A castle is at the center of the several flat paths which su

Question

CastlePaths1.gif A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths—see the diagram to the right. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat? A) q / 4(2+2*sqrt(2)+pi) km B) q / 2(2+2*sqrt(2)+pi) km C) q / (2+2*sqrt(2)+pi) km D) 2q / (2+2*sqrt(2)+pi) km E) 4q / (2+2*sqrt(2)+pi) km

kundankshrivastava · Accepted Answer

Let's take distance from castle to circular moat is r.
side of square = sqrt2 * r
Perimeter of Circle = 2* Pi * r

Attachment:

File comment: Castle Paths

gmatcastle.jpg [ 32.43 KiB | Viewed 13307 times ]

Total length of paths=q = 4* r + 4 *sqrt2 *r +2*pi*r

q = (4+4*sqrt2 + 2* pi) r
or , r = q/(4+4*sqrt2 + 2* pi)
or, r = q/2(2+2*sqrt2+Pi)

Hence, Answer is B

crejoc · Answer

i couldnt figure out how to find the perimeter of the square in terms of r...

bhushan252 · Answer

OA B

let distance from the castle to the circular moat is x

distance from 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat ( addition of lenght of diagonals of sqr)= 4x

Thus side of inscribed sqr = x*sqrt2 (use pythagoras 2x^ = (side of sqr)^2.....hence perimeter of sqr = 4*sqrt2*x

circumference of perimeter = pi * (2*x)
 
Thus our eqn becomes => q= pi * (2*x) + 4*sqrt2*x + 4x

on simplification we get x = q / 2(2+2*sqrt(2)+pi) km... OA B

crejoc · Answer

Thanks for the solution...

rathoreaditya81 · Answer

A castle is at the center of the several flat paths which surround it: 4 straight paths that travel from the castle to its circular moat, where they meet up with a perfectly circular path which borders the moat; that circular path circumscribes a square path which has its corners at the ends of the 4 straight paths. If the total length of all of the pathways is q kilometers, then which expression represents distance from the castle to the circular moat?

1) q/4(2+2\sqrt{2}+pie)
2) q/2(2+2\sqrt{2}+pie)
3) q/(2+2\sqrt{2}+pie)
4) 2q/(2+2\sqrt{2}+pie)
5) 4q/2(2+2\sqrt{2}+pie)

Attachments

CastlePaths1.gif [ 2.33 KiB | Viewed 16063 times ]

1) q/4(2+2\sqrt{2}+pie)
2) q/2(2+2\sqrt{2}+pie)
3) q/(2+2\sqrt{2}+pie)
4) 2q/(2+2\sqrt{2}+pie)
5) 4q/2(2+2\sqrt{2}+pie)

rathoreaditya81 · Answer

Strange, this problem seems to have a different answer when we take the side of the square to be a ( or r) or may be I am doing something wrong!
Here, what I am doing -

Let the side of the sq be a, then sum of four sides = 4a;
Using formula, diag of a sq is sqrt(2)a, we get sum of the paths four paths from castle to moat ( i.e, sum the diagonals) as = 2a sqrt(2) ;
Circumference of the circle as 2pie a sqrt(2)/2

Putting it all together we 
4a +2asqrt(2) +2pie sqrt(2)/2 = q
=> 4a +2asqrt(2) +pie asqrt(2) =q
=> sqrt(2)a (2 sqrt(2)+2 +pie) = q
=> a = q/sqrt(2) (2 sqrt(2)+2 +pie)

Can anybody shed some more light on this..?

swatirpr · Answer

Let take radius of circular path is R
so distance from the castle to the circular moat = R

total length of all of the pathways = total length of 4 straight paths that travel from the castle to its circular moat + length of circular path which borders the moat + total length of square path which has its corners at the ends of the 4 straight paths

(Diagonal of square is 2R, (90-45-45 right triangle with hypotenuse 2R) so side of square would be  R\sqrt{2})

q = 4R + 2R*pie + 4 (R \sqrt{2}) 
 q = 2R (2+pie+2 \sqrt{2}) 
 R = q/2 (2+2 \sqrt{2}+ pie)

Answer B

rathoreaditya81 · Answer

Hi Swatipr, have a look at - castle-paths-82093.html#p653668

rathoreaditya81 · Answer

Thanks a lot swatipr.. oblivious to obvious..strange manifestation of alcohol after effect! :twisted:

ruturaj · Answer

can we solve this in more simpler way..say plugging smart nos

TooLong150 · Answer

This problem is not that difficult once you understand the question; it just takes too long. Can anyone explain how to do this under 3 minutes?

KarishmaB · Answer

You can actually do it in a minute after you understand that all you need is the radius of the circle given that sum of lengths of all lines and curves here is q.

So q is made up of the circle, the inscribed square in it and the 4 radii of the circle. Assume radius of the circle is r.
Each side of the square will be  \sqrt{2}r  because each side makes an isosceles right triangle with two radii. Each radius is r and sides of isosceles right triangle are in the ratio  1:1:\sqrt{2}

q = Circumference of circle + Perimeter of square + 4*radius
 q = 2 \pi r + 4\sqrt{2}r + 4r

So  r = q/2( 2 + 2\sqrt{2} + \pi)

Answer (B)

russ9 · Answer

Hi Karishma,

The part that threw me off was the  \sqrt{2}r

I understood that the hypotenuse was  \sqrt{2}  but I couldn't connect the R there, and more importantly, I couldn't tell that the problem was asking us to solve for R.

Any thoughts on how I would go about this in the future?

mvictor · Answer

wow..the wording is very confusing...
total pathways : circumference, the square, and 4 radii.
circumference is 2*pi*r
square - since we have 4 equilateral triangles, and all right triangles, we can apply 45-45-90 property, and get one side of the square equal to 4*r*sqrt(2)
then we have 4r
we are then told that:
q=2*pi*r + 4*r*sqrt(2) + 4r
factor r
q=r(2pi+4*sqrt(2)+4)
r= q/(2pi+4*sqrt(2)+4)

we can eliminate D and E right away
from the above, we can factor out a 2;
q/2(pi+2*sqrt(2)+2)

so B.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

A castle is at the center of the several flat paths which su

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

A castle is at the center of the several flat paths which su

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards