1st clock gains 5 mins in every 1 hour, 2nd clock loses 5 mins in every 1 hour, after how many hours both will show same time again? At 3:00 PM, Monday clock was showing the right time
1st clock gains 5 mins in every 1 hour.
2nd clock loses 5 mins in every 1 hour.
After how many hours both will show same time again?
I know that the time difference between the two clock will be 10 minutes. 1st clock will show the time 4:05 PM and the 2nd clock will show the time 3:55 PM. But how come after exactly 12 hours both will show the same time again? I didn't understand this 
 A: It's equivalent to think of one clock going at a normal speed and the other clock gaining 10 mins every hour. The question then becomes how long before those 10 minute gains add up to 12 hours? After 6 hours the second clock will have gained an hour, so 12 lots of this will put the second clock 12 hours ahead. So the answer is $6\times 12 = 72$ hours. Double this if you're using a 24 hour clock.
A: Clock1 - Clock2
mon:
3:00pm - 3:00pm
4:05pm - 3:55pm
 ....
tue morning:
3:00am - 2:00am
...
so, through this progression we can see that every 12 hours there will be a variation of 1 hour. therefore the clocks will meet exactly at 3:00 o'clock, 3 days later.
Its supposed to meet after 6 days if you want them in 24hour clock.
A: Look at this in tho eother pespective:
1st clock is $5$[ticks/h] faster, than normal clock. 2nd clock is  $5$[ticks/h] slower, than normal clock.
We know, that the whole cycle of clock lasts $24\cdot 60$ [tics].
Let's place the boundaries of this cycle on points A and B. Now we have the question:
Cities A and B are $24\cdot 60$[ticks] away from each other. From A to B travels a train with velocity $5$[ticks/h], from B to A travels another train with velocity $5$[ticks/h]. Both trains have started at 3PM, Monday. When the trains will meet each other?
A: Divide the $12$-hour clock into $144$ five-minute intervals (there are $12$ hours on the clock and $12$ five-minute intervals per hour; $12\cdot12=144$). The first clock advances $13h$ intervals every $h$ hours, and the second advances $11h$.
We want to find the smallest positive $h$ for which $13h \equiv 11h \pmod {144}$. In other words,
$$2h \equiv 0 \pmod {144},$$
so $h=72$.
