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

 It is currently 28 Jul 2016, 12:09

### 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

# A searchlight on top of the watch-tower makes 3 revolutions

Author Message
TAGS:

### Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 408 [0], given: 0

A searchlight on top of the watch-tower makes 3 revolutions [#permalink]

### Show Tags

17 Feb 2008, 06:34
8
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

60% (02:07) correct 40% (01:20) wrong based on 165 sessions

### HideShow timer Statistics

A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?

(A) 1/4
(B) 1/3
(C) 1/2
(D) 2/3
(E) 3/4
[Reveal] Spoiler: OA
Senior Manager
Joined: 20 Dec 2004
Posts: 255
Followers: 7

Kudos [?]: 105 [0], given: 0

### Show Tags

17 Feb 2008, 12:46
Is it 3/4....I am not very confident abt the solution but I think that is the answer.

If this was a real GMAT test I would have picked that.

If this is correct then I will explain.
_________________

Stay Hungry, Stay Foolish

Senior Manager
Joined: 20 Dec 2004
Posts: 255
Followers: 7

Kudos [?]: 105 [0], given: 0

### Show Tags

17 Feb 2008, 23:13
marcodonzelli wrote:
A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?

 1/4
 1/3
 1/2
 2/3
 3/4

Here is my attempt to solve this one:

3 revolutions per minute = 1 revolution every 20 seconds

So no matter what anybody appearing at the tower cannot stay in the dark for more than 20 seconds. This will be our total number of possibilities i.e the denominator.

P(man in dark for at least 5 seconds) = 1 - P (man in dark for max of 5 seconds) = 1 - 5/20 = 1 - 1/4 = 3/4

or the other way would be:
P(man in dark for at least 5 seconds) is like saying he can be in dark for 5,6,7...all the way to 20 seconds because that is the max. In this approach it would be 15/20 seconds = 3/4.

Its kind of a weird approach to solve the problem but I could not think of anything better in the 2mins....
_________________

Stay Hungry, Stay Foolish

VP
Joined: 05 Jul 2008
Posts: 1431
Followers: 40

Kudos [?]: 321 [0], given: 1

### Show Tags

01 Sep 2008, 18:52
Initially I solved the problem for staying in dark for atmost 5 sec. I should read the Q better. Shyt

3 rev- 60 sec means 1 rev - 20 sec

That means the man will become visible under the searchlight with in 20 seconds

Probability that he can stay in the dark for max 5 sec is 5/20 =1/4

Probability that he can stay atleast 5 sec in the dark is 1-1/4 =3/4
Current Student
Joined: 11 May 2008
Posts: 555
Followers: 8

Kudos [?]: 135 [0], given: 0

### Show Tags

01 Sep 2008, 19:14
yes ... 3/4 is the right ans. such problems are stunners initially, but if properly understood are a cakewalk.
Director
Joined: 04 Jan 2008
Posts: 914
Followers: 64

Kudos [?]: 460 [0], given: 17

### Show Tags

25 Feb 2009, 22:45
1 min=3*360
1 s=18 degree
5 s=90 degree
so probability =1-(90/360)=75%

A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?

_________________

http://gmatclub.com/forum/math-polygons-87336.html
http://gmatclub.com/forum/competition-for-the-best-gmat-error-log-template-86232.html

Manager
Joined: 26 Dec 2008
Posts: 58
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)
Followers: 2

Kudos [?]: 11 [0], given: 0

### Show Tags

25 Feb 2009, 23:44
A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?

3 rpm --> each interval of darkness = 60/3 = 20 s

P(atleast 5 s) = 1 - P(less than 5 secs) = 1 - P(<= 4 s) = 1 - (1/20+2/20+3/20+4/20) = 1/2

intuitively this answer seems off, but we can do sanity check by thinking that the lower bound for p(atleast 5 s) is 5/20 = 1/4, additional cases will only increase this value.
Manager
Joined: 27 Feb 2009
Posts: 62
Followers: 1

Kudos [?]: 15 [0], given: 0

### Show Tags

27 Feb 2009, 13:17

1-(5/20) atleast 5 includes 5 too. What are the choices?
Manager
Joined: 17 Dec 2008
Posts: 177
Followers: 2

Kudos [?]: 105 [0], given: 0

### Show Tags

28 Feb 2009, 10:59
60 sec => 3 revolutions
5 sec => 1/4 revolution.

Hence 1/4 of the time => Light.
Hence P(darkness) = 3/4.
Manager
Joined: 07 Feb 2009
Posts: 50
Followers: 0

Kudos [?]: 13 [0], given: 1

### Show Tags

02 Mar 2009, 04:39
probabilty to stay in light = 5/20 (3 rev per minutes means 20 sec required for 1 rev).
so, probability to stay in dark is 1-5/20. i.e 3/4.
Manager
Joined: 26 Dec 2008
Posts: 58
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)
Followers: 2

Kudos [?]: 11 [0], given: 0

### Show Tags

02 Mar 2009, 19:41
abhishekik wrote:
probabilty to stay in light = 5/20 (3 rev per minutes means 20 sec required for 1 rev).
so, probability to stay in dark is 1-5/20. i.e 3/4.

