# Twisted Timekeeping [closed]

Two office workers need to arrive at work at the exact same time. However that's not likely to happen because the only thing they use for telling the time is their watches and their timekeeping is a little bit off...

One worker's watch gains 5 minutes every hour and the other's watch loses 5 minutes every hour.

So here's the question:

If they need to arrive at 7:30 AM, how many minutes will they be off by once they arrive?

I've added the tag because someone pointed out a second solution to my puzzle. One that uses algebra (the intended solution) and another that uses lateral thinking (the unintended solution).

• Doesn't this depend on what time it is 'now'? Or when the two watches were last in sync? Am I missing something, or don't we need more information to solve this?
– Stiv
Commented Sep 19, 2021 at 13:15
– Taco
Commented Sep 19, 2021 at 15:58

Assuming they synchronise their watches when leaving the office, they’ll have a 10 min difference accumulating every hour.

Assuming a standard 5-day week and 8-hour day,

• Tuesday through Friday, they’ll be separated for a nominal 16h (by each watch), so they will out by 160 min: one early by 80 min, the other late by 80 min;

• Monday, they’ll be separated for a nominal 48+16=64h, so it would be a whopping 640 minutes (5h 20min early/late).

After the first week, though, you’d expect either the late worker to have been fired, or else both to clue to their watches’ idiosyncrasies and adjust their time of arrival accordingly (or both).

• With the known arrival time of 0730, you're forgetting half an hour. However, one could argue that this point is moot without knowing whether or not the sync time was 30 minutes past the hour. For all we know it could be quarter 'til two, and then it would be even worse. Great answer though! +1
– Taco
Commented Sep 19, 2021 at 16:36
• @Tacoタコス Thanks! I used the term "8-hour day" deliberately - the end of the working day would be 8 hours after the actual start of the day, 7:30 start in this case, measured by the company's accurate clock :) They might start the day at odd times, but they are guaranteed to finish each day at the same time. Commented Sep 19, 2021 at 17:02

Without a time provided for when their watches were last in sync, their drift can be expressed as:

0.166666666666667 * n

Where n is the total number of minutes between the time when their watches were last in sync and their expected arrival time of 0730.

How?

By dividing their drift of 10 minutes per hour by an hour, then multiplying by the total number of minutes between sync and arrival.

Proof:

Assume the last time their watches were in sync was 1300 (1pm). This gives us 18.5 hours until they need to arrive at work again (0730 the next day). So, we translate this to 1,110 minutes and then multiply by 0.166666666666667 to get the total drift of 185.00000000000037 minutes. So, by time 0730 rolls around, the employee with the fast watch will have been at work for approximately 3 hours and 5 minutes. Meanwhile, the employee with the backwards watch will not be at work for another approximately, 3 hours and 5 minutes.

Assuming that their watches are in sync and that they’re both aware of the issues with their respective watches:

They won’t be off at all. Simply because the same amount of time has passed for both employees, so drift doesn’t matter. One goes backwards the same amount that the other goes forwards by. Assuming that the employee whose watch goes backwards is aware of their misdirection, then they would still arrive on time.

Question seems incomplete, but I will post 2 answers using 2 assumptions:

1. Simple practical assumption:

Both the workers know that the watches have to be adjusted, hence both know what the watches will show at 7:30 AM, hence both will arrive at the correct time, though the watches might show some other times.
Delta Time = 0

1. Syncing point assumption:

I will assume that neither of the workers know that the watches have to be adjusted, but they did sync their watches at a midnight Syncing Point (Maybe they had a late night "SYNC-UP" meeting!)
When the slower watch shows 7:30 AM (=7x60+30 in minutes), one worker arrives, but the time is T1 (in minutes) and that watch will show 7x60+30 = T1 - 5xT1/60
(55xT1) = 60x(7x60+30)
T1 = 490 minutes = 8:10 AM
When the faster watch shows 7:30 AM (=7x60+30 in minutes), the other worker arrives, but the time is T2 (in minutes) and that watch will show 7x60+30 = T2 + 5xT2/60
(65xT2) = 60x(7x60+30)
T2 = 415 minutes = 6:55 AM
Delta Time = 8:10 AM - 6:55 AM = 490 minutes - 415 minutes = 75 minutes
We can calculate Delta Time given some other Syncing Point, only the total time elapsed will change.