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Difficulty: MediumClocks and Calendars

A mechanical watch, which gains time at a continuous and uniform rate, is observed to be exactly 5 minutes slow at 8:00 AM on a Sunday. By 8:00 AM on the immediately following Tuesday, the same watch is exactly 7 minutes fast. How many hours after 8:00 AM on that Sunday did the watch display the true, correct time?

Answer: 20 hours

Answer

20
The correct answer is derived by mapping the total error shift over the total elapsed time. The watch gains a total of 12 minutes over a span of 48 hours. This establishes a constant gain rate of 1 minute every 4 hours. Because the watch started exactly 5 minutes slow, it requires 5×4=205 \times 4 = 20 hours of elapsed true time for it to catch up and display the correct, synchronized time.

Step-by-Step Solution

1
Calculate the total true time elapsed between the two observations.
48 hours
To find the rate at which the watch gains time, we first need to determine the total duration of the period. From Sunday 8:00 AM to Tuesday 8:00 AM is precisely two full days, which equals 48 hours.
2
Calculate the total amount of time the watch gained over this period.
12 minutes
The watch transitions from being 5 minutes slow (-5) to being 7 minutes fast (+7). The total change in its display relative to true time is 7−(−5)=127 - (-5) = 12 minutes.
3
Determine the uniform rate at which the watch gains time.
1 minute gained every 4 hours
By dividing the total elapsed time by the total minutes gained (48÷1248 \div 12), we discover that the watch gains exactly 1 minute for every 4 hours of true time that pass.
4
Calculate the hours needed to make up the initial 5-minute deficit.
20 hours
In order to display the correct time, the watch must gain precisely the 5 minutes it was initially lagging behind. At a steady rate of 1 minute per 4 hours, this process will take 5×4=205 \times 4 = 20 hours.

Key Concept

Uniform time gain and synchronization of faulty clocks
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Topics

Clocks and Calendars

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