Question

Four different electronic devices make a beep after every 30 minutes, 1 hour, $1\dfrac{1}{2}$ hour and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. They will again beep together at :
(a) 12 midnight
(b) 3 a.m.
(c) 6 a.m.
(d) 9 a.m.

Answer 1

Hint: We can calculate the respective time in minutes after which each devices beep and the least number that is a multiple of all the times (L.C.M) will give the time after which all of them will together beep and that can be added to the time they last beeped i.e. 12 noon as given in the question and we will get the required answer.
Remember:
1 hour = 60 minutes

Complete step-by-step answer:
Electronic devices make a beep respectively after following minutes:
First = 30 mins
Second = 1 hr
 = 60 mins
Third = $1\dfrac{1}{2}$ hour $\left(
 \because 1hr = 60\min s
 \dfrac{1}{2}hrs = \dfrac{{60}}{2} = 30\min \right)$
 = ( 60 + 30 ) mins
 = 90 mins
Fourth = 45 mins
These devices will beep together after the interval of minutes which is their least common multiple (L.C.M).
Calculating the L.C.M of these times using prime factorization:
30 = 2 X 3 X 5
60 = 2 X 2 X 3 X 5
90 = 2 X 3 X 3 X 5
105 = 3 X 5 X 7
L.C.M = 15 X 2 X 2 X 3 X 7
= 1260 mins
The devices will beep together after 1260 mins
In hours this time can be written as:
$
 60\min s = 1hr
 1260\min s = \dfrac{1}{{60}} \times 1260
 $
= 21 hours
The devices will beep together after 21 hours.
Now, according to the question, the devices beeped together at 12 noon.
They will beep together again after 21 hours (calculated)
12 noon + 21 hrs = 9 am.
Therefore, the given devices will again beep together at 9 am, (option d)
So, the correct answer is “Option D”.

Note: The L.C.M of the given numbers will not be less than any of the numbers for whom the L.C.M is calculated. In prime factorization, the prime numbers (whose multiples are only 1 and the number itself) and then check for the common numbers and if more than one, they get multiplied.
If in case, one number is the factor of another, then the one which is larger will be the L.C.M