Question

Three bells ring once every \(12\) minutes, \(20\) minutes and \(36\) minutes respectively. If all of them ring together at \(6{:}30\) a.m., at what time will they next ring together on the same morning?

• A. \(8{:}30\)
• B. \(9{:}30\)
• C. \(10{:}30\)
• D. \(11{:}30\)

Hint:

For synchronization problems (bells, traffic lights, gears), use the LCM of the individual periods to find the next alignment time, then add it to the given start time.

Answer 1

The Correct Option is B

Solution:
Step 1 (Why LCM?).
They start together and then repeat together after a time that is a common multiple of their individual intervals. The least such time is the LCM} of \(12,\ 20,\ 36\) minutes.
Step 2 (Prime factorization).
\[ 12 = 2^2 \cdot 3,\quad 20 = 2^2 \cdot 5,\quad 36 = 2^2 \cdot 3^2. \] Take the highest powers of each prime: \[ \text{LCM} = 2^2 \cdot 3^2 \cdot 5 = 4 \cdot 9 \cdot 5 = 180\ \text{minutes}. \] Step 3 (Convert minutes to hours).
\(180\) minutes \(= 3\) hours.
Step 4 (Add to the start time).
Starting time \(= 6{:}30\) a.m. Next common ringing time \(= 6{:}30 + 3\ \text{hours} = 9{:}30\) a.m.
Step 5 (Quick check).
\(180\) is a multiple of each interval: \(12 \times 15=180,\ 20 \times 9=180,\ 36 \times 5=180\) all align at \(9{:}30\) a.m.
\[ {9{:}30\ \text{a.m. (Option (b)}} \]