Professor Freeman likes to to think that he has more time to finish his work than he really has. So, he has set the four clocks in his study to be an average of exactly 2 hours slow. If 2 of the professor's clocks read 6:55, a third reads 6:42 and the fourth reads 6:28, what is the the correct time? Thanks for help.
+129839 Professor Freeman likes to to think that he has more time to finish his work than he really has. So, he has set the four clocks in his study to be an average of exactly 2 hours slow. If 2 of the professor's clocks read 6:55, a third reads 6:42 and the fourth reads 6:28, what is the the correct time? Thanks for help.
Let the elapsed minutes [since 12:00] for the 4 clocks be 415, 415, 402 and 388 respectively
And the average of these minutes =
[415 + 415 + 402 + 388] / 4 = 405 minutes
So......405 minutes after 12:00 = 360 + 45 = [ 360 / 6] : [45mod 60] = 6:45
So.......the correct time is actually 2 hours later = 8:45
Professor Freeman likes to to think that he has more time to finish his work than he really has. So, he has set the four clocks in his study to be an average of exactly 2 hours slow. If 2 of the professor's clocks read 6:55, a third reads 6:42 and the fourth reads 6:28, what is the the correct time? Thanks for help
First, will convert the minutes to decimal fractions:
1) 6:55 =6 + 55/60 =6.9167..... hours
2) 6:55 =6 + 55/60 =6.9167..... hours
3) 6:42 =6 + 42/60 =6.7 hours
4) 6:28 =6 + 28/60 =6.467.... hours
5) Since the professor set them an AVERAGE of 2 hours too slow, we therefore have:
6.9167 + 6.9167 + 6.7 + 6.467 =27
27/4 = 6.75 hours = 6:45 hours. But, since he set then 2 hours too slow, therefore we have:
6:45 + 2 = 8:45, which is the correct time.
