Fast Clock Slow Clock

When 60 minutes elapse on a correct clock, 62 minutes register on clock F (fast), and only 56 minutes register on clock S (slow). If later in the day, clock F reads 8\!:\!00 and clock S reads 7\!:\!00, what was the correct time when the two clocks were originally set?
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
Clock F reads 8\!:\!00 while clock S reads 7\!:\!00 implies that clock F is 60 minutes ahead of clock S. Since clock F is ahead of clock S\: 6 minutes per hour, 10 hours must have passed from the time they were set
10 hours\times 6 minutes/hour = 60 minutes
10 hours before 8\!:\!00 is 10\!:\!00. Clock F is 2 minutes fast per hour. We need to make clock F go back an extra 20 minutes (10\times 2=20) to pass 10 hours exactly.
10\!:\!00-20 minutes = 9\!:\!40

Alternatively, from the perspective of clock S\: 10 hours before 7\!:\!00 is 9\!:\!00. Clock S is slow 4 minutes per hour. We need to spring it forward 40 minutes (10\times 4=40) to pass 10 hours exactly.
9\!:\!00+40 minutes = 9\!:\!40
9\!:\!40 is the correct time when the two clocks were originally set.

Answer: 9\!:\!40

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About mvtrinh

Retired high school math teacher.
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