Question

In a school, the duration of a period in the primary section is 40 minutes and that of the secondary section is 45 minutes. If the bell for each section rings at 7 a.m., when will the two bells ring together again?

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 two equations which state that the time when the bells toll together is divisible by the time taken by the first bell and time taken by the second bell. Using the concept of LCM we find the value of the variable.
* 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.

Complete step by step answer:
We are given a bell for primary section tolls after 40 minutes.
The bell for the secondary section tolls after 45 minutes.
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.
PRIMARY SECTION BELL:
\[n \times \] time taken by the primary section bell to complete a cycle \[ = \] total time taken for primary section bell to complete n cycles.
Let the time taken for primary section bell to complete n cycles is \[{T_1}\]
\[ \Rightarrow n \times 40 = {T_1}\]
This states that \[{T_1}\] is divisible by 40.
SECONDARY SECTION BELL:
\[n \times \]time taken by the secondary section bell to complete a cycle \[ = \] total time taken for secondary section bell to complete n cycles.
Let the time taken for secondary section bell to complete n cycles is \[{T_2}\]
\[ \Rightarrow n \times 45 = {T_2}\]
This states that \[{T_2}\] is divisible by 45.
We have to find after how many minutes the two bells toll together, so we take the time \[{T_1} = {T_2} = T\]
So, T is divisible by 48, 72 and 108.
Now we find the LCM of the numbers 40 and 45 which will give us the value of time after which the bells toll together.
We can write prime factorization of numbers as:
\[40 = 2 \times 2 \times 2 \times 5\]
\[45 = 3 \times 3 \times 5\]
Collecting the terms of same base
\[40 = {2^3} \times {5^1}\]
\[45 = {3^2} \times {5^1}\]
From the prime factorization we see that the highest power of prime number 2 is 3, 3 is 2 and 5 is 1
\[ \Rightarrow \]LCM \[ = {2^3} \times {3^2} \times {5^1}\]
\[ \Rightarrow \]LCM \[ = 2 \times 2 \times 2 \times 3 \times 3 \times 5\]
\[ \Rightarrow \]LCM \[ = 8 \times 45\]
\[ \Rightarrow \]LCM \[ = 360\]

So, the time after which the three bells toll together is 360 minutes.

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.