Question

Q.6 Three traffic lights at three different crossings change after 48
seconds , 72 seconds and 100 seconds respectively. If they all change
simultaneously at 8 am, at What time will they again change
simultaneously? *
O O
10 am
O 9 am
09pm
O 11 am​

Answer 1

Solution :

clearly the time taken for a simultaneously change

=> {L.C.M. of 48, 72, 100} second

=> L.C.M 48, 72, 100

=> 2 x 2 x 2 x 2 x 3 x 3 x 5 x 5

=> 3600

time taken simultaneously change

=> 3600 second

=> $\sf\Large\frac{3600}{60\:x\:60}$ hour = 1hour

$\Large{ \begin{array}{c|c} \tt 2 & \sf{ 48 , 72 , 100} \\ \cline{1-2} \tt 2 & \sf { 24 , 36 , 50} \\ \cline{1-2} \tt 2 & \sf{ 12 , 18 , 25} \\ \cline{1-2} \tt 2 & \sf{ 6 , 9 , 25} \\ \cline{1-2} \tt 3 & \sf{ 3 , 9 , 25 }\\ \cline{1-2} \tt 3 & \sf{ 1 , 3 , 25 }\\ \cline{1-2} \tt 5 & \sf{ 1, 1 , 25} \\ \cline{1-2} \tt 5 & \sf{ 1 , 1, 5 }\\ \cline{1-2} & \sf{ 1 , 1 , 1} \end{array}}$

=> Hence they will be Change again simultaneously at 9 am

Answer 2

Answer:

Given

The traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds respectively.

So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds

⇒ 48 = 2 × 2 × 2 × 2 × 3

⇒ 72 = 2 × 2 × 2 × 3 × 3

⇒ 108 = 2 × 2 × 3 × 3 × 3

Hence LCM of 48, 72 and 108 = (2 × 2 × 2 × 2 × 3 × 3 × 3)

LCM of 48, 72 and 108 = 432

So  after 432 seconds they will change simultaneously

We know that

60 seconds = 1 minute

so on dividing 432 / 60 we get 7 as quotient and 12 as reminder

Hence, 432 seconds = 7 min 12 seconds

∴ The time  = 7 a.m. + 7 minutes 12 seconds

Hence the lights change simultaneously at  7:07:12 a.m