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Is a < 2?

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Is a < 2?  [#permalink]

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New post Updated on: 09 Oct 2013, 10:17
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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

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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New post 09 Oct 2013, 10:27
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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. This circle is centered at the origin and has the radius of \(\sqrt{3}\approx{1.7}\) (check here: math-coordinate-geometry-87652.html). No point inside that circle has x-coordinate 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?  [#permalink]

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New post 03 Sep 2016, 06:18
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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,

\((x-a)^2 + (y-b)^2 = r^2\), 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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New post Updated on: 10 Apr 2015, 23:42
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Hi All,

As complex as this graphing-based 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) co-ordinate (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 Y-co-ordinate 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.

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Originally posted by EMPOWERgmatRichC on 28 Mar 2015, 22:34.
Last edited by EMPOWERgmatRichC on 10 Apr 2015, 23:42, edited 1 time in total.
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Re: Is a < 2?  [#permalink]

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New post 10 Apr 2015, 16:19
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EMPOWERgmatRichC wrote:
Hi All,


2(4) + 4(A) = 10
8 + 4A = 10
A = 2
A = 1/2
The answer to the question is YES.
Fact 2 is SUFFICIENT.

Final Answer:

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Rich

Hi Rich!
Nice solution as usual! But please correct a typo. The bold part should be "4A" not A=))
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New post 10 Apr 2015, 23:42
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Hi Konstantin1983,

Good eye. I've made the correction. Thanks for catching it.

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New post 13 Apr 2015, 04:12
EMPOWERgmatRichC wrote:
Hi Konstantin1983,

Good eye. I've made the correction. Thanks for catching it.

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Ah Rich no problem=)). We are human being and even assassins sometimes make typos=))
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New post 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

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Re: Is a < 2?  [#permalink]

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New post 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?  [#permalink]

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New post 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/math-coordina ... 87652.html). No point inside that circle has x-coordinate 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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New post 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 S-2 because it seems simpler:

2) If (a,4) lie on the line \(2y + 4x = 10\), then \(2(4) +4x=10\) ---> \(x=\frac{1}{2}\). Sufficient

1) 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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New post 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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Is a < 2?   [#permalink] 26 Aug 2019, 10:14
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