Three boys A, B and C a]can go round a circular pathway in $3 m i n 20 s e c, 3 m i n 40 s e c a n d 4 m i n$ respectively. If they start simultaneously from the same place and at the same time, when will they all meet next?

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Solution

We need to find a number which is bigger than both of the given numbers
Step 1 : First we need to convert all the time in seconds.
$3 m i n 20 s e c = (3 \times 60) + 20 = 200 s e c 3 m i n 40 s e c = (3 \times 60) + 40 = 220 s e c 4 m i n = (4 \times 60) = 240 s e c$
Step 2 : Lets find the LCM of both the numbers
$2$ $200, 220, 240$
$2$ $100, 110, 120$
$2$ $50, 55, 60$
$5$ $25, 55, 30$
$5, 11, 6$
LCM of $200, 220$ and $240$ is $2 \times 2 \times 2 \times 5 \times 5 \times 11 \times 6 = 13200$
So they will all meet after $13200 s e c$ , converting it to minutes $\frac{13200}{60} = 220 m i n s$
Hence they all will meet next after $220 m i n s$

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$2$	$200, 220, 240$
$2$	$100, 110, 120$
$2$	$50, 55, 60$
$5$	$25, 55, 30$
	$5, 11, 6$

Three boys A, B and C a]can go round a circular pathway in 3min20sec,3min40secand4min respectively. If they start simultaneously from the same place and at the same time, when will they all meet next?

Three boys A, B and C a]can go round a circular pathway in $3 m i n 20 s e c, 3 m i n 40 s e c a n d 4 m i n$ respectively. If they start simultaneously from the same place and at the same time, when will they all meet next?