This is not related to programming.This was asked during test by one of the company: The minute and the hour hand of a watch meet every 65 minutes. How much does the watch lose or gain time and by how much?

Hi Capablanca I think that it's fast by a minute. I guess that the clock hands(hours and mins) are supposed to meet every 66 mins. Now that they meet every 65 mins so the clock is fast by a min. Secondly you can delete your thread by editing the thread(click on the icon with a pencil mark on it) and when in the edit page click the Delete Post checkbox. [ June 05, 2003: Message edited by: Anupam Sinha ]

they are supposed to meet every 12/11 hour = 720/11 minutes. so if they meet every 65 minutes, then the clock gain time (720/11 -65) minutes, during 65 minutes. so the clock is ~0.7% faster.

In 60 minutes: Minute hand "gains" 55 minutes over Hour hand. Hence, in 65 min. Minute hand should gain (55/60)*65 i.e. 59.583333 minutes over Hour hand. but in question it says its meeting i.e. minute hand gaining 60 minutes, so its faster.

- Varun

John Lee
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yes, see my previous post. it is 0.7% faster.

Anupam Sinha
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Hi all In my earlier post I was wrong. Hoping this time I am right. When an hour hand completes 60 mins the min hands moves 5 mins so for every min the min hand moves, the hour hand would move 5/60 mins. Which is 0.083. So in 65 mins it would move 65 * 0.083 = 5.4167. So the clock is actually faster by 5.4167 -5 = .4167 mins. If this time I am agin wrong please tell me why. [Edited the message after Don's post] [ June 05, 2003: Message edited by: Anupam Sinha ]

John Lee
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Originally posted by Anupam Sinha: I guess my earlier post was wrong.

yes, you are wrong earlier.

Varun Khanna
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Originally posted by Don Liu: yes, see my previous post. it is 0.7% faster.

you are close, actually it should be 0.69444% Ermm... depends what you mean. The exact answer is that clock is fast by a ratio of 144/143, which means that every hour (every true hour) the defective clock completes an extra 1/143 of a revolution, which is 0.00699300699300..., or ~0.69901%. This works out to 25.175 seconds every hour. I believe the .69444% answer could be explained this way: every "hour" as measured by the defective clock, the true time is 143/144 of the apparent time, so the true time is lagging by 1/144 of the apparent time - which is .69444%, or exactly 25 seconds. So, imagine we start the defective clock at 12:00 noon and set it to the correct time, exactly. Then later when the defective clock shows 1:00:00 P.M. exactly, the true time is 12:59:35 exactly. And five seconds later when the true time is 1:00:00 P.M. exactly, the defective clock shows 1:00:25.175 (rounded to milliseconds). Is the clock ahead by 25 seconds, or 25.175? I would say the latter, but it depends on your reference point.

"I'm not back." - Bill Harding, Twister

John Lee
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i am glad someone else gets the same result!

Arjun Shastry
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The answer they gave was clock gains 5/11 minutes per hour. [ June 06, 2003: Message edited by: Capablanca Kepler ] [ June 06, 2003: Message edited by: Capablanca Kepler ]

John Lee
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thanks for the good question!

Arjun Shastry
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All except Morris got the answer 5/12 minutes.Correct answer is 5/11 minutes(given bye Morris in doubly posted thread).My apprach was : Let the clock be x minutes ahead .At 60 minutes after the clock starts,position of minute hand is at x.After 5 minutes position, minute hand will move by (60+x)/12.Hence at the end of 65 minutes,position of minute hand is at x + (60+x)/12-------------(1). For hour hand,in 12 minutes it moves by one minute.hence at 60 minutes its position will be 5 + (x/12).After 5 minutes ,it will move by (5+x)/12.Hence at 65 minutes, its position will be 5 + x/12 +(5+x)/12-----------(2). Equating 1 & 2 ,we get x=5/11 minutes. [ June 08, 2003: Message edited by: Capablanca Kepler ]

All except Morris got the answer 5/12 minutes.Correct answer is 5/11 minutes(given bye Morris in doubly posted thread) You need to look at my answer again.

Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius - and a lot of courage - to move in the opposite direction. - Ernst F. Schumacher

John Lee
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Originally posted by Capablanca Kepler: The answer they gave was clock gains 5/11 minutes per hour.

i believe clock gains 5/12 minutes per hour.

Arjun Shastry
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Originally posted by Michael Morris: You need to look at my answer again.

27 3/11 seconds = (27*11+3)/11 seconds = (27*11+3)/(11*60) minutes = 0.454545.. minutes. 5/11 minutes = 0.4545454... How did you arrive at 27 3/11 seconds? [ June 10, 2003: Message edited by: Capablanca Kepler ]

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How did you arrive at 27 3/11 seconds? 5/11 * 60 = 27 3/11. If you wanted my reasoning, the minute and hour hands cross 11 times in 12 hours, the minute, hour and second hands cross only once in 12 hours. A correct clock should have the hour and minute hands crossing every 1hr 5min 27 3/11 sec. Your clock's hour and minute hands are crossing every 1hr 5min. So the deficit is 27 3/11 sec fast. At that rate your clock gains five minutes every 12 hours. [ June 10, 2003: Message edited by: Michael Morris ]

Varun Khanna
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The basic logic to be used is: - In a normal clock, "minute hand" gains 55 min. over "hour hand" in 60 min. -In the mentioned watch, "minute hand" is gaining 60 min. over "hour hand" in 65 minutes. these two are sufficient to get the answer

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Originally posted by Don Liu:

i believe clock gains 5/12 minutes per hour.

I will disagree. again ... in 65 min., minute hand is gaining 60 min. over hour hand i.e its gaining 55.384615 minutes In normal clock the gain 55 min. in 60 min. so this watch is (55.384615 -55) minutes faster i.e. 0.384615 .... minutes in every hr. but 5/12 = 0.41666666667 ...

Hum, this begs some questions: Is the clock one with a sweep motion or step motion? Is the clock a standard 12 hour clock or a 24 hour Military issue clock? What device was used to measure the time lapse? How accurately was this time reported? A standard 12 hour watch with sweep motion, the minute hand moves at a rate of .1 degrees a second where the hour hand moves at a rate of .008333 degrees a second. A second on the clock is 6 degrees. Now, if both moved correctly for 65 minutes. The hour hand moved 32.5 degrees and the hour hand moved 390 degrees which is 30 degrees past where it was last. So, if your time is reported correctly the minute hand move 2.5 degrees further than it should have. so the standard rate of .1 degrees a second (Which is now slightly off but, close enough for government work) show the clock to gain 25 seconds each hour. Well, ok then the second hand is now moving at a rate of .1005 degrees a second which means it's actually only about 24.8 seconds fast [ June 11, 2003: Message edited by: Carl Trusiak ]

I'm really starting to get confused on this one. I thought Capa was saying I was wrong, but rereading it would seem he agrees with me. I remember doing this same sort of problem in High School Algebra 100 years ago. Please, what the hell is the real right answer? I need peace of mind or a bottle of Cuervo.

I'd look into buying a digital watch about this time and throw the other one away.

Arjun Shastry
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Originally posted by Michael Morris: Please, what the hell is the real right answer? I need peace of mind or a bottle of Cuervo.

