# Continuous moving buses puzzle

A bus route has a total duration of 40 minutes.

Every 10 minutes, two buses set out, one from each end.

How many buses will one bus meet on its way from one end to the other end?

2 busses. One after 20 min and the other after 30 min. 3 if we can include the bus pulling out when this Bus reaches its destination.

**5 buses** including the one starting when the bus reach its destination.

In simple words the bus had to pass all the buses from opposite direction before its reach its destination.

So during 40 minute interval there are 5 buses from opposite direction.

**bus1** which start at **0 minute** which our bypasses at **20 minute**

**bus2** which starts at **10 minute** which our bypasses at **30 minute**

**bus3** which starts at **20 minute** which our bypasses at **35 minute**

**bus4** which starts at **30 minute **which our bypasses at **37.5 minute**

and **bus5** which starts at **40 minute**(at the destination) which our bus bypasses at **40 minute**

None!!

Here t denotes time in minutes.

X denotes path not visited by any bus yet.

b1,b3,b5……..denotes the buses which start from left side of the road

while b2,b4,b6……denotes the buses which start right side of the road!!!

(number)…..denotes how many buses they meet on the way…….

when t=0; b1(0)———X———-X———-X———b2(0)

when t=10; b3(0)———b1(0)———-X———-b2(0)———b4(0)

when t=20; b5(0)———b3(0)———-b1(1)b2(1)———-b4(0)———b6(0)

when t=30; b7(0)———b5(1)b2(2)———-b3(1)b4(1)———-b1(2)b6(1)———b8(0)

when t=40; ……………………………………..every bus will meet at least one bus during its path from one end to another.

**Answer** – The total number of buses met will be **9**

**Explanation-**

EXCEPT for the FIRST 3 buses starting from either end, every other bus will meet another bus from the opposite site every **5** minutes. This is because buses from both ends are moving simultaneously (covering equal distances, assuming buses moving at uniform speed).

Let us denote the distance covered in every minute as a point. i.e, considering one end of the route as Po**int 0,** the other end will be P**oint 40**. Consider a bus (Say, A) starting from point **0**. The moment it starts, another bus B which had started 30 minutes earlier from the other end (Point **40**) will be already at 10 min. away (**Point 10**) from reaching its destination. By the time bus A covers the first 5 points of its journey, the bus B also will have covered the same distance (from Point 10) and hence will meet bus A. By that time, yet another bus C (started from Point 40 following bus B)will be at 10 points away, and this bus will meet A in the next 5 minutes, just like bus B did. Thus, bus A will meet buses B, C… etc. every five minutes. This will continue till it reaches **Point 40**. Thus, the number of buses it meets en route will be **7**. Counting the buses it meets at the starting point (the bus which had left Point 40, at 40 min. earlier) and at the terminating point (the bus which leaves Point 40 at the 40th minute of Bus A’s journey), the total number of buses met will be **9**.

(For the **first** 3 buses starting at either end, this calculation will not apply, because, when the first bus starts from Point 0, there is no bus which had started earlier from Point 40 on the way to meet it at Points 5, 10 and 15. It will meet the first bus from the opposite end at **Point 20** only. Thereafter, it will meet subsequent buses as described above. Similarly, when the 2nd bus leaves Point 0, there will be **one** bus at Point 30 (which was the first bus from Point 40, left 10 min, earlier) which will be the first bus it meets en route at Point **15**. followed by subsequent buses as above.)

man you put a lot of effort into explanation .Nicely done

Shouldn’t the answer be 10 ? Won’t the bus meet a bus when it is just about to start i.e., Won’t the bus starting at point 0 meet the one bus which left point 40 , 40 mins earlier , at the station or starting point ?

### Your Answer

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