Three electronic devices make a beep repeatedly at fixed intervals of 48 seconds, 72 seconds, and 108 seconds respectively. They all beep together exactly at 10:00 a.m. At what time will all three devices next beep together at the same instant after 10:00 a.m.?

Introduction / Context:

This problem is a classic example of using the concept of the least common multiple in time and work situations. Here, the repeated actions are beeps from three devices at different regular intervals. Finding when they next beep together is equivalent to finding the smallest time interval that is a common multiple of all three given intervals. This is a common pattern in exam questions involving bells, alarms, and periodic processes.

Given Data / Assumptions:

• Device 1 beeps every 48 seconds.
• Device 2 beeps every 72 seconds.
• Device 3 beeps every 108 seconds.
• All three beep together at exactly 10:00 a.m.
• We assume perfect regularity in intervals with no delay or drift.

Concept / Approach:

The time after which they all beep together again is the least common multiple of 48, 72, and 108 seconds. The least common multiple represents the smallest positive time that is simultaneously a multiple of all three individual periods. Once this number of seconds is found, it is converted into minutes and seconds and then added to the starting time of 10:00 a.m. to get the required clock time.

Step-by-Step Solution:

Verification / Alternative Check:

Check that 432 is a multiple of each interval. Dividing 432 by 48 gives 9, by 72 gives 6, and by 108 gives 4. All are integers, so 432 is a common multiple. To confirm that it is the least common multiple, we note that we used prime factorisation with the highest powers of each prime, which by definition yields the smallest positive number that is a multiple of all three, so there is no smaller positive time when they beep together again.

Why Other Options Are Wrong:

All other options correspond to different second values that are not equal to the least common multiple. For example, 10:07:24 corresponds to 444 seconds, which is not divisible evenly by all three periods. Similar checks show that 10:07:36 and 10:07:48 do not yield integer multiples for every device, so they cannot be the first common beep time.

Common Pitfalls:

Common mistakes include taking the greatest common divisor instead of the least common multiple, adding the intervals directly, or performing incorrect factorisation. Another error is to forget to convert between seconds and minutes correctly or to add the interval incorrectly to the starting clock time. Careful factorisation and unit handling avoid these errors.

Final Answer:

The three devices will next beep together at 10:07:12 hrs after the initial beep at 10:00 a.m.