Solve the Puzzle: Couple of Puzzles #1 & #2

[dontdisturbmycircles]

Number 1
A guy who likes watching trains enjoys walking to a nearby railroad track to wait for a train to go by. Every time, upon his return, he makes a note of whether the one he saw was a passenger train or a freight train. Over the years, his figures show that 90 per cent of the trains have been passenger trains. One day he meets an official of the railroad and is surprised to learn that in fact passenger trains and feight trains are precisely equal in number. Why did the man, whom we may presume to have made random trips to the railroad tracks, see such a disproportionate number of passenger trains?

Number 2
Bob owns a couple of hour glasses-- One is a four minute hourglass and the other is a seven minute hour glass. If Bob wants to measure 9 minutes, how could he do it?

 

[Number2Pencil]

I think I have question 2...you have to immediately flip things when asked

1)flip both at the same time
2) when the 4er runs out...3 is left on the 7er...flip the 4er
3) when the 3 runs out of the 7er...1 is left on the 4er (start counting)...
4) when the 1 runs out of the 4er...flip the 4er (1 min)
5) let the 4er run out twice more (9 min)

 

[verty]

2: flip both, flip 4 when it runs out, flip 7 when it runs out, flip 7 when 4 runs out, 7 runs out at 9 minutes

 

[verty]

1: the ingredient is that he waits for a train, so it means the intervals between trains are sized differently. In order to see a freight train, he must arrive in an interval 1:10 smaller than the interval required to see a passenger train.[/color]

 

[dontdisturbmycircles]

2: flip both, flip 4 when it runs out, flip 7 when it runs out, flip 7 when 4 runs out, 7 runs out at 9 minutes

That is Correct.

 

[DaveC426913]

1: I'm not sure I understand verty's answer but I get the idea. Freight trains immediately follow passenger trains on the tracks.

In any given time period (could be ten minutes or ten hours):
no train
no train
no train
no train
no train
no train
no train
no train
no train
no train
passenger train
freight train

9 trips out of 10, he will arrive and wait and see a passenger train and then leave. Only 1 trip in 10 will he arrive in time to have missed the passenger train but see the freight train.

 

[dontdisturbmycircles]

1: I'm not sure I understand verty's answer but I get the idea. Freight trains immediately follow passenger trains on the tracks.

In any given time period (could be ten minutes or ten hours):
no train
no train
no train
no train
no train
no train
no train
no train
no train
no train
passenger train
freight train

9 trips out of 10, he will arrive and wait and see a passenger train and then leave. Only 1 trip in 10 will he arrive in time to have missed the passenger train but see the freight train.

Correct, the freight trains came 6 mins after the passenger trains. Thus the man's odds of arriving after a freight and before a passenger are 9 to 1, since for fifty four out of every sixty minutes it is a pssenger train that is expected.
(assuming a time period of 1hr, the solution can be adapted to any time period)

Verty I had a hard time understanding what you said, but I think you had the right idea.

 

[powergirl]

Number 2
Bob owns a couple of hour glasses-- One is a four minute hourglass and the other is a seven minute hour glass. If Bob wants to measure 9 minutes, how could he do it?

Here's the solution:

Take the 2 hour glasses;
Time 0 :turn both the hour glasses;
Time 4 :4 mint hour glass is done-turn it,(3 minutes remaining for 7 mint timer);
Time 7 :8 mint hour glass is done.
Time 8 :4 mint hour glass is done,6 mint remaining for 7 mint hr glass;
So completed 1 mint is there on the other side-Turn it;
Time 9 :That 1 mint is done;
Thus he could measure 9 mints.


I think this's correct.

 

[Gokul43201]

Alternative solution to 2:

9=4n+7m has a solution at (4,-1), so flip both initially and subsequently flip only the 4-clock thrice (as soon as it runs out). The time from when the 7-clock runs out to the time when the 4-clock runs out for the fourth time is 9 minutes long.

The train problem reminds of an "anecdote" from one of my favorite books from when I was little - Yakov Perelman's Mathematics can be Fun. The anecdote involves an informal gathering of the author's friends at someone's place. The host casually offers a wager of (paraphrasing heavily) 10 of his Roubles against anyone's 1000 Roubles that the next 100 people that pass by his front window will all be male. A mathematician in the group takes up the wager, confident that his 1000 Roubles are in no danger of leaving him, given the odds of this happening. Shortly after he accepts, the sounds of drums and trumpets are heard, followed by the thump-thump-thump of the boots of the Russian Army marching in formation down the street.

Moral: Beware - it's very easy to fall for the trap that lies in assuming a random distribution of events.

 

[atthepointe]

[STRIKE][/STRIKE]

Number 1
A guy who likes watching trains enjoys walking to a nearby railroad track to wait for a train to go by. Every time, upon his return, he makes a note of whether the one he saw was a passenger train or a freight train. Over the years, his figures show that 90 per cent of the trains have been passenger trains. One day he meets an official of the railroad and is surprised to learn that in fact passenger trains and feight trains are precisely equal in number. Why did the man, whom we may presume to have made random trips to the railroad tracks, see such a disproportionate number of passenger trains?

Number 2
Bob owns a couple of hour glasses-- One is a four minute hourglass and the other is a seven minute hour glass. If Bob wants to measure 9 minutes, how could he do it?

Spoiler

#1.He did not see all of the trains;he only saw them when the passenger trains were coming by.