Question

Four bells toll at an interval of 8, 12 , 15 and 18 seconds respectively. All the four begin to toll together. How many times will they toll together in one hour excluding the one at the start ?

Answer 1

The nest time they toll together is given by the LCM of  8,12,15 and 18 
8= 2*2*212=2*2*315 = 3*518=2*3*3 
LCM = 2*2*2*3*3*5 = 360
so they will toll together every 360 seconds
So in hour hour they will toll 3600/360 = 10 times Answer
.

Answer 2

Answer:

The number of times they toll together excluding the one at the start = 10 times

Step-by-step explanation:

Given,

Four bells toll at an interval of 8,12,15, and 18 seconds and begin to toll together.

To find,

The number of times the four bells toll together in one hour excluding the one at the start.

Solution:

Since the four bells toll together at the beginning, then next they toll together at an interval equal to LCM of 8,12,15 and 18

The prime factorization of 8 = 2×2×2

The prime factorization of 12 = 2×2×3

The prime factorization of 15 = 3×5

The prime factorization of 18 = 2×3×3

LCM of (8,12,15,18) = 2×2×2×3×3×5 = 360

The four bells toll together in every 360seconds.

Since there are 3600 seconds in an hour,

The number of times they toll together, excluding the one at the start = $\frac{3600}{360} = 10$

The number of times they toll together excluding the one at the start = 10 times

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