On Friday morning two clocks are correctly set at 6 a.m. One clock gains 1 minute every 2 hours. The other clock loses 1 minute every hour. When the real time is 10 p.m. that evening, the fast clock is ______ minutes ahead of the slow clock.
Please explain the answer
Thanks
30 mins ahead
real time passed from 6:00 am to 10:00 pm is 16 hrs
time gained by fast clock = 8 minute
will read: 10:08
time lost by slow clock = 8 minutes
clock will read: 9:52 pm
so fast clock is 16 minutes ahead of slow clock
a clock shows correct time at 12:00 noon.it gains 2.5 minutes every hour.what time will it show at 6:00 pm
Pls helped
To find out the number of minutes the fast clock is ahead of the slow clock at 10 p.m., we need to determine the time difference between the two clocks from 6 a.m. to 10 p.m.
Let's start by looking at the fast clock. It gains 1 minute every 2 hours. To find out how many minutes it gains in 16 hours (from 6 a.m. to 10 p.m.), we divide the total hours by 2 and multiply the result by 1. So:
16 hours / 2 = 8
8 x 1 minute = 8 minutes
Therefore, by 10 p.m., the fast clock is 8 minutes ahead.
Now, let's analyze the slow clock. It loses 1 minute every hour. Between 6 a.m. and 10 a.m., the slow clock loses 1 minute per hour. So it loses:
4 hours x 1 minute = 4 minutes
From 10 a.m. to 10 p.m., the slow clock loses 1 minute every hour. So it loses:
12 hours x 1 minute = 12 minutes
Therefore, by 10 p.m., the slow clock is 4 minutes + 12 minutes = 16 minutes behind.
Now, to determine the time difference between the two clocks, we subtract the number of minutes the slow clock is behind from the number of minutes the fast clock is ahead:
8 minutes - 16 minutes = -8 minutes
Therefore, at 10 p.m., the fast clock is 8 minutes ahead of the slow clock.
It's important to note that the negative sign (-8) indicates that the fast clock is ahead of the slow clock, and the magnitude of 8 represents the time difference in minutes.