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Cunning Circle!

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by KICKGMATASS123 » Thu Jul 30, 2009 10:13 am
In the figure shown, the circle has center O and radius 50, and point P has coordinates (50,0). If point Q (not shown) is on the circle, what is the length of line segment PQ ?

1) The x-coordinate of point Q is – 30.
2) The y-coordinate of point Q is – 40.

OA is A

I fell for C can someone explain???

Thanks much in anticipation!
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by m&m » Thu Jul 30, 2009 11:11 am
in

a) point can be (-30, +x) or (-30, -x) -- we can figure out x by sqrt(50^2-30^2), but we don't care of the actual number... we care about the length from each point to (0, 50)... the length is same from each point
--> sufficient

b) point (+y, -40) or (-y, -40) -- again we can figure out y, but length will be different for each point (+ and -) so unless we know which point it is (+ or -) we cannot find out length -- insufficient

so ans is A

-- btw if you graph this on scrap paper it's much easier to see
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by KICKGMATASS123 » Thu Jul 30, 2009 1:28 pm
m&m wrote:in

a) point can be (-30, +x) or (-30, -x) -- we can figure out x by sqrt(50^2-30^2), but we don't care of the actual number... we care about the length from each point to (0, 50)... the length is same from each point
--> sufficient

b) point (+y, -40) or (-y, -40) -- again we can figure out y, but length will be different for each point (+ and -) so unless we know which point it is (+ or -) we cannot find out length -- insufficient

so ans is A

-- btw if you graph this on scrap paper it's much easier to see
I dont get your explanation for A.. Can you provide details as to how x is figured by sqrt (50^2-30^2)??
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by raghavsarathy » Thu Jul 30, 2009 5:21 pm
Point (50,0) is on the circle. We don't know that centre of the circle. The centre can be (0,0) or (100,0)
Looking at the statements 1 and 2 , we see negative values. Hence we can consider the centre at (0,0)

Statement 1

sqrt((-30)^2 + (y)^2) = 50

solving this y = -40 or y=40

So, the points can be (-30, -40) or (-30, 40)

When (-30,-40) or (-30,40) distance PQ = sqrt(80^2 + 40^2)

Hence sufficient

From Statement 2

sqrt(x^2 + (-40)^2) = 50
x= 30 or x= -30

Points are (30,-40) or (-30,-40)
When point =(30,-40) dist PQ = sqrt( 20^2 + 40^2)
When point =(-30, -40) dist PQ = sqrt( 80^2 + 40^2)

Hence not sufficient

Ans A

The same problem would have ans as B if in the main statement the x-coordinate of point P is 0 and y-coordinate is non-zero.

Cool problem..
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by vinayakdl » Thu Jul 30, 2009 6:44 pm
Agree with A.

Good Problem.

I am assuming the center to be at (0,0)

1) the two possible points are (-30,y) & (-30,-y) you can see the distance between (50,0) and these is same sqrt(80^2 + y^2)

2) possible points are (x,-40) & (-x,-40)

d1 = sqrt( (50-x)^2 + 40^2 )
d1 = sqrt( (50+x)^2 + 40^2)


Vinayak
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by pink_08 » Thu Jul 30, 2009 7:39 pm
Just curious, where is this question from ? Good one
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by fruti_yum » Sat Aug 01, 2009 5:03 am
pink_08 wrote:Just curious, where is this question from ? Good one
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by real2008 » Fri Aug 07, 2009 11:49 am
where is the figure?
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Re: Cunning Circle!

by shahdevine » Fri Aug 07, 2009 1:34 pm
KICKGMATASS123 wrote:In the figure shown, the circle has center O and radius 50, and point P has coordinates (50,0). If point Q (not shown) is on the circle, what is the length of line segment PQ ?

1) The x-coordinate of point Q is – 30.
2) The y-coordinate of point Q is – 40.

OA is A

I fell for C can someone explain???

Thanks much in anticipation!
p is a point that is on the x-axis, which splits the circle in half. so as long as point q is on the other side of the y-axis the length formed is measurable.

statement 1)

q has x coordinate -30. that means if you start from (-30, 0) and trace up (above the x axis) there is a point, Q, on the circle at (-30,y), which will form line PQ. Similarly if you trace down below the x-axis, there is a point, Q at (-30, -y) on the circle, which will form line PQ. We don't need to solve for y. All we need to know is that we can solve for it. both lines from (-30,y) to (50,0) and (-30,-y) to (50,0) are equal in length. Sufficient.

statement 2)

q has y coodinate -40. that means if you start from (0,-40) and trace to the left of the y-axis, there is a point Q, on the circle at (-x,-40) which will form line PQ. Similarly if you trace to the right of y-axis, there is a point Q, on the circle at (x,-40) which will form line PQ. However, both of these lines have different lengths because one Q is closer to P than the other Q. Insufficient.

A is answer.

you got this, man!
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by coderversion1 » Mon Sep 17, 2012 11:56 pm
Anyone who thinks that answer should be E as O is not mentioned to be origin?

It would be improper to assume so.
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