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

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

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17 Feb 2008, 06:34
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59% (02:17) correct 41% (01:19) wrong based on 189 sessions

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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
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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.
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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....
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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
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01 Sep 2008, 19:14
yes ... 3/4 is the right ans. such problems are stunners initially, but if properly understood are a cakewalk.
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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?

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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.
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27 Feb 2009, 13:17

1-(5/20) atleast 5 includes 5 too. What are the choices?
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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.
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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.
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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?
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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)
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20 Sep 2010, 11:49
Bunnel pls explain this...
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20 Sep 2010, 13:36
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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 3327 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.
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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?
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20 Sep 2010, 21:05
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.
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20 Sep 2010, 23:15
Thanks Bunnel I got it now...Its certainly not a Gmat question.
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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..
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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?
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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

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