Analog clock algebra word problem

1. Aug 22, 2016

late347

1. The problem statement, all variables and given/known data
define the time in an analog clock up to the second. The time needs to be such that the minute hand, is at the exact same position as the hour hand. This position of the hands needs to be between 10th and 11th hour.

2. Relevant equations
3. The attempt at a solution

I suppose that in the analog watch. If the time had been exactly 11:00. Then the minute hand is too far forward, but the hour hand would still be in the tolerances. I suppose that the answer is somewhere between [50-55] minutes. and I think it must necessarily be more towards the end part. Because already at the hypothetical time 10:45, the hour-hand is 3/4 finished at the hours interval.

I'm definitely having ttrouble formulating an equation about this problem though. This was one of the tougher exercises in my book, and I just couldn't seem to crack it. I'm a bit stumped on this final problem and it has been nagging my brain a little bit.

I suppose one could get a watch and try to rewind the time from 11:00 until the hands are on top of each other but that would not get the correct seconds count. :)

2. Aug 22, 2016

Staff: Mentor

You'll need some coordinate for the position of the hands - the hours are the most convenient choice (so 0 is directly up, 3 is to the right, and so on). For an integer T, at which position is the hour hand T hours after midnight/midday? At which position is the minute hand? Where are both if T is not an integer?

3. Aug 22, 2016

SammyS

Staff Emeritus
Certainly the correct answer can be obtained with Algebra. However, you can narrow the possible correct result fairly simply.

Where are two hands at 10:50 ?

Where are the two hands at 10:55 ?

Last edited: Aug 22, 2016
4. Aug 22, 2016

late347

Well, if you choose midday as the basis. 12:00 noon

Then you have domain for T between (-2, -1) for the hour-hand itself
I dunno how that is relevant to the minute-hand,though.

Of course when T= -2
At that point the minute hand is exactly at 12 :00 noon. ( the time is exactly 10:00)

everybody knows that, when the hour-hand is exactly at an even hour (doesn't matter which hour, but an exact hour). Then logically therefore the minute hand must be at 12 o'clock together with the seconds-hand.

I guess that when the minutes hand shows 50 minutes past 10:00. Then the hour hand will have moved about to about 5/6 of the interval in the hours.
Then again in an alternative perspective.The hour hand has moved from the starting point at 12:00 (650minutes/60) = 10 hours + 0.833...hours
(10.833... hours/ 12 hours)*(360deg) = 325 degrees moved forwarrd. The minute-hand is about 300 degrees moved forward so it is still catching up to the hour hand.

at 10:55 hours...

The hour hand has moved to 327.5deg

the minute hand has passed over the hour hand at already 330 deg

Looks like it will be about 54 minutes necessary plus some amount of seconds for the correct time.

I still have little ideaa how this can be done algebraically. It's just not really clicking for me at all.

5. Aug 22, 2016

haruspex

Good start.
When next will the two hands align? Suppose the minute hand has travelled an angle theta. How far has the hour hand moved?

6. Aug 23, 2016

late347

Well I suppose that the minute hand is obviously faster in turn-rate in any case

When 60 minutes have passe then the hourhand moved 5 minutes in that time (1h interval in analog clocks is same interval as 5 min)

So the hourhand moves at 1/12 rate of the minute hand. You could call it turn rate I suppose.

Minute hand turns 6deg/minute and hourhand turns 0.5deg/minute

Essentially 360 deg travelled by the minutehand in 60 mins.

. Hourhand should be like (5÷60)×(360deg)=30 degrees

I did not get exactly when the hands meet again. But if you start the clock from 12:00.
Then the minute hand requires full 60minutes and then some extra.

If the time is like 65 minutes +(1min/3)
Then the minute hand has gone 32deg into the new circle which begins at 12:00 position.

Inbthat same time the hourhand moved 5 min already (already accounted for) +(1/12 × 5.333min)= 49min/9

(49min/9) ÷60 ×360=32.666...deg

So at that point the hour hand is only slightly ahead of the minute hand. I couldnt get an easy fixation somehow as to where exavtly the hands would meet.

