Question

Five bells begin to ring together and ring respectively at intervals of 6, 5, 7, 10 and 12 seconds. How many times will they ring together in one hour excluding the one at the start?

• A. 7 times
• B. 8 times
• C. 9 times
• D. 11 times

Hint:

When multiple periodic events start together, their next simultaneous occurrence time is the LCM of their periods. In a time window, count the multiples of that LCM.

Answer 1

The Correct Option is B

Step 1: Find the LCM of the intervals.
\[ \mathrm{LCM}(6,5,7,10,12)=\mathrm{LCM}(2\cdot3,\ 5,\ 7,\ 2\cdot5,\ 2^2\cdot3)=2^2\cdot3\cdot5\cdot7=420\ \text{s}. \] Step 2: Count how many multiples of 420 s occur in one hour.
One hour \(=3600\) s. The bells coincide at \(t=420, 840, \ldots\) up to \( \le 3600\).
Number of coincidences (excluding \(t=0\)) is \[ \left\lfloor \frac{3600}{420} \right\rfloor = 8. \] \[ \boxed{8\ \text{times}} \]

Answer 2

Build the LCM step by step by combining two intervals at a time, then check each option against the resulting count of coincidences in one hour.

1. Combine 6 and 5: \(\mathrm{LCM}(6,5)=30\).
2. Bring in 7: \(\mathrm{LCM}(30,7)=210\).
3. Bring in 10: \(\mathrm{LCM}(210,10)=210\) (already a multiple of \(10\)).
4. Bring in 12: \(\mathrm{LCM}(210,12)=420\), so the bells coincide every \(420\) seconds. In one hour (\(3600\) s), the number of coincidences after the start is \(\left\lfloor 3600/420\right\rfloor=8\).

Checking the options: \(7\) times would need coincidences every \(3600/7\approx514\) s (not \(420\)); \(9\) times or \(11\) times would require an interval smaller than \(420\) s, which contradicts the computed LCM. Only \(8\) times matches.

the correct answer is 8 times.