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Is a < 2?
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Updated on: 09 Oct 2013, 10:17
Question Stats:
68% (01:37) correct 32% (01:47) wrong based on 591 sessions
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Is a < 2? (1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3 (2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10
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Originally posted by ranjitaarao on 09 Oct 2013, 10:15.
Last edited by Bunuel on 09 Oct 2013, 10:17, edited 1 time in total.
RENAMED THE TOPIC.




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Re: Is a < 2?
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09 Oct 2013, 10:27
Is a < 2?(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3. This circle is centered at the origin and has the radius of \(\sqrt{3}\approx{1.7}\) (check here: mathcoordinategeometry87652.html). No point inside that circle has xcoordinate greater than \(\sqrt{3}\approx{1.7}\), thus \(a<1.7<2\). Sufficient. (2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10 > 2*4+4x=10 > x=1/2. Sufficient. Answer: D.
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Re: Is a < 2?
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03 Sep 2016, 06:18
narendran1990 wrote: Bunuel: Was able to understand your solution, have a query as to how the circle is centered at the origin. Is it a default assumption or its inferred from the equation.? Equation of the circle is given by the formula, \((xa)^2 + (yb)^2 = r^2\), where a,b are the center coordinates and r is the radius. Now, in the question we are given the form \((x)^2 + (y)^2 = r^2\), implying a and b are zero. Or the circle is centered at origin. I think you should go through the basic of Circle once. That would help you in such questions.
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Is a < 2?
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Hi All, As complex as this graphingbased DS question looks, the given Facts focus on two equations, so they're relatively easy to deal with/solve. We're asked if A is less than 2. This is a YES/NO question. Fact 1: In an XY plane, point (A,1) lies inside the circle whose equation is X^2 + Y^2 = 3 With the given equation and the given (X,Y) coordinate (A,1), we can plug in and solve for A.... A^2 + 1^2 = 3 A^2 = 2 Since we're dealing with a circle, there are two possible points that have a Ycoordinate of 1: (A, 1) and (A, 1). Since A^2 = 2, A = Root(2). Root(2) is < Root(4) Root(2) is < 2, so whether we're dealing with A = Root(2) or A = Root(2), the answer to the question is ALWAYS YES. Fact 1 is SUFFICIENT. Fact 2: In an XY plane, point (A,4) lies on the line whose equation is 2Y + 4X = 10 Here, we have a simple line to work with, so we can substitute in the value of A (for X) and 4 (for Y), giving us.... 2(4) + 4(A) = 10 8 + 4A = 10 4A = 2 A = 1/2 The answer to the question is YES. Fact 2 is SUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Is a < 2?
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10 Apr 2015, 16:19
EMPOWERgmatRichC wrote: Hi All, 2(4) + 4(A) = 10 8 + 4A = 10 A = 2A = 1/2 The answer to the question is YES. Fact 2 is SUFFICIENT. Final Answer: GMAT assassins aren't born, they're made, Rich Hi Rich! Nice solution as usual! But please correct a typo. The bold part should be "4A" not A=))
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Re: Is a < 2?
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10 Apr 2015, 23:42
Hi Konstantin1983, Good eye. I've made the correction. Thanks for catching it. GMAT assassins aren't born, they're made, Rich
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Re: Is a < 2?
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13 Apr 2015, 04:12
EMPOWERgmatRichC wrote: Hi Konstantin1983,
Good eye. I've made the correction. Thanks for catching it.
GMAT assassins aren't born, they're made, Rich Ah Rich no problem=)). We are human being and even assassins sometimes make typos=))
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Re: Is a < 2?
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15 Aug 2016, 07:34
1. We are told that a is the x coordinate for a point within a circle with radius sqrt(3) its center is origo. The largest x coordinate possible would be if y=0, otherwise x would have to decrease(if +) or increase (if ) to make sure the distance from origo does not exceed sqrt(3). If y=0, a can be at most sqrt(3). This is less than 2. As y is 1, a has to be even smaller. Sufficient. 2. We can solve for x. Sufficient. Answer D Posted from my mobile device
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Re: Is a < 2?
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03 Sep 2016, 06:02
Bunuel: Was able to understand your solution, have a query as to how the circle is centered at the origin. Is it a default assumption or its inferred from the equation.?



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Re: Is a < 2?
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20 Oct 2018, 04:49
Bunuel wrote: Is a < 2?(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3. This circle is centered at the origin and has the radius of \(\sqrt{3}\approx{1.7}\) (check here: http://gmatclub.com/forum/mathcoordina ... 87652.html). No point inside that circle has xcoordinate greater than \(\sqrt{3}\approx{1.7}\), thus \(a<1.7<2\). Sufficient. (2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10 > 2*4+4x=10 > x=1/2. Sufficient. Answer: D. Bunuel  Statement 1 says point lies inside the circle so I get a = + and  2 as 2 values I can confirm that since point is inside the circle then it has to be less than 2 and greater than 2 [i.e. 2 <a<2 ] does that make sense ?



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Re: Is a < 2?
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25 Aug 2019, 19:11
ranjitaarao wrote: Is a < 2?
(1) In an xy plane, point (a,1) lies inside the circle whose equation is x^2 + y^2 = 3 (2) In an xy plane, point (a,4) lies on the line whose equation is 2y + 4x = 10 \(a<2\)? Looking at S2 because it seems simpler: 2) If (a,4) lie on the line \(2y + 4x = 10\), then \(2(4) +4x=10\) > \(x=\frac{1}{2}\). Sufficient1) If (a,1) lies inside the circle defined by \(x^2 + y^2 = 3\), a simple way to check whether a<2 at this point on the x,y plane, would be to sub value of y and check value of x. \(x^2 + 1^2= 3\)> \(x^2=2\)> \(x=\sqrt{2}<2\). If (a,1) lies within the circle and value of \(x<2\) when \(y=1\) this definitively means that \(a<2\) Sufficient.



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Is a < 2?
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26 Aug 2019, 10:14
Statement 1:Assume the point lies on the circle, we can plug in the point to get \(a^2 + 1^2 = 4\), or \(a^2 = 3\). Then the points \((\sqrt{3}, 1)\)and \((\sqrt{3}, 1)\) are on the circle. For the point to be within the circle, we must have roughly\(1.7 < a < 1.7\), therefore \(a < 2\) is true. Sufficient. Statement 2:We are allowed to solve for the coordinates of \((a, 4)\)with this line, sufficient.
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