Not as easy as it first appeared. Many of you sent in attempts at this very popular problem . The most common mistakes in your arguments involved rounding to the nearest minute and assuming that the only time the hands cross is at twelve o'clock. One of you pointed out that on a standard clock the hour hand never passes over the minute hand - it is always the other way around - a bit of a red herring!

Two correct solutions were received from Hannah of the School of St. Helens and St. Katharine and Andrei of School 205, Bucharest. It is Hannah's solution that is given below. Well done Hannah.

This took me a couple of attempts; it wasn't as simple as it first seemed because of all the fractions of time that are involved.

Firstly - 8 miles is approximately a quarter of the distance travelled in an hour at a speed of 33mph, so we are looking at a journey time of just less than 15 minutes.

More precisely:-

Secondly - it does not seem unreasonable to assume that the train leaves on a whole number of minutes past (or to) the hour.

This means that the time when the hands cross that we are looking for will probably end in 0.545454...; or (6/11) of a minute so that when we take the journey time off we will end up with a precise departure time (no seconds or part seconds left over).

We know that the hands cross at 12 o clock and that they cross 11 times in every 12 hours so that is every

11 12 = 1.090909... = 1  1 11 hours.

In terms of time this is 1 hour 5 minutes  5 11 minute. = 1  1 11 ×60

If this is so then the tenth time the hands pass by each other after 12:00 will be 10h 54 mins 6/11 minute.

If we subtract the journey time of 14 6/11 off this then we will end up with an exact departure time for John's train of 10:40 or 22:40.