Two watches are set at 4:00 A.M. One watch is 2 min fast each hour. The second watch is 1 min late each hour. The faster watch will be one hour ahead of the slower watch at ____ A.M the next morning. I have no clue how to do this problem so please give me some hints. Thanks!
5 Am
After the first hour the two watches will be three minutes apart and three minutes more. 60/3 which is after 20 hr, and since it is an hour and 3 apart you would have to do 9pm + 8 hr= 5Am
Thank you so much!
Your welcome!!
To solve this problem, we need to understand the rate at which each watch gains or loses time.
The first watch gains 2 minutes each hour, which means it gains 2 minutes every 60 minutes.
The second watch, on the other hand, loses 1 minute each hour, which means it loses 1 minute every 60 minutes.
Now, let's think about how long it will take for the faster watch to be one hour ahead of the slower watch.
Since the first watch gains 2 minutes every 60 minutes, and we want it to gain 60 minutes (1 hour), we can set up a proportion:
2 minutes / 60 minutes = x minutes / 1 hour
Cross-multiplying, we get:
2 minutes * 1 hour = 60 minutes * x minutes
2 hours = 60x
Now, we can solve for x:
x = 2 hours / 60 = 1/30 of an hour
So, the faster watch gains 1 hour on the slower watch in 1/30 of an hour.
To determine the time at which this happens, we need to add this time to the initial time of 4:00 A.M.
4:00 A.M + 1/30 hour = 4:02 A.M
Therefore, the faster watch will be one hour ahead of the slower watch at 4:02 A.M the next morning.