## 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: $9\!:\!40$