Question:
There are five machines A, B, C, D, and E situated on a straight line at distances of 10 metres, 20 metres, 30 metres, 40 metres and 50 metres 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 material is 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 received.
Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5th second and B at the beginning of the 6th second, and E at the beginning of the 10th second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.
Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that is, 10 metres 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?
There are five machines A, B, C, D, and E situated on a straight line at distances of 10 metres, 20 metres, 30 metres, 40 metres and 50 metres 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 material is 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 received.
Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5th second and B at the beginning of the 6th second, and E at the beginning of the 10th second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.
Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that is, 10 metres 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?
Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5th second and B at the beginning of the 6th second, and E at the beginning of the 10th second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.
Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that is, 10 metres 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?
Show Hint
After each trip, the robot goes back to whichever station is nearer: from D (40 m) that is the new station at 60 m, not the origin.
Updated On: Nov 4, 2026
- 120
- 140
- 340
- 70
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Trip 1 (serves A and D).
The robot leaves the origin at \(t=0\) and goes to D (40 m), serving A on the way. Now the origin is 40 m away, while the second station (at 60 m) is only 20 m away, so it goes to the second station: 20 m. Trip 1 = \(40+20=60\) m, ending at \(t=6\) s at the 60 m station.
Step 2: Pending messages.
C (sent at \(t=4\)) and B (sent at \(t=5\)) have already arrived, and both stations receive them, so the robot notices them at \(t=6\). E's message is sent at \(t=9\), later.
Step 3: Trip 2 (serves C and B).
From the 60 m station the robot moves toward the origin: C at 30 m (30 m away), then B at 20 m (10 m further), serving them in the order encountered: 40 m. It then returns to the nearest station: the origin is 20 m from B and the far station is 40 m from B, so it goes to the origin: 20 m. Trip 2 = \(40+20=60\) m, ending at \(t=12\) s at the origin.
Step 4: E's message.
E sent its message at \(t=9\), while the robot was away, so it is noticed when the robot gets back at \(t=12\).
Step 5: Total distance.
\(60+60=120\) m.
Final Answer:
The correct option is (1): 120
The robot leaves the origin at \(t=0\) and goes to D (40 m), serving A on the way. Now the origin is 40 m away, while the second station (at 60 m) is only 20 m away, so it goes to the second station: 20 m. Trip 1 = \(40+20=60\) m, ending at \(t=6\) s at the 60 m station.
Step 2: Pending messages.
C (sent at \(t=4\)) and B (sent at \(t=5\)) have already arrived, and both stations receive them, so the robot notices them at \(t=6\). E's message is sent at \(t=9\), later.
Step 3: Trip 2 (serves C and B).
From the 60 m station the robot moves toward the origin: C at 30 m (30 m away), then B at 20 m (10 m further), serving them in the order encountered: 40 m. It then returns to the nearest station: the origin is 20 m from B and the far station is 40 m from B, so it goes to the origin: 20 m. Trip 2 = \(40+20=60\) m, ending at \(t=12\) s at the origin.
Step 4: E's message.
E sent its message at \(t=9\), while the robot was away, so it is noticed when the robot gets back at \(t=12\).
Step 5: Total distance.
\(60+60=120\) m.
Final Answer:
The correct option is (1): 120
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