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On Time




This problem was more complicated than it seemed - we had over 50 solutions suggesting that when a bus travelled from one end to another it would meet buses travelling from the other direction every 10 minutes. Unfortunately this is not correct!

When the bus sets off there will be a bus arriving from the other end, a bus due to arrive 10 minutes later, a bus due to arrive 20 minutes later, a bus due to arrive 30 minutes later, and a bus just leaving at the other end which is due to arrive in 40 minutes. The bus will meet all these buses and all the buses that leave while it is travelling, that is, the buses that leave 10, 20 and 30 minutes after it has set off. It will arrive as another bus prepares to leave.

So the bus will set off as a bus arrives from the other end, meet a bus every 5 minutes while travelling (7 in all) and arrive at the other end as another is leaving.

If it is the first bus of the day, it will only meet 4 buses on its way, the second bus of the day will meet 5, the third bus will meet 6, the fourth bus will meet 7, the buses after the fourth one will all meet 7 except the last three - they will meet 6, 5 and 4 respectively.

Jessica from West Herts College sent a correct solution:

A bus in the middle of the day will either meet 7 or 9 depending on whether you count the buses it meets at either end of its journey. However the first and last bus will only meet 4 or 5 depending on whether you count the bus leaving as it arrives (in the case of the 1st bus) or leaves (in the case of the last bus).

Anna, PD and Emily from Ardingly College Junior School thought of it like this:

... if it left the station at 3.40pm it would meet the ones that left the station within 40 minutes before and after it and the one that left at the same time as it.

Heather from Stow Heath Junior School noticed that:

If we start counting as the first bus leaves for the day then it will not see another bus until it is half way along it's route.

Laura and Harriet from The Mount School explained it as follows:

Bus A and B set out at the same time, then 10 minutes later buses C and D set out. At this point none of the buses have advanced enough to meet another. Then, after 20 minutes, two more buses, E and F set out, and A and B meet. After 25 minutes A will meet D. 30 minutes into the cycle, A will meet F. G and H will set off. After 35 minutes A will meet H. Then at 40 minutes, the cycle will be complete, and A will meet one final bus, J, which is setting out. If bus A had been the first bus to set off, it would meet 5 other buses. If bus A had set off while the cycle was taking place, it would meet 8 other buses.

Well done to all of you who managed to crack this problem.