Bell intervals and tank flow on a school day
The Government Senior Secondary School at a block headquarters works from 8:00 a.m. to 6:00 p.m. Three bells serve different wings. The office bell rings every 18 minutes, the primary-wing bell every 24 minutes and the workshop bell every 40 minutes. All three were tested together at exactly 8:00 a.m. on the first day of the session. The caretaker, Ramkishan, says the bells will never again ring together during the school day and wants the workshop timer changed. Meanwhile the overhead tank is fed by a pump whose flow meter is marked 21/1600 kilolitre per minute; the pump runs throughout the interval between two successive simultaneous rings, and the peon must record the water delivered in that interval. The physics teacher wants the flow rate written as a decimal on the board, so that students can compare it with the 0.014 kL/min reading displayed at a neighbouring school.
1. Is the caretaker's claim correct? Find when all three bells next ring together and count how many simultaneous rings fall within the 8:00 a.m. to 6:00 p.m. school day.
Read the worked answer
Prime factorisation
- 18 = 2 × 3²
- 24 = 2³ × 3
- 40 = 2³ × 5
LCM = 2³ × 3² × 5 = 360 minutes = 6 hours.
So simultaneous rings occur at 8:00 a.m., 2:00 p.m., 8:00 p.m., and so on.
Within a day ending at 6:00 p.m., the only simultaneous ring after the opening test is at 2:00 p.m. (8:00 p.m. is outside the day). Counting the 8:00 a.m. test as well, two simultaneous rings occur in the school day.
Verdict: Ramkishan is wrong - one more simultaneous ring does happen, so the 40-minute timer need not be changed on this ground.
2. Express 21/1600 kL per minute as a decimal, decide with reasoning whether it is terminating, and find the water delivered between two successive simultaneous rings.
Read the worked answer
Terminating check: 1600 = 2⁶ × 5². A fraction in lowest terms has a terminating decimal only when the denominator has no prime factor other than 2 and 5. Here 21 and 1600 are coprime and the denominator is 2⁶ × 5², so the decimal terminates.
21/1600 = (21 × 625)/(1600 × 625) = 13125/1,000,000 = 0.013125 kL/min (= 13.125 L/min).
Water between two simultaneous rings:
0.013125 × 360 = 4.725 kL = 4725 L.
Comparison: 0.014 kL/min > 0.013125 kL/min, so the neighbouring school draws slightly more water per minute; over 360 minutes the difference is 0.000875 × 360 = 0.315 kL, i.e. about 315 L.
3. If the office bell's interval were changed to 20 minutes while the other two stayed the same, would the bells ring together more often or less often? Support your answer with a calculation.
Read the worked answer
With intervals 20, 24 and 40 minutes:
- 20 = 2² × 5
- 24 = 2³ × 3
- 40 = 2³ × 5
LCM = 2³ × 3 × 5 = 120 minutes = 2 hours, against 360 minutes earlier.
So the bells would coincide every 2 hours (10:00 a.m., 12:00 noon, 2:00 p.m., 4:00 p.m.) instead of every 6 hours - more often. The shorter common interval arises because 20 introduces only a factor 5 and drops the 3² required by 18; reducing the LCM shortens the waiting time. In the 10-hour day this would give five simultaneous rings instead of two, which is why the caretaker's proposed change is not as harmless as it looks.