Three runners complete one lap of a circular track in 2 h, 4 h, and 5.5 h respectively. After how many hours will they all be together again at the starting point?

Difficulty: Easy

Correct Answer: 44 h

Explanation:


Introduction / Context:
When multiple periodic events start together and repeat at different periods, the first time they coincide again is given by the least common multiple (LCM) of their periods. Here, periods are times to complete one lap. Because one time is fractional in hours, converting to minutes simplifies the LCM computation.


Given Data / Assumptions:

  • Times per lap: 2 h, 4 h, 5.5 h.
  • All start together at time 0.


Concept / Approach:
Convert hours to minutes to avoid decimals, find the LCM in minutes, then convert back to hours. The LCM ensures each runner has completed an integer number of laps at the meeting instant.


Step-by-Step Solution:

Convert: 2 h = 120 min; 4 h = 240 min; 5.5 h = 330 min.Prime factors: 120 = 2^3*3*5; 240 = 2^4*3*5; 330 = 2*3*5*11.LCM = 2^4 * 3 * 5 * 11 = 16 * 165 = 2640 minutes.Convert to hours: 2640 / 60 = 44 h.


Verification / Alternative check:

Check multiples: 44/2 = 22 laps, 44/4 = 11 laps, 44/5.5 = 8 laps; all integers, so they meet together at 44 h.


Why Other Options Are Wrong:

  • 40 h, 20 h, 22 h are not common multiples of all three periods in hours; some runner would not be at the start at those times.


Common Pitfalls:

  • Taking the greatest common divisor instead of the least common multiple, or mishandling the 5.5 h decimal.


Final Answer:

44 h

More Questions from Problems on H.C.F and L.C.M

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