Question

a watch which gains uniformly was observed to be 5 minutes slow at 12 noon on monday.It was noticed 10 mins fast at 6pm on the nxt day. When did the show correct time?

Answer 1

Answer:

This is your answer

Step-by-step explanation:

Since the watch is 5 minutes slow at 12 noon on a Sunday and 5 minutes fast on the subsequent Wednesday at 6:00 p.m, it will show the right time at a time exactly mid way of theses 2 times.

hours passed=24+24+24+6=78hrs

mid way will be 39hrs after Sunday noon, which will be 3:00 a.m. tuesdsay.

Answer 2

Total time from 10 am on Friday to 10 pm on following Friday

= 7 days 12 hours = (7x24) + 12 hours

= 180 hours.

According to the question, the watch gains total time = (5 min. + 5 min. 48 sec.)

$(5+5\frac{4}{5})$ min = $54/5$ min

Now, $54/5$ min is gained in 180 hours.

As we know in the starting, the watch was 5 min. slow from the actual time. It means, that to get the correct time, the watch needs to gain 5 min.

So,

5 min is gained in = $(180*5*5)/54$ hours = $83\frac{1}{3}$ hours

=83 hours 20 min after 10 am on Friday.

= 3 days 11 hours 20 min after 10 am on Friday.

Hence, the watch was correct at 9:20 pm on Monday.

#SPJ2