Question

Four bells begin to toll together and toll respectively at intervals of 5, 6, 8 and 12 seconds. How many times will they toll together in an hour excluding the one at the start?​

Answer 1

Step-by-step explanation:

L.C.M. of 6, 5, 7, 10 and 12 is 420.

So, the bells will toll together after every 420 seconds i.e. 7 minutes.

Now, 7 x 8 = 56 and 7 x 9 = 63.

Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.

Answer 2

Answer:

The number of times the four bells toll together  = 29times

Step-by-step explanation:

Given,

Four bells begin to toll together and toll respectively at intervals of 5, 6, 8, and 12 seconds.

To find,

The number of times the bells toll together in an hour

Recall the formula

The number of terms of an AP = $\frac{t_n - t_1}{d} +1$, ------------(1)

where tₙ is the last term, t₁ is the first term and d is the common difference of the AP

Solution:

The intervals in which the four bells toll together = LCM (5,6,8,12)

The prime factorization of 5,6,8,12 are

5, 3×2, 2×2×2, 2×2×3

LCM  (5,6,8,12) =  2×2×2×3×5 =120

That is the four bells toll together every 120 seconds

That is the four bells toll together every 2 minutes

Let us  take the starting time as 0 minutes

Then the minutes in which they toll together in an hour are

2,4,6,8,........................58

This is an AP with first term 2 and common difference 2 and last term 58

The number of times the four bells toll together will be equal to the number of terms of this AP

The number of terms of the AP  = $\frac{58 - 2}{2} +1$= 29

∴The number of times the four bells toll together  = 29times

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