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### A Train and Clock Problem

```
Date: 04/29/2000 at 07:33:32
From: Stan
Subject: VERY Confusing Word Problem

Here's my problem:

A train leaves the station at the top of the minute. If it averages 66
miles per hour for the first 10 miles it will pass milepost 10 exactly
at the moment that the minute hand of the clock is directly over the
hour hand. At what time does the train leave the station?

How in the world am I supposed to understand and solve this? What
to the right answer; I'm really desperate to solve this.

Thanks a lot for previous help, Dr. Math!
Stan
```

```
Date: 05/02/2000 at 10:16:57
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi, Stan.

This is certainly a tough problem. It calls for attacks from several
directions.

First, you can find the time that elapses between the two events: when
the train leaves (on an exact minute), and when it reaches the
milepost (when the minute hand coincides with the hour hand).

Then you need to find two events of the specified types that are
separated by exactly this time. There are lots of exact minutes in a
day, but there are only 11 times in 12 hours when the minute hand and
the hour hand coincide. Try to find these times. This is a good
problem in itself!

Once you have found the 11 possible times for the train to reach the
milepost, subtract the elapsed time to find when the train would have
left the station in each case. Which of these times is an exact
minute?

There is an outline of your plan of attack. There is still plenty of
work to do.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/02/2000 at 17:22:35
From: Stan
Subject: Re: VERY Confusing Word Problem

Okay, I partially get this, but how can you get an exact minute when
you subtract, for example, 9.09 minutes. Are you saying that there is
a time like 1:06:09?

```

```
Date: 05/03/2000 at 08:43:04
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi again, Stan.

Yes, time is continuous. We're assuming that the minute hand turns
smoothly. There are dial clocks that are run digitally with stepper
motors, so the hand will jump from minute to minute. This problem
would have no solution with that sort of clock.

The time you have written, though, is 1 hour, 6 minutes, and 9 seconds
(or 9/60 minute). It's easier to use decimal minutes, like 1:06.09.
Better yet, though, is to stick with exact times like 9 1/11 minutes,
rather than 9.09, which is only an approximation to 2 decimal places.

You did step 1 successfully. Now on to step 2: finding all the times
when the minute hand and the hour hand coincide. This will take some
algebra - as I said, it's a good tough problem in itself. I'll get
you started.

One obvious time when the two hands are together is 12:00, when both
hands are straight up. The other times are not going to be even
minutes. The next one is sometime after 1:05, as I suspect you already
see. At 1:00 the hour hand is at the 5-minute mark, but by the time
the minute hand gets there, the hour hand will have moved on a bit.
where the hour hand is now, then advancing the hour hand again, but
this will go on forever. Algebra is better.

Let's say that the hands are together at n minutes past 1:00. The
minute hand is at the n-minute mark (remember, it won't be an exact
minute, so there won't really be a mark there). Where is the hour
hand, in terms of minutes? When n = 0, the hour hand is at the
5-minute mark, as I said above. When n = 60, it's 2:00, so the hour
hand is at the 10-minute mark. Can you write an expression that says
where the hour hand is for values of n between 0 and 60?

When you have this expression, you can quickly write an equation that
says that the two hands point to the same minute. Solve this equation.
Then go on to the time between 2:10 and 2:15 when the hands are
together again, and so on.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/03/2000 at 16:52:35
From: Stan
Subject: Re: VERY Confusing Word Problem

Hi:

Boy am I confused! I understand what you said, but...

I tried to write an expression in all different kinds of ways, but
couldn't get one that worked. (I thought of proportions, but that
didn't make sense.) I know when one minute passes, the hour hand moves
1/60 of its way between the 5-minute mark and 10-minute mark (I

Thank you, Dr. Rick, for coping with me  :)
Stan
```

```
Date: 05/04/2000 at 08:22:11
From: Doctor Rick
Subject: Re: VERY Confusing Word Problem

Hi again, Stan.

You're right, the hour hand moves 1/60 of the distance between 1:00
and 2:00 each minute. In terms of minutes on the dial, that's 1/60 of
5 minutes, since 1:00 is at the 5-minute mark and 2:00 is at the
10-minute mark. So in one minute, the hour hand advances 5/60 minute
on the dial. That fraction reduces to 1/12 minute.

Now we're ready to write an equation. The hour hand starts at 5
minutes when n = 0 (at 1:00). It advances 1/12 minute when n increases
by 1. Can you write a linear equation expressing the hour-hand
position h in terms of the minute n?

When you have this equation, you can easily change it into an equation
that states that the hour-hand position is the same as the minute-hand
position. Then you're on your way!

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/04/2000 at 17:30:03
From: Stan
Subject: Re: VERY Confusing Word Problem

I tried. Is

h + n = 5 + (1/12n)

correct? It may not be.

Thanks, Dr. Rick!
Stan
```

```
Date: 05/05/2000 at 15:29:35
From: Doctor Peterson
Subject: Re: VERY Confusing Word Problem

Hi, Stan. I'm going to take over from my brother Dr. Rick for now,
because he's not available to answer you today. It sounds like it's
time for a change of direction anyway. I'd like to go back to the
original question:

A train leaves the station at the top of the minute. If it
averages 66 miles per hour for the first 10 miles it will pass
milepost 10 exactly at the moment that the minute hand of the
clock is directly over the hour hand. At what time does the
train leave the station?

Let's try to work through the whole problem; I won't give you all the
details you have to fill in, but will try to guide you through the
process. I'm not taking the time to go through all your interactions
with Dr. Rick, but will try to focus on what has to be done, even if
it repeats some of what he has said, and I will ignore any specific
problems you have dealt with. You need a fresh start.

What do we know? The train started at some time that is an exact
number of minutes, then went 66 miles per hour for 10 miles, at which
time the hour and minute hand were together.

We can first look at the rate aspect of the problem. If he went 66
miles per hour for 10 miles, you need to determine how long this took;
I'd recommend finding it in minutes, and writing it as an exact
fraction. This gives you the elapsed time of the trip.

We'll be looking for two times that are exactly that far apart, the
first being an exact minute, the second being one where the hour and
minute hand are in the same position. That's the second part of the
problem, for which we have to think about how a clock works.

The minute hand moves 360 degrees in 60 minutes, or 6 degrees per
minute. The hour hand moves 360/12 = 30 degrees in 60 minutes, or 1/2
degree per minute, 1/12 as fast as the minute hand. The position of
the minute hand at M minutes after any hour is 6M degrees; the
position of the hour hand at M minutes after hour H is (60H+M)/2
degrees. (Notice that I'm assuming that H is a whole number, but M
doesn't have to be.) The hands will be together when these two angles
are equal, or

6M = (60H+M)/2

You should solve this to find M in terms of H; H could be any integer
from 0 to 11, so this will give you 12 (actually, it will turn out,
11) different values of M. From this you can find all the times at
which the hour and minute hand are in the same position. Work them
out, again, exactly, giving a fractional number of minutes where
necessary.

Now, does the fractional part of any of these values coincide with
that of the number of minutes you calculated for the elapsed time?
That's what you are looking for, since you know that the starting time
was an exact number of minutes.

Once you find the time that fits, you will know the ending time, and
can subtract the elapsed time to find when the train started.

Of course, this problem is terribly unrealistic; we're depending on
the knowledge that the minute and hour hands are together exactly,
down to a fraction of a minute (and no mention has been made of the
second hand!); any slight error could make a huge difference in the
answer. But it does bring in some interesting math.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra

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