Q7 ARITHMETIC PROBLEMS

There are five machines-A, B, C, D, and E-situated on a straight line at distances of 10 m, 20 m, 30 m, 40 m and 50 m respectively from the origin of the line. A robot is stationed at the origin of the line. The robot serves the machines with raw material whenever a machine becomes idle. All the raw materials are located at the origin. The robot is in an idle state at the origin at the beginning of a day. As soon as one or more machines become idle, they send messages to the robot-station and the robot starts and serves all the machines from which it received messages. If a message is received at the station while the robot is away from it, the robot takes notice of the message only when it returns to the station. While moving, it serves the machines in the sequence in which they are encountered, and then returns to the origin. If any messages are pending at the station when it returns, it repeats the process again. Otherwise, it remains idle at the origin till the next message(s) is (are) received.

Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 m from the origin, i.e. 10m from E. After finishing the services in a trip, the robot returns to the nearest station. If both stations are equidistant, it chooses the origin as the station to return to. Assuming that both stations receive the messages sent by the machines and that all the other data remains the same, what would be the answer to the above question?

Since the machine recieves the message from A and D, it will cater to them in the same trip.In return journey,it will go to the point which is 10m away from E

Distance travelled=40+20-60m

Now it recieves the message from B and C and caters them in the single journey.In return journey, it will go to origin.

Distance travelled=40+20=60m

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