Question

In a shop, there are three clocks which chime at intervals of $15,20$ and $30$ minutes respectively. If they all chime together at $10$ a.m, at what time they all chime together again in the morning?

Answer 1

Hint: We can solve this question by taking the Lowest common multiple of time intervals i.e LCM of the time intervals, because the LCM of the time intervals will give us the minimum time after which all the three clocks will chime together.

Complete step by step solution:
 Here in the question we know that there are three different clocks having different time intervals to chime.
$ \Rightarrow $ Let we take three different clocks named Clock1, Clock2 and Clock3.
So, according to question Clock1 has a time interval of $15$ minutes to chime,
Clock 2 has time interval of $20$ minutes to chime,
And Clock3 has a time interval of $30$ minutes to chime.
So, firstly while solving this question we need to know that these three clocks will chime together when their Lowest common multiple i.e LCM of time intervals will occur.
So, Lowest common multiple of $15$ minutes,$20$minutes and $30$ minutes will be calculated as
\[
  5\left| \!{\underline {\,
  {20,15,30} \,}} \right. \\
  3\left| \!{\underline {\,
  {4,3,6} \,}} \right. \\
  2\left| \!{\underline {\,
  {4,1,2} \,}} \right. \\
  2\left| \!{\underline {\,
  {2,1,1} \,}} \right. \\
  1\left| \!{\underline {\,
  {1,1,1} \,}} \right. \\
 \]
$ \Rightarrow LCM\left( {15,20,30} \right) = 5 \times 3 \times 2 \times 2 \times 1$
$ \Rightarrow LCM\left( {15,20,30} \right) = 60$
This implies that all the 3 clocks will chime together after $60$minutes or $1$ hour of first chime.
So,
Time at which first chime occurred together $ = 10$ a.m
$ \Rightarrow $ Time at which second chime will occur together $ = \left( {10 + 1} \right) = 11$ a.m
$ \Rightarrow $All the three clocks will chime together at $11$ a.m together for the second time after $10$ a.m

Note:
In these types of questions the simplest method to solve the question is by finding the LCM or lowest common multiple of the entries.
The other method would be by adding the time intervals in$10$a.m one by one and then finding the common time at which all the clocks will chime together, but in this case this method could appear short but if the LCM would be a bigger value then this alternative method would be very lengthy.