In a plane, points P and Q are 20 inches apart. If point R is randomly

saik1993 wrote:

IMO OA is 0
All the points of R form a circle with P as the center of the circle. Point Q and any point R always lie on the circle. so the Point P is always at equal distance from R & Q and there is no chance where R can be closer to P than Q. So the combinations are 0 hence the probability is 0

saik1993, Your all explanation is wrong.

Please see the above post, explained very good by nick1816

Hope you will read his answer, if you didn't get it. we will explain you.

Answer is E i.e. 2/3

rajatchopra1994
I read the solution from nick1816

I have a few questions,

1. If point R is on the circle, with P as the center, Isn't the distance between them is now the radius of the circle =20
and Q is also 20 inches away from P?

2. From the diagram, Let us consider the other two points be A & B. I can see that PA= 20 & PB= 20 as they are radius,
but how do we get the other side (QA & QB length as 20? why did we assume the angle BPA as 120 and line segment PQ bisects the angle BPA? Why can't the angles be 180 or 60 only?

Please explain in detail,
Thank you.

Bunuel or nick1816 can you please in more detail, please?
If I choose a point on the circle which is not highlighted in red, it is 20 inches away from P just as in the red highlighted area?

In the non-highlighted potion, R is closer to Q (less than radius of circle) than it is to P. We are only looking for the portion where R is closer to P than it is to Q.

Thank you all, read the question wrong.

Hey nick1816

how have u calculated 60 degrees?? & logic for quadrilateral inside circle

i think it should be 45 Degrees
Pl explain


thx

same question, Please help me to understand calculation of 60 degree Bunuel

All the points behind P will be closer to P than Q. So the answer has to be greater than 50% . Only option E satisfies that .

E.

The line Y = X creates a 45 degree angle with the X Axis.


For the point R to be equidistant from P and Q and be on that Line Y = X such that the angles should be 45 degrees and not 60 degrees:


Make P the Origin

Make Q (20 , 0)



For the point R to be closer to Point P (and further from point Q) we are first looking for the point at which R will be equidistant from both points P and Q. Let’s make this in Quadrant I.

If we used 45 degrees as the Angle/Slope of elevation from Point P, the slope would = +1.

then for Point R to be equidistant from P and Q, the slope from Point Q would have to be the negative reciprocal of this slope and be downward sloping into Point Q. Where these 2 lines intersect in quadrant I——-> that point will be equidistant from P and Q (label it R)


The resulting triangle we have made is a 45 - 45 - 90 isosceles right triangle PQR.

The Angle between P and the X axis is 45 deg.

The Angle between Q and the X axis is 45 deg.

At point R where the 2 lines intersect, the angle must be 90 degrees.

This makes PQ across from the 90 degree angle = hypotenuse = 20 units

Thus, point R will lie (20) / (sqrt(2)) units away from point P and Q.

Although point R is equidistant from P and Q, this is in violation of the question stem. The question stem states that point R must lie 20 inches from P at all times, not 20 divided by root(2)


Furthermore, using the 45 degree angle to cut up the quadrants would give you an answer of 6/8 = 3/4


Satisfying the question stem:

Point R must be 20 inches from P at all times.

Since Q is 20 inches from P, we can start point R immediately on top of Point Q at (20 , 0) where P is the center at the Origin.

Then the Circle with Center P and radius of 20 will represent all the points that R can lie on in which R is exactly 20 inches from P.

In such a case, point Q would be on the circumference of this circle and PQ would be the radius of 20.

Again, let’s use Quadrant I to see which points on the circumference of this circle will be closer to point Q rather than point P.

For Point R to be equidistant from P and Q (and ALSO be 20 inches from P at all times), point R must lie on a point that is:

—On this circle

And

—20 inches from Q

Again, if we draw a triangle to represent this point of equidistance, you will have an equilateral triangle in which point R lies on the circumference of the circle with radius 20.

QR will be equal to 20.

And

PR will be equal to 20 (another radii)

We also know that PQ length is 20 from the question stem, and we can see this since we place P at the Origin and Q at (20 , 0)

Thus, we’ve created an equilateral triangle with central angle 60 degrees.

If point R lies anywhere to the right on the circumference up until point Q, it will be closer to Point Q.

You can reflect this figure through through the X Axis.

The result is that if point R lies anywhere on the minor arc measured by a central angle of 120 degrees, it will be closer to Q and not P.


Probability that R will be closer to Q and not P = (120 deg. / 360 deg.) = 1/3


Therefore the probability that R will be closer to P and not Q (the complement) is:

1 - (1/3)


2/3





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