M13 Q12 : GMAT Problem Solving (PS)
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# M13 Q12

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Intern
Joined: 20 Aug 2010
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Schools: Duke,Darden,Chicago University
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04 Dec 2011, 18:56
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Question Stats:

33% (01:03) correct 67% (01:05) wrong based on 6 sessions

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What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ?

$$1.4$$
$$\sqrt{2}$$
$$1.7$$
$$\sqrt{3}$$
$$2.0$$

Can anyone Please explain this Question??
[Reveal] Spoiler: OA
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04 Dec 2011, 23:45
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Expert's post
CracktheGmat2010 wrote:
What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ?

$$1.4$$
$$\sqrt{2}$$
$$1.7$$
$$\sqrt{3}$$
$$2.0$$

Can anyone Please explain this Question??

The question has been discussed before. This is my take on it.

Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? (shown by the bold line)

Attachment:

File.jpg [ 17.76 KiB | Viewed 1534 times ]

Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer.

Then, let's try to find the length of the bold line, x.
Since hypotenuse is 5,
(1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line
x = 2.4
So minimum distance is 2.4 - 1 = 1.4
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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 07 Nov 2011 Posts: 31 Followers: 0 Kudos [?]: 2 [0], given: 9 Re: M13 Q12 [#permalink] ### Show Tags 07 Dec 2011, 07:15 VeritasPrepKarishma wrote: CracktheGmat2010 wrote: What is the least possible distance between a point on the circle $$x^2 + y^2 = 1$$ and a point on the line $$y = \frac{3}{4}x - 3$$ ? $$1.4$$ $$\sqrt{2}$$ $$1.7$$ $$\sqrt{3}$$ $$2.0$$ Can anyone Please explain this Question?? The question has been discussed before. This is my take on it. Look at the diagram below and forget the circle for the time being. What is the minimum distance from the center (0,0) to the line? It will the perpendicular from the center to the line, right? (shown by the bold line) Attachment: File.jpg Now think, what will be the shortest distance from the circle to the line? It will be 1 unit less than the distance from the center to the line. Can we say it will be the least in case of the bold line which is perpendicular to the given line? Yes, it will be because in all other cases, the lines are longer than the perpendicular and hence (line - 1) will also be longer. Then, let's try to find the length of the bold line, x. Since hypotenuse is 5, (1/2)*3*4 = (1/2)*5*x = Area of triangle made by the co-ordinate axis and the given line x = 2.4 So minimum distance is 2.4 - 1 = 1.4 i have one serious doubt if we find the perpendicular distance (d) given by the formula |ax+by+c|/(a^2+b^2)^1/2 the value comes to be 4 against your value which comes out to be 2.4 tell me where i am going wrong. Intern Joined: 07 Nov 2011 Posts: 31 Followers: 0 Kudos [?]: 2 [0], given: 9 Re: M13 Q12 [#permalink] ### Show Tags 07 Dec 2011, 07:17 sorry i did some calculation mistake ... i am also getting 2.4 Manager Joined: 26 Apr 2011 Posts: 228 Followers: 2 Kudos [?]: 3 [0], given: 14 Re: M13 Q12 [#permalink] ### Show Tags 12 Dec 2011, 23:15 what is the difficulty level of this question? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7191 Location: Pune, India Followers: 2173 Kudos [?]: 14058 [1] , given: 222 Re: M13 Q12 [#permalink] ### Show Tags 13 Dec 2011, 02:28 1 This post received KUDOS Expert's post sandeeepsharma wrote: what is the difficulty level of this question? 700+, actually closer to 750. It takes a degree of imagination to figure out the solution quickly. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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13 Dec 2011, 03:37
thanks karishma..
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13 Dec 2011, 10:00
That's an awesome question.
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Re: M13 Q12   [#permalink] 13 Dec 2011, 10:00
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# M13 Q12

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