# The hands of a faulty clock meet every 66 minutes; which solution is correct?

The minute hand of a clock overtakes the hour hand at intervals of $66$ minutes of the correct time. How much time does the clock gain or lose in $4$ hours?

Solution 1 The normal clock time it takes for the minute hand to overtake the hour hand is $65\cfrac{5}{11}$. So the time $\underline{\text{gained or lost}}$ in $4$ hours will be $\cfrac{5}{11}\ \text{minutes per hour}$:

\begin{align}\dfrac{5}{11}\times \dfrac{1}{66}\times 4\times 60=1\cfrac{76}{121}\ \text{minutes in 4 hours}\end{align}

or

Solution 2 The normal clock time it takes for the minute hand to overtake the hour hand is $65\cfrac{5}{11}$. So the time $\underline{\text{gained or lost}}$ in $4$ hours will be

\begin{align}66-65\cfrac{5}{11}=\dfrac{6}{11} \ \text{minutes per hour} \\~\\~\\\to\dfrac{6}{11}\times \dfrac{1}{66}\times 4\times 60=1\cfrac{119}{121}\ \text{minutes in 4 hours}\end{align}

Please explain which solution is correct and why (step by step).

Also throw light on the underlined words, that is if it loses or gains time and why.

The "gains or loses" question is fairly straightforward: a normal clock takes $60\cdot \frac {12}{11}=65\frac 5{11}$ minutes for the minute hand to catch up to the hour hand, while our faulty clock takes $66$ minutes which is $\frac 6{11}$ minutes longer. This indicates that the faulty minute hand is moving slower than the normal minute hand, which means that the faulty clock is "losing" time.
How much time is lost is then $\frac 6{11}$ minutes in $66$ minutes, which is the solution $2$ calculation you have already done, but does not correspond to $6\over 11$ minutes per hour, which would result in $2\frac3{11}$ minutes lost over $4$ hours.
• But how did u get $\large 2\cfrac{3}{11}$. – R K Mar 18 '15 at 19:54
• $2\frac 3{11}=4\cdot \frac 6{11}$ which is "$6\over 11$ minutes lost per hour over the course of four hours". The point is that we are losing $6\over 11$ minutes over the course of $66$ minutes which is one "cycle", not over the course of an hour. – abiessu Mar 18 '15 at 20:42
• So does the answer $\large 1\cfrac{119}{121}$ shown by book is wrong ? – R K Mar 18 '15 at 22:04
• The book's answer is correct. What I am saying is that your statement "$6\over 11$ minutes are lost per hour" is incorrect, as that is the rate per hand-matching cycle, not per hour. – abiessu Mar 19 '15 at 1:37
• $\quad\quad$Oh ok – R K Mar 19 '15 at 10:38