A, B and C start at the same time in the same direction to run around a…

Key idea: Find the least common multiple (LCM) of the three times (in seconds). The LCM gives the smallest time when all three return to the starting point together.

Step 1 — Convert and factorize each time:

• 252 seconds = 2^2 × 3^2 × 7

• 308 seconds = 2^2 × 7 × 11

• 198 seconds = 2 × 3^2 × 11

Step 2 — Take the highest power of each prime appearing in any factorization:

• Highest powers: 2^2, 3^2, 7, 11

• LCM = 2^2 × 3^2 × 7 × 11 = 4 × 9 × 7 × 11 = 2772 seconds

Step 3 — Verify divisibility by each runner's period:

• 2772 ÷ 252 = 11

• 2772 ÷ 308 = 9

• 2772 ÷ 198 = 14

Step 4 — Convert to minutes and seconds: 2772 seconds = 46 minutes and 12 seconds (because 46 × 60 = 2760, remainder 12).

Conclusion: They will all again be at the starting point after 2772 seconds, i.e., 46 minutes 12 seconds.