# How long will it take two clocks to show the same time again?

2 clocks A and B are now showing 3'o clock. Clock B is 2 mins/hr faster than clock A. How long will it take them to show the same time again

My solution: hour hand has speed 1/2 degree per minute so Clock B will run 1 degree faster then clock B , for showing same time clock B have to move 360 degree so time = 360hrs

Am I correct?

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Your answer is correct. However, I'm not sure if it's okay to simply assume clock A has perfect timekeeping. You can reason without geometry as follows: the error between the two clocks' time is increasing linearly, and the clocktimes will be equal once the error reaches 12hrs, so you need 12hrs/(2min/hr) = 360 hours. Also, if this is homework, you should tag it as such. –  anon Aug 7 '11 at 18:08
The speed of the hour hand is 360 deg/12 hr=30 deg/hr. –  Ross Millikan Aug 7 '11 at 18:13
@anon its not homework i am preparing for aptitude test so i encounter this problem –  rohit Aug 7 '11 at 18:18
@ross i have edit my mistake hour to minute thanks to point out. –  rohit Aug 7 '11 at 18:20
I don't see anything wrong with this question. A new user posts a question and their solution. (+1), maybe for as much reason as the (-1)s... –  The Chaz 2.0 Aug 7 '11 at 18:33

I am interpreting this question as follows: the hour hand of A rotates through a full 30 degrees in an hour, and the minute hand of B rotates through a full 30 degrees plus the equivalent of two minutes (which is 1 degree) in an hour. If this is how you've interpreted it too, your answer's right: my thoughts are detailed below.

So, A goes at 30 deg/h and B goes at 31 deg/h. I'm safe to ignore the minute and second hands here: if the hour hands are in the same position, they show the same time. So the question is: how many hours have to elapse before A and B show the same time? If the number of hours that elapse is $x$, then the hands of A and B have rotated $30x$ and $31x$ degrees respectively. We want these to "show the same time". By this we mean that they should be equal, up to adding or subtracting integer multiples of 360 (because 360 degrees = a full revolution). So we want a solution to $30x = 31x + 360k$, for some integer k, and some x > 0 - in fact, we want $x$ to be as small as possible, because $x$ is in hours, and the question says "how long will it take?".

(However, clock A might not run correctly. But that's fine. If it goes at 53 deg/h, then B goes at 54 deg/h. This doesn't affect the solution.)

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One minute is 6 degrees so 2 minutes are $24$ degres therefore it will be increasing $12$ degrees per hour and will need $30$ hours to be $360$ degrees and marking the same hour or $9:00$ pm of next day. Assuming I understood the instructions.

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Where is the $24$ coming from? –  Cameron Buie Nov 2 '12 at 20:16

I think answer 360 hours is not correct. Two options are there

Assume Clock A (run correctly) travels at 60 minutes/ Hr & B travel at 62 minutes per Hour. B travels at 2 minutes faster than A per hour. So Answer should be LCM of 62 Minutes and (12 Hrs x 60 minutes = 720 minutes) which is equal to 22320 minutes which is equal 372 Hours

Assume Clock A (run correctly) travels at 60 minutes/ Hr & B travel at 58 minutes per Hour.A travels at 2 minutes faster than B per hour. So Answer should be LCM of 58 Minutes and (12 Hrs x 60 minutes = 720 minutes) which is equal to 20880 minutes which is equal 348 Hours

Answer can be either 348 hours or 372 hours

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The values you're finding seems to be the amount of time that has elapsed according to clock B, whereas 360 is how much time passes according to clock A. But you reasoning doesn't really work -- you have a hidden assumption that the next agreement has to come after an integral number of half-days, but by that reasoning if clock B had been $\sqrt 2$ minutes faster per hour would never again agree with clock A, which is absurd. –  Henning Makholm Sep 25 '12 at 19:53