* Question does not ask probability to stay in dark, it asks prob of staying in dark for atleast 5 seconds

* why is probability to stay in light = 5/20?
Manager
Joined: 26 Dec 2008
Posts: 58
Schools: Booth (Admit R1), Sloan (Ding R1), Tuck (R1)
Followers: 2

Kudos [?]: 11 [0], given: 0

### Show Tags

02 Mar 2009, 19:42
ConkergMat wrote:
60 sec => 3 revolutions
5 sec => 1/4 revolution.

Hence 1/4 of the time => Light.
Hence P(darkness) = 3/4.

why is 1/4 of time = light??

also question asks p(darkness for >=5 secs) and not P(darkness)
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2797
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 220

Kudos [?]: 1479 [0], given: 235

### Show Tags

20 Sep 2010, 11:49
Bunnel pls explain this...
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Math Expert
Joined: 02 Sep 2009
Posts: 34106
Followers: 6099

Kudos [?]: 76761 [2] , given: 9981

### Show Tags

20 Sep 2010, 13:36
2
KUDOS
Expert's post
1
This post was
BOOKMARKED
gurpreetsingh wrote:
Bunnel pls explain this...

A searchlight makes 1 revolution in 20 seconds. Consider the diagram below:
Attachment:

Circle1.gif [ 2.42 KiB | Viewed 3064 times ]
A man randomly appears at some point M. Now, if beam of light is somewhere in the dark quarter, then the beam will need less than 5 seconds to reach a man and if beam of light is somewhere in the white 3 quarters then it'll need more than 5 seconds to reach a man (so he'll be in the dark more than 5 seconds).

So P=3/4.

Hope it's clear.
_________________
Senior Manager
Joined: 20 Jul 2010
Posts: 269
Followers: 2

Kudos [?]: 64 [0], given: 9

### Show Tags

20 Sep 2010, 19:00
My doubts - We are not told how much area is in light by light house. We are assuming a quadrant is in light. If light house lights which circles at one revolution per 20 secs only lights an area of narrow beam (30 degree), will the probability be same?
_________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

Math Expert
Joined: 02 Sep 2009
Posts: 34106
Followers: 6099

Kudos [?]: 76761 [0], given: 9981

### Show Tags

20 Sep 2010, 21:05
Expert's post
saxenashobhit wrote:
My doubts - We are not told how much area is in light by light house. We are assuming a quadrant is in light. If light house lights which circles at one revolution per 20 secs only lights an area of narrow beam (30 degree), will the probability be same?

We are not assuming that a quadrant is in light. We are assuming that lighted sector has 0 degrees. Anyways not a GMAT question so don't worry about it.
_________________
CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2797
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 220

Kudos [?]: 1479 [0], given: 235

### Show Tags

20 Sep 2010, 23:15
Thanks Bunnel I got it now...Its certainly not a Gmat question.
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned

Jo Bole So Nihaal , Sat Shri Akaal

GMAT Club Premium Membership - big benefits and savings

Gmat test review :
http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html

Intern
Status: fighting hard..
Joined: 12 Jul 2010
Posts: 29
Schools: ISB, Hass, Ross, NYU Stern
Followers: 0

Kudos [?]: 10 [0], given: 24

### Show Tags

24 Nov 2010, 23:47
A searchlight on top of the watch-tower makes 3 revolutions per minute. What is the probability that a man appearing near the tower will stay in the dark for at least 5 seconds?

I also could not do it in the first go. But here's what I figured out after some thought.

Consider someone standing within the imaginary circle lit by the search-light as it revolves.

What is the probability that the man will stay in dark for 20 seconds (3 revs/min)? zero or (1 - (20/20))
i.e. (1 - P(coming under focus))
What is the probability that the man will stay in dark for 19 seconds? (1 - (19/20))
What is the probability that the man will stay in dark for 18 seconds? (1 - (18/20))
.
.
What is the probability that the man will stay in dark for 5 seconds? (1 - (5/20)) i.e. 3/4

but somehow I am not able to fit in the 'at-least' part. For 'at least' cases, we add the probabilities of all the possible elements (OR). In this case, that would amount to summing up the probabilities of 5 sec, 4 sec, 3 sec, 2 sec and 1 sec.

But I guess the problem is with the discrete approach that I, and many others above have taken.
Shouldn't we approach the problem like the summation of velocity-time graph to find the total distance that we used to do in physics? That is finding the area under the graph..

To be edited, if the bulb in my mind glows, discerning an elegant solution..
Manager
Joined: 03 Jun 2010
Posts: 183
Location: United States (MI)
Concentration: Marketing, General Management
Followers: 6

Kudos [?]: 51 [0], given: 40

### Show Tags

20 May 2011, 23:29
Don't know the OA, but...
3 revs in 60 sec -> 1 rev in 20 sec.
5 seconds is one fourth of 20 seconds.
1 revolution is 360 degrees (just to remind: in 20 seconds).
We need one forth, means 360/90=4.
So 1/4 will be an answer.
(A)
What's the OA?
Manager
Joined: 03 Jun 2010
Posts: 183
Location: United States (MI)
Concentration: Marketing, General Management
Followers: 6

Kudos [?]: 51 [0], given: 40

### Show Tags

21 May 2011, 01:13
For sure!
It is said "at least"!
I've found the minimal value.
We need to subtract: $$1-1/4=3/4$$
Re: m13#37   [#permalink] 21 May 2011, 01:13

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
6 Six students - 3 boys and 3 girls - are to sit side by side for a make 5 08 Dec 2014, 06:23
1 During the break of a football match, the coach will make 3 3 12 Jun 2014, 01:28
During the break of a football match, the coach will make 3 2 01 Jun 2014, 00:38
94 3 cooks have to make 80 burgers.They are known to make 20 32 20 Aug 2009, 15:22
12 A coach will make 3 substitutions. The team has 11 players, 16 19 Dec 2007, 13:58
Display posts from previous: Sort by