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A circular racetrack is 3 miles in length and has signs post [#permalink]

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13 Dec 2010, 00:26

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A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile, what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles?

A circular racetrack is 3 miles in length and has signs posted to indicate each 1/10 mile increment. If a race car starts at a random location on the track and travels exactly one half mile,what is the probability that the car ends within a half mile of the sign indicating 2 1/2 miles? (A) 1/6 (B) 3/10 (C) 1/3 (D) 1/2 (E) 2/3

Answer: C

Could somebody explain how to work this out?

The car ends within a half mile of the sign indicating 2 1/2 miles means that the car will end in one mile interval, between the signs indicating 2 and 3 miles.

Now, it doesn't matter where the car starts or what distance it travels, the probability will be P=(favorable outcome)/(total # of outcomes)=1/3 (as the car starts at random point end travels some distance afterwards we can consider its end point as the point where he randomly appeared, so the probability that the car appeared within 1 mile interval out of total 3 miles will be 1/3).

Bunuel, Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Bunuel, Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Look at the diagram:

Attachment:

untitled.PNG [ 5.9 KiB | Viewed 5034 times ]

The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the car appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Don't we somehow need to use the information 1/10 mile increment signposts in calculating the probability? Approach mentioned above is logical but there must be simple mathematical solution using the increment signposts?

Don't we somehow need to use the information 1/10 mile increment signposts in calculating the probability? Approach mentioned above is logical but there must be simple mathematical solution using the increment signposts?

Stem gives us the information about the signs only to fix the point of 2.5 miles on the track and thus fix the 1 mile interval the car should end within. 1/10 mile increment is totally irrelevant.

Consider the following: A circular racetrack is 3 miles in length. If a race car starts at a random location on the track and travels exactly Y miles, what is the probability that the car ends within a half mile of some point X on the track?

The answer will be the same: as the interval for the endpoint of the car is 1 mile (from x-0.5 to x+0.5) then the probability will be 1/3.

Note that a car starts at a random location on the track and travels exactly Y miles means that the endpoint of the car will also be at random location on the track (travel part is also to confuse us: random location plus Y miles=random location). So the question basically ask what is the probability that the car ends within the particular 1 mile interval on the track of 3 miles.

Re: A circular racetrack is 3 miles in length and has signs post [#permalink]

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07 Jun 2013, 07:33

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Here is a rewording of the original question: What is the probability that the car starts between the 1.5 mile and 2.5 mile mark on a 3 mi racetrack?

2.5 - 1.5 = 1 that is 1/3 of the racetrack

Answer is C

Everything else in this question is irrelevant. Although, if you really wanted you could rephrase the question to account for the 1/10th mile sign posts: What is the probability that the car stars between the 15th and 25th sign posts if there are 30 sign posts total. But that is unnecessary extra work.

Bunuel, Sorry, I could not understand the language of the question and hence probably the answer is not clear to me. If possible , could you reframe the question or explain as to what the question is asking?

Look at the diagram:

Attachment:

untitled.PNG

The car ends within a half mile of the sign indicating 2.5 miles means that the car should end in one mile interval, between the signs indicating 2 (2.5-0.5=2) and 3 miles (2+0.5=3), so within the red segment on the diagram.

Now if the cars appears somewhere between the blue dots, between 1.5 and 2 miles signs then after traveling 0.5 miles the car will be in the red segment. So in order after traveling 0.5 miles the car to end within the red segment it should appear between 1.5 and 2.5 miles, so within 1 mile interval, as the circumference of the track is 3 miles then the probability of that will be P=favorable/total=1/3. As I mentioned in my previous post actually it doesn't matter where the car appears or what distance it travel, as long as favorable interval in the end is 1 mile and total interval is 3 miles then the probability will be 1/3 miles.

Hope it's clear.

Hi Bunnel,

What about if car starts between 0.5 and 2.5 miles?

Re: A circular racetrack is 3 miles in length and has signs post [#permalink]

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09 Jun 2014, 07:14

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Re: A circular racetrack is 3 miles in length and has signs post [#permalink]

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