Question

Two neon lights are turned on at the same time . One blinks every 4 seconds and the other blinks every 6 seconds.When will they blink together

Answer 1

Answer:

At 24th second

Step-by-step explanation:

The first one will blink at 4th second then at 4+4=8th second then +4 +4... respectively

While the second one will blink at 6th second then 6+6=12th then +6 +6 ... respectively

They will blink together at 24 second because it is the first number where multiples of 6 and 4 coincide.

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Answer 2

Delay time of first neon, $\delta_1 = 4 s$

Delay time of second neon, $\delta_2 = 6 s$

Time after n-th blinking of first neon, $t_1(n)=\delta_1\times n$

Time after k-th blinking of second neon, $t_2(k)=\delta_2\times k$

Assuming that they will, at some point, blink together,

$t_1(n)=t_2(k)$

$\implies 4n = 6k \:\:\: \forall n,k\in \mathbb{N}\\\implies 2n = 3k\\\implies n = \frac{3}{2}k$

Since $n\in\mathbb{N}$,

$\frac{3}{2}k\in\mathbb{N}$

$\implies k=2t,\:\:\:t\in\mathbb{N}$

$\implies n=3t$

$\therefore\:the\:set\:of\:solutions\:are,\:S=\{(2t,3t)\mid t\in \mathbb{N}\}\\= \{(2, 3), (4,12),(6,9),(8,12),...\}$

So, It is sufficient to say that they blink together every time the number of blinks of the 1st neon is odd.

Now, the measure of time every time the blinking of the 1st neon is even is,

$t_1(2t)=\delta_1\times3t=12t$

∴They will blink together after every multiples of 12 i.e., 12 s, 24 s, 48 s, etc.