Question

3 bells in a bhootnath temple toll at the interval of 48, 72 and 108 seconds. If they have tolled all together at 6:00 AM then after how many seconds all the bells will tolled together.

Answer 1

Hint:Here we take the time when the bells toll together as a variable. Using the information given in the question we form three equations which state that the time when the bells toll together is divisible by the time taken by the first bell, time taken by the second bell and the time taken by the third bell. Using the concept of LCM we find the value of the variable.

Complete step-by-step answer:
LCM is the least common multiple of two or more numbers. We write each number in the form of its prime factors and write the LCM of the numbers as multiplication of prime factors with their highest powers.
We are given bell 1 which takes 48 seconds to toll.
The second bell takes 72 seconds to toll.
The third bell takes 108 seconds to toll.
Let us assume the T as the time when both the three bells toll together.
Since, the bells toll in a cycle one after the other. Let n be the number of cycles to be completed by each bell to toll together.

BELL 1:
\[n \times \]time taken by the first bell to complete a cycle \[ = \]total time taken for the first bell to complete n cycles.
Let the time taken for bell 1 to complete n cycles is \[{T_1}\]
\[ \Rightarrow n \times 48 = {T_1}\]
This states that \[{T_1}\]is divisible by 48.

BELL 2:
\[n \times \]time taken by the second bell to complete a cycle \[ = \]total time taken for the second bell to complete n cycles.
Let the time taken for bell 2 to complete n cycles is \[{T_2}\]
\[ \Rightarrow n \times 72 = {T_2}\]
This states that \[{T_2}\]is divisible by 72.

BELL 3:
\[n \times \]time taken by the third bell to complete a cycle \[ = \]total time taken for the third bell to complete n cycles.
Let the time taken for bell 3 to complete n cycles is \[{T_3}\]
\[ \Rightarrow n \times 108 = {T_3}\]
This states that \[{T_3}\]is divisible by 108.
We have to find after how many seconds the three bells toll together, so we take the time \[{T_1} = {T_2} = {T_3} = T\]
So, T is divisible by 48, 72 and 108.
Now we find the LCM of the numbers 48, 72 and 108 which will give us the value of time after which the bells toll
We can write prime factorization of numbers as:
\[48 = 2 \times 2 \times 2 \times 2 \times 3\]
\[72 = 2 \times 2 \times 2 \times 3 \times 3\]
\[108 = 2 \times 2 \times 3 \times 3 \times 3\]
Collecting the terms of same base
\[48 = {2^4} \times {3^1}\]
\[72 = {2^3} \times {3^2}\]
\[108 = {2^2} \times {3^3}\]
From the prime factorization we see that the highest power of prime number 2 is 4 and 3 is 3
\[ \Rightarrow \]LCM \[ = {2^4} \times {3^3}\]
\[ \Rightarrow \]LCM \[ = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3\]
\[ \Rightarrow \]LCM \[ = 16 \times 27\]
\[ \Rightarrow \]LCM \[ = 432\]
So, the time after which the three bells toll together is 432 seconds.

Note:Students might get confused of why we take LCM of two given time values and not HCF, keep in mind whenever we have to find the common value where the bells toll together or any other object intersects, we always take LCM because that gives us the least or the minimum value of time when the two objects meet.