Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Aug 2016, 09:52

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In the diagram above, WZ = XZ, and circular arc

Author Message
TAGS:

### Hide Tags

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3309
Followers: 1128

Kudos [?]: 4937 [1] , given: 54

In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

20 Feb 2013, 17:48
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

73% (03:17) correct 27% (02:25) wrong based on 136 sessions

### HideShow timer Statistics

Attachment:

right isosceles triangle with sector.JPG [ 14.66 KiB | Viewed 2089 times ]

In the diagram above, WZ = XZ, and circular arc XY has a center at W. If a point is selected from anywhere within this figure, what is the probability that it is selected from the shaded region?

(A) $$\frac{{\pi}-2}{{\pi}}$$

(B) $$\frac{{\pi}-1}{2{\pi}}$$

(C) $$\frac{2{\pi}-3}{4{\pi}}$$

(D) $$\frac{4{\pi}-1}{8{\pi}}$$

(E) $$\frac{6{\pi}-3}{8{\pi}}$$

For the complete solution to this problem, as well as a detailed discussion of geometric probability, please see this post:
http://magoosh.com/gmat/2013/geometric- ... -the-gmat/

Mike
[Reveal] Spoiler: OA

_________________

Mike McGarry
Magoosh Test Prep

 Magoosh Discount Codes Economist GMAT Tutor Discount Codes Optimus Prep Discount Codes
Manager
Joined: 24 Jan 2013
Posts: 79
Followers: 5

Kudos [?]: 122 [0], given: 6

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

20 Feb 2013, 18:17
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

By the way, solution is A because we have to find the shadowed area (1) and divide by the total area (2).

How to find (2): the angle of the arc is 45º (there is a right-angled isosceles triangle), therefore the area of this part must be pi*(radius^2) divided by 8 (because we have 45º and not 360º). In this case, the area is $$\frac{1}{8}*pi*r^2$$ --> $$\frac{1}{8}*pi*(L*\sqrt{2})^2$$ --> $$\frac{1}{8}*pi*(L^2*2)$$.

How to find (1): there is a right-angled isosceles triangle. Therefore the side of the radius must be $$L*\sqrt{2}$$, and the equal sides have length L. The right angle is obviously 90º and the other two angles are 45º each. Therefore the area of this triangle is $$\frac{1}{2}*L*L$$ --> $$\frac{1}{2}*L^2$$. Therefore the shadowed area (1) is $$\frac{1}{8}*pi*(L^2*2)$$ minus $$\frac{1}{2}*L^2$$.

Finally, we divide (1) by (2) and the solution is A.
Intern
Joined: 16 Jan 2013
Posts: 23
Followers: 0

Kudos [?]: 4 [2] , given: 17

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

21 Feb 2013, 08:43
2
KUDOS
wz=xz
therefore, angle xwz = angle wxz = 45 degree
let, wz = xz = k
therefore, wx = (\sqrt{2})k
therefore, area of sector wxyw = 45/360 * pi*square of (k sqrt(2) )= ( pi *square(k))/4
area of shaded portion = area of sector wxyw - area of triangle
area of triangle = 1/2 * square (k)
probability = ( shaded area )/ (total area)
={ ( pi *square(k))/4 - 1/2 * square (k) }/ ( pi *square(k))/4
= 1- 2/pi
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3309
Followers: 1128

Kudos [?]: 4937 [1] , given: 54

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

21 Feb 2013, 12:21
1
KUDOS
Expert's post
johnwesley wrote:
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

I would say, this is probably at the outer limit of what could appear on the GMAT, in terms of difficulty. I would imagine the CAT would throw something like this at you only if you were getting almost every other math question right. It's true, most people cannot solve this in under 3 minutes, but most people would not ever see this question --- the CAT would have no reason to give them something this hard. Someone very skilled in math, someone acing the Quantitative Section, could see something like this and probably could dispatch this question in under a minute. It's all a matter of perspective.

BTW, I love your founder-of-the-Methodists screenname.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Moderator
Joined: 10 May 2010
Posts: 823
Followers: 25

Kudos [?]: 388 [0], given: 192

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

21 Feb 2013, 12:55
I agree Mike, that GMAT might throw this to someone who has aced 15 questions in a row. Someone in higher rungs of Q50 and lower rungs of Q51. Still now harm in flexing your maths muscle. I was on the right track to solving this in 90 seconds but miscalculated the area of the entire arc by twice. Hence took 3 mins. The graph question on the link you shared is a bit too tough and needs very accurate graphing (one needs to verify where the line cuts the 4 sides of the square, which is not an easy task without graph).
_________________

The question is not can you rise up to iconic! The real question is will you ?

Manager
Joined: 24 Jan 2013
Posts: 79
Followers: 5

Kudos [?]: 122 [0], given: 6

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

21 Feb 2013, 14:08
mikemcgarry wrote:
johnwesley wrote:
I can bet this is not a real GMAT question. Too much complicated, and I always say GMAT is complicated... but not that much. It is not possible to have a question that will take you more than 3 minutes for sure. And I did GMAT two years ago.

I would say, this is probably at the outer limit of what could appear on the GMAT, in terms of difficulty. I would imagine the CAT would throw something like this at you only if you were getting almost every other math question right. It's true, most people cannot solve this in under 3 minutes, but most people would not ever see this question --- the CAT would have no reason to give them something this hard. Someone very skilled in math, someone acing the Quantitative Section, could see something like this and probably could dispatch this question in under a minute. It's all a matter of perspective.

BTW, I love your founder-of-the-Methodists screenname.

Mike

Many thanks... you are right: this is a very difficult one. Many thanks for posting it.

And are you the Michael McGarry (born 17 May 1965)? The successful association footballer who frequently represented New Zealand in the 1980s and 90s?

Best!
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11112
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: In the diagram above, WZ = XZ, and circular arc [#permalink]

### Show Tags

07 Jul 2014, 01:45
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: In the diagram above, WZ = XZ, and circular arc   [#permalink] 07 Jul 2014, 01:45
Similar topics Replies Last post
Similar
Topics:
2 If O is the center of the circle above and the length of arc RSP 2 10 Mar 2015, 11:20
3 In the figure above, AC = BC = 8, angle C = 90°, and the circular arc 3 10 Mar 2015, 06:07
6 In the diagram above, BC is parallel to DE 5 05 Mar 2013, 15:40
10 In the diagram above, <PQR is a right angle, and QS is 13 05 May 2012, 19:37
11 In the diagram above, the line y = 4 is the perpendicular bisector of 7 25 May 2011, 12:09
Display posts from previous: Sort by