You may also like

problem icon

Speedy Sidney

Two trains set off at the same time from each end of a single straight railway line. A very fast bee starts off in front of the first train and flies continuously back and forth between the two trains. How far does Sidney fly before he is squashed between the two trains?

problem icon

Illusion

A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?

problem icon

Walk and Ride

How far have these students walked by the time the teacher's car reaches them after their bus broke down?

John's Train Is on Time

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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:-

$$ \frac{8}{33} \times 60 = 0.242424\dots \times 60 = 14.545454\dots = 14 \frac{6}{11} \mbox{minutes.} $$

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\dots; 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

$$ \frac{11}{12} = 1.090909\dots = 1 \frac{1}{11} \mbox{hours.} $$

$$ \mbox{In terms of time this is } 1 \mbox{ hour } 5 \frac{5}{11} \mbox{minutes} = 1 \frac {1}{11} \times 60 $$

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.