meet this condition both inside and outside the circle. For instance, $$(1, 0)$$ is within the circle, but $$(101, 100)$$ is not.

Statements (1) and (2) together: INSUFFICIENT. The case from statement 1 in which both $$a$$ and $$b$$ equaled 7 is no longer valid, but the case of $$a = 8$$ and $$b = 6$$ still works. Thus, we have a point satisfying both statements that lies on the circle. In fact, some suitable points lie within the circle, such as $$(7.5, 6.5)$$. However, we can still find suitable points that lie outside the circle - for instance, $$(14, 0)$$.

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##### General Discussion
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Joined: 10 Sep 2014
Posts: 96
Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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11 Nov 2014, 09:40
2
1

So basically what we are working with is (a+b)(a-b) < 100

statement 1: not sufficient
If A is -7 and B is 21, then (a+b)(a-b) would be much larger than 100 - not in region Q
If A is 7 and B is 7, then (a+b)(a-b) would be 0 - yes in region Q

statement 2: not sufficient
If A is 40 and B is 30, then (a+b)(a-b) would be 700 and this is not less than or equal to 100 - not in region Q
If A is 10 and B is 5, then (a+b)(a-b) would be 75 which is less than 100 - yes in region Q

So now I am down to answer choices C and E to choose from.

Combining both statements:
If A is 8 and B is 6, then (a+b)(a-b) would be 28 - yes in region Q
If A is 21 and B is -7, then (a+b)(a-b) would be much larger than 100 - not in region Q

Manager
Joined: 22 Sep 2012
Posts: 129
Concentration: Strategy, Technology
WE: Information Technology (Computer Software)
Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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11 Nov 2014, 18:59
1
x^2+y^2 <= 100 represents a circular region.

Statement 1 : a + b = 14

If a = 14 , b= 0, the point lies outside the region
If a= 7 , b = 7, then the point lies within the circle.
Insufficient

Statement 2 : a > b

If a = 15 , b= 0, the point lies outside the region
If a = 2, b= 2, the point lies within the circle.
Insufficient

Combining both , we can say
If a = 14 , b= 0, the point lies outside the region
If a = 8, b= 6, the point is on the circle.

Still it is not conclusive whetherthe point is outside or inside the circle. Therefore E) is the answer
Manager
Joined: 21 Mar 2017
Posts: 54
Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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20 Jul 2017, 05:15
Bunuel wrote:
Bunuel wrote:

Tough and Tricky questions: Coordinate Geometry.

In the $$xy$$-plane, region $$Q$$ consists of all points $$(x, y)$$ such that $$x^2 + y^2 \le 100$$. Is the point $$(a, b)$$ in region $$Q$$?

(1) $$a + b = 14$$

(2) $$a \gt b$$

Kudos for a correct solution.

Official Solution:

The first step is to figure out what region $$Q$$ represents. Let's consider the boundary of region $$Q$$ by ignoring the "less than" part: $$x^2 + y^2 = 100$$. This equation represents a circle in the $$xy$$-plane, centered on the origin, with a radius of 10 (the square root of 100). Thus, region $$Q$$ consists of all points on or inside this circle. We are asked whether point $$(a, b)$$ lies inside this region. We can rephrase the question by substituting $$a$$ for $$x$$ and $$b$$ for $$y$$: do the variables $$a$$ and $$b$$ always satisfy the inequality $$a^2 + b^2 \le 100$$?

Statement (1): INSUFFICIENT. Relatively quickly, we can find a point on the line $$a + b = 14$$ that does not satisfy the inequality. Choose $$a = 0$$. Then $$b = 14$$, and the sum of the squares is 196, which is greater than 100. Thus, in this case, $$(a, b)$$ would not fall within region $$Q$$.

However, can we find any point on or within the circle? If we make both $$a$$ and $$b$$ equal 7, then the sum of their squares is $$49 + 49 = 98$$, which is less than 100. We could also choose $$a = 8$$ and $$b = 6$$, which gives us the sum of squares $$64 + 36 = 100$$. Either case satisfies the inequality, and so $$(a, b)$$ in these cases would fall within region $$Q$$.

Statement (2): INSUFFICIENT. The condition that $$a$$ is greater than $$b$$ is not very restrictive. We can find points that meet this condition both inside and outside the circle. For instance, $$(1, 0)$$ is within the circle, but $$(101, 100)$$ is not.

Statements (1) and (2) together: INSUFFICIENT. The case from statement 1 in which both $$a$$ and $$b$$ equaled 7 is no longer valid, but the case of $$a = 8$$ and $$b = 6$$ still works. Thus, we have a point satisfying both statements that lies on the circle. In fact, some suitable points lie within the circle, such as $$(7.5, 6.5)$$. However, we can still find suitable points that lie outside the circle - for instance, $$(14, 0)$$.

Bunuel
I understood this method. Would you plz help me to do the same DS using graphic method? I have tried but couldn't.
walker You can also help, I believe
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Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +  [#permalink]

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14 Oct 2018, 00:53
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Re: In the xy-plane, region Q consists of all points (x,y) such that x^2 +   [#permalink] 14 Oct 2018, 00:53
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