My watch is 1 second fast each hour, and my friend's is 1.5 seconds slow each hour. Right now they show the same time. When will they show the same time again?
There is an accumulating difference of 2.5 seconds per day. You want that to equal 43,200 sec (one-HALF day) before the watches agree again (forget about AM/PM differences).
43,200 sec/2.5 (sec/day) = 17,280 days (47.3 years!).
How did you get that?
What's the way ?
The formula and the logic are explained in my answer. Each day one watch gets 2.5 seconds ahead of the other. You have to calculate how many days it takes for the difference to equal 12 hours. Divide 12 hours (in seconds) by 2.5 seconds per day.
To find out when your watch and your friend's watch will show the same time again, you need to determine the time it takes for the time difference between the two watches to equal a full hour.
Let's break down the situation:
Your watch is 1 second fast each hour, which means it gains 1 second every hour.
Your friend's watch is 1.5 seconds slow each hour, which means it loses 1.5 seconds every hour.
To find the time it takes for the time difference between the two watches to equal a full hour, you should calculate the reciprocal of the difference in their timekeeping speeds.
In this case, the reciprocal of 1 second is 1/1 = 1, and the reciprocal of 1.5 seconds is 1/1.5 = 2/3.
Now, you can set up an equation to solve for the time it takes for the watches to show the same time again:
1 * t = (2/3) * t
Here, 't' represents the time it takes for the watches to show the same time again.
By solving this equation, we find that 't' is equal to 0. Hence, the watches will show the same time again in 0 hours, which means they are already showing the same time right now.