A light blinks after every $3$ seconds, another light blinks after every $5$ seconds. How many times do they blink together in half an hour?

Hint: In this problem we need to calculate the number of times where the two lights will blink in the given period of time. For this we will first calculate the LCM (Least Common Multiples) for the both the given times. Now we will consider the given time period which is half an hour. We have given the blinking times of the lights in seconds, so we will convert the given half an hour time period into seconds and we will divide it by the calculated LCM of the given blinking times to get the required result.

Complete step-by-step answer:
Given that, A light blinks after every $3$ seconds, another light blinks after every $5$ seconds.
Considering the given blinking times which are $3$ seconds, $5$ seconds.
We can observe that the both the given blinking times are prime numbers and do not have any factors, so the LCM of the two number will be product of the two numbers, then the LCM of the blinking times will be
$\begin{align}
  & \text{LCM}=3\times 5 \\
 & \Rightarrow \text{LCM}=15 \\
\end{align}$
In the problem they have mentioned the time period as the half an hour. We know that $30$ minutes is taken as half an hour and one minute has $60$ seconds, so the number of seconds for a half an hour is given by
$\text{half an hour}=30\times 60\text{ seconds}$
In order to find the number of times that the both the lights will blink at same time we are going to divide the above value with the LCM of the given blinking times, then we will get
$\begin{align}
  & \text{no}\text{.of times}=\dfrac{30\times 60}{15} \\
 & \Rightarrow \text{no}\text{.of times}=120 \\
\end{align}$

Hence the both the lights blink $120$ times in the given period of half an hour.

Note: In this problem we have given only two lights, so we have considered the two numbers for calculating the LCM. In some cases, they may give more than two numbers of lights, then we need to calculate the LCM of all the given numbers by using their factors.