Daily Math Puzzle: 2026-05-24
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2026-05-24
A robot factory has two assembly lines, A and B, that each produce a robot and then send it down a straight hallway to the loading dock. The hallway is 120 meters long. Robots from line A walk at 1.5 m/s, while robots from line B ride a conveyor that moves at 2 m/s (the robot itself does not walk). Both lines start producing robots at the same moment, and each line produces a new robot every 30 seconds. However, the conveyor for line B can carry only one robot at a time; a robot must wait at the start of the conveyor until the previous robot has cleared the hallway. After the first robot from each line reaches the dock, the next robot from each line is released immediately (i.e., production continues regardless of travel). How many robots from line A will reach the dock before the third robot from line B reaches the dock?
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Solution
4 — First find travel times:
- Line A robot: 120 m ÷ 1.5 m/s = 80 s.
- Line B robot on conveyor: 120 m ÷ 2 m/s = 60 s.
Production interval for both lines is 30 s.
Line A departures: 0 s, 30 s, 60 s, 90 s, …
Arrivals = departure + 80 s → 80 s, 110 s, 140 s, 170 s, 200 s, …
Line B departures are limited by the conveyor:
- B1 ready at 0 s, departs 0 s, arrives 60 s.
- B2 ready at 30 s but must wait until B1 clears at 60 s, departs 60 s, arrives 120 s.
- B3 ready at 60 s but must wait until B2 clears at 120 s, departs 120 s, arrives 180 s.
We need the number of A‑robots that arrive **before** B3’s arrival at 180 s. The A‑arrival times before 180 s are 80 s, 110 s, 140 s, and 170 s – four robots. The next A‑robot arrives at 200 s, which is after B3.
Therefore, 4 robots from line A reach the dock before the third robot from line B.
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