7. Aug 23, 2016

haruspex

That's because you keep plugging in numbers instead of working with algebra.
How much faster is the minute hand, as a ratio? I ask again,

8. Aug 24, 2016

late347

well, if we are given that angle theta exists for minutehand. This means that in the same period of time which has passed. The hourhand has moved angle theta/12

if the minute hand travels 360 degrees then it must quite necessarily be, that the hour hand travels one twelfth of that =30deg.
so the minutehand travels theta degrees
hour hand travels theta/12

In clocks the turn-rate (or speed?) of the hands, is constant speed, ideally speaking. Maybe not in practical mechanical watches, but in theory.

9. Aug 24, 2016

haruspex

Right. If the minute hand has lapped the hour hand, how much further has it travelled? What equation does that give you?

10. Aug 25, 2016

late347

I dont quite follow.

I suppose if I knew the equation already so easily then I would have solved the problem already.

How can you make the one side of the equation equate to the other side when the hourhand and minutehand are not actually equal...(in degrees of travels from 12:00 until the next overlapping position)

I guess when the minutehamd cranks the full circle + (then some degrees). It would be over 360deg.

Essentially theres more degrees which have been cranked by the minutehand. Or is that the wrong way to think about the minutehand?

For the actual problem the required overlapping time was between 10th and 11th hours though)

(Assuming start location at 12:00: with 0 seconds)

11. Aug 25, 2016

haruspex

Yes. When it catches the hour hand up the first time after 12 midday, how many extra degrees has it travelled?

12. Sep 9, 2016

late347

Well this question looked like it was not easily solvable at least for myself and with the tips given to me.

I think without further input I will abandon this problem because I havent made any progress and I have other homework and projects inthe pipeline so to say.

I asked my stepfather for a little bit of help but he failed to achieve the correct answer also. The problem's difficulty level was ostensibly high school level math textbook.

My stepdad is somewhat of a math whiz and he's a mathematics major from Helsinki university and a retired insurance analyst/specialist.

13. Sep 9, 2016

Staff: Mentor

haruspex gave multiple hints how to approach that problem, you did not follow those so far. It is quite a basic problem if you take the right approach.

14. Sep 9, 2016

late347

Apparently not quite basic enough, the problem that is.

Alternatively I myself am too basic in my brain to solve the problem. Honestly I was a little surprised that my stepfather couldnt solve the problem because he is one of the mathematically smartest people I know in real life. I took a photo directly of the problem assignment so it's not because of a misunderstanding of the problem statement.

I suppose I will do my official homework instead because those problems give credit to the course grade. And theyre easier haha

15. Sep 9, 2016

haruspex

I'm not quite ready to give up.
A hare and a tortoise race around a 100m circular track. When the hare laps the tortoise for the first time, how much further has the hare run than the tortoise has?

16. Sep 9, 2016

Staff: Mentor

The hour hand moves 1/12 as fast as the minute hand. So if the clock starts at 10 o'clock, the hour hand starts at 50 minutes and the minute hand starts at 0 minutes. If the minute hand is later located at m minutes, at what minute is the hour hand located?

17. Sep 10, 2016

ehild

Watch this video, and show it to your stepfather, too. Note, at what times are minute hand and the hour hand at the same position. There are such events in every hour. You need to find that time between 10h and 11h.

18. Sep 10, 2016

late347

I'm not sure I can answer that question because the clockface resets itself at the 0 minute. Like... it is only possible to have 60 minutes at maximum.
Or is this irrelevant to the issue. I read that post as though you wanted to know the exact clockface.

but if m means number of minutes travelled forwards.
the minutes travelled with the hour-hand will have been
50+m(1/12)
because 50 was the starting point for the hour hand. And the second part is the added minutes.

19. Sep 10, 2016

Staff: Mentor

Excellent. So, when the two hands are at the same location,$$50+\frac{m}{12}=m$$ From this equation, what is the value of m?

20. Sep 10, 2016

late347

m = 54.5454...

So do you suppose that 4 minues and 32.72 seconds time (the time which is past the starting time of 10:50: 0 seconds)

the hands should be in equal position measured in e.g. degrees from 0deg= 12 o clock

Can you explain in precise English wording what does the variable m ostensibly represent in that equation. I think that was the confusing part in the end for myself.

m represents what with regards to the word problem?