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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 11 Nov 2014, 09:14
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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) \(a \gt b\)

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E
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Bunuel
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 12 Nov 2014, 04:47
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Bunuel

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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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, \((14, 0)\).


Answer: E
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 11 Nov 2014, 10:40
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I chose answer choice E.

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

Answer choice E!
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 11 Nov 2014, 19:59
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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
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 20 Jul 2017, 06:15
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Bunuel
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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.

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, \((14, 0)\).


Answer: E
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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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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 07 Dec 2020, 11:02
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Shaded Region in Red represents area in the question stem i.e. \(x^2+y^2<=100\)
Shaded Region in Blue Represents a>b
Green line Represents a+b=14

I found this to be the simplest way of solving the mentioned type of question (perhaps with the logic of line a+b=14 that it will cut through the circle)
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 24 Feb 2022, 15:03
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Please could someone let me know where I messed up in my attempt to answer this


(1) tells us that " a+b = 14", thus (a+b)^2 = a^2 + b^2 + 2ab. We know what a+b is, so we know that (a+b)^2 = 14^2 = 196

Therefore (via substitution into the first line): 196 = a^2 + b^2 + 2ab. For (a,b) to lie in the region, we require a^2 + b^2 <= 100.

This is the same as requiring 196 - 2ab <= 100 (via rearranging the above equation)

Thus, -2ab <= -96, therefore, ab >= 48. But Statement 1 tells us that a+b = 14, it does not tell us whether the product of 'a' and 'b' is greater than or equal to 48. E.g. 'a' could be 10 and 'b' could be 4; so a+b = 14 but a*b < 48 (equals 40) so the point does NOT lie in the region. But on the other hand, we could have 'a' = 8 and 'b' = 6, so a+b = 14 AND a*b = 48 (so the point lies in the region). We are getting mixed answers depending on the values of 'a' and 'b' so Statement 1 is INSUFFICIENT.

Statement 2 tells us that a > b, which is next to useless as we can easily pick numbers for 'a' and 'b' that result in the point (a,b) lying in the region (like 'a' = 5 and 'b' = 3) and easily pick numbers for 'a' and 'b' such that the point does NOT lie in the region (e.g 'a' = 30 and 'b' = 20 as this causes a^2 + b^2 > 100). So Statement 2 is INSUFFICIENT as well.


Combining both statements - From Statement 1, we require a*b >= 48 (and a+b = 14) and from Statement 2, we are told a>b. Well, we could have a = 8 and b = 6 - which works. But we could also have a = 7.5 and b = 6.5 - which also satisfies both requirements. So wouldn't this imply that both statements together are sufficient as any pair 'a' 'b' where a > b and a*b >= 48 and a+b = 14 will be such that a^2 + b^2 will lie in the region?

Please could someone let me know where I went wrong, thank you!
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In the xy-plane, region Q consists of all points (x,y) such that x^2 + 03 May 2023, 19:35
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