answer is "Clock gains 5/11 minutes/hr"Enjoy now

Jim Yingst
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[Capablanca Kepler]: All except Morris got the answer 5/12 minutes. Well, this is probably the first sign that your math is not to be trusted, as it's fairly obvious that my previous answer of 1/143 revolution is not 5/12 minutes - it's 60/143 which is approximately .41958 minutes. 5/12 would be .41667; 5/11 would be .45455. I still maintain that 60/143 minutes is correct, so I will try to follow your post to see where we disagree... Correct answer is 5/11 minutes(given bye Morris in doubly posted thread).My apprach was : Let the clock be x minutes ahead .At 60 minutes after the clock starts,position of minute hand is at x.After 5 minutes position, minute hand will move by (60+x)/12.Hence at the end of 65 minutes,position of minute hand is at x + (60+x)/12-------------(1). I agree up to here. For hour hand,in 12 minutes it moves by one minute. No, in the time it takes the minutes hand to move an angle which looks like 12 minutes, the hours hand will appear to move exactly 1 minute. Remember that it's not really 12 minutes of actual elapsed time, since the clock is wrong. However the effect of this error is minor compared you a later error... hence at 60 minutes its position will be 5 + (x/12). Agreed - this line is exactly true for 60 minutes of real time. After 5 minutes ,it will move by (5+x)/12. Uh, what? Where did this come from? This is your main problem. After 5 more minutes (real minutes) it should move exactly 5/60 as far as it did in the preceding hour. Which means it should move (5 + x/12)/12. Hence at 65 minutes, its position will be 5 + x/12 +(5+x)/12-----------(2). No. At 65 minutes, its position will be ( 5 + (x/12) ) + ( (5 + x/12)/12 ) Which simplifies to 13/12 * (5 + x/12) ------ (2) Equating 1 & 2 ,we get x=5/11 minutes. Using the correct version of 2, we get x + (60+x)/12 = 13/12 * (5 + x/12) Multiply both sides by 12*12: 144x + 720 + 12x = 13 * (60 + x) 156x + 720 = 780 + 13x 143x = 60 x = 60/143 Hmmm, that number looks somehow familiar... [ June 18, 2003: Message edited by: Jim Yingst ]

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[MM]: If you wanted my reasoning, the minute and hour hands cross 11 times in 12 hours, the minute, hour and second hands cross only once in 12 hours. A correct clock should have the hour and minute hands crossing every 1hr 5min 27 3/11 sec. Your clock's hour and minute hands are crossing every 1hr 5min. Agreed up to this point. So the deficit is 27 3/11 sec fast. OK, that's the difference after 65 minutes of elapsed time. (Remember, that's when the seconds and minutes are intersecting.) But in this thread you said: The watch is gaining(running fast) 27 3/11 seconds per hour. How did we get from "per 65 minutes" to "per hour"? This would seem to account for why your answer is exactly 13/12 times my answer. I tried giving you an extra week to work on it and correct your error... Ah well... As for other answers: 5/12 minutes per hour is pretty close, but still wrong. varun Khanna: sorry, I can't follow this post. Too much work now. Carl Trusiak: geez, what a troublemaker. I assume it's a 12-hour sweep-motion clock, and the figure of 65 minutes is exact. (It couldn't possibly have come from the defective clock, and so other source of time measurement is posited, so what else is there but to assume it's 65 min exactly?) As for the answer, Carl acknowledges his is probably a little off due to his method of estimation, but 24.8 seconds isn't bad - the correct answer works out to 25.17 sec. Much closer than some of the other answers here. [ June 18, 2003: Message edited by: Jim Yingst ]

Michael Morris
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How did we get from "per 65 minutes" to "per hour"? Yep. Math is one of those use or lose it propositions. I've spent the last twenty plus years using only trig and analytic geometry. Though, the 65 minutes should have been obvious. The Area covered by a path problem made me realize how much I had forgotten about simple calculus. Jason suggested that I get my hands on a copy of The Java Number Cruncher which I did and am thorougly enjoying every page of it. The light clicked back on when I just saw the terms "Trapezoidal Rule" and "Simpson's Rule". Anyway, this forum has made me realize that I need to get back on top of my Math game again.

Arjun Shastry
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{ Uh, what? Where did this come from? This is your main problem. After 5 more minutes (real minutes) it should move exactly 5/60 as far as it did in the preceding hour. Which means it should move (5 + x/12)/12. } Correct,thanks for correction.I am also going for that Java Number Cruncher

Varun Khanna
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Originally posted by Jim Yingst: [QB sorry, I can't follow this post. Too much work now. [/QB]

too much work?? ... and till today i was under the impression tht u must be in software industry

Jim Yingst
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Sorry, what I meant was "sorry my brain is already fried from parsing some of the other answers". I am also going for that Java Number Cruncher Dang, I saw the author doing a book signing at JavaOne; I should have grabbed a copy. But they probably would've made me pay for it.