Four bells tolls at an interval of 8,12,15 &18 second a respectively. How many times will they toll together in one hour excluding the one at the start?
This is the same as asking for the highest common multiple
8= 4x2
12 = 4x3
15 = 3x5
18 = 3x3x2
so we need: 2x2x2x3x3x5 = 360
they will toll together every 360 seconds
in 1 hour there are 3600 seconds
so they will toll at
360 s, at 720 s, ... at 3600 s
that would be 10
(we are supposed to exclude the toll right now)
8=4*2
12=4*3
15=3*5
18=3*3*2
HCF: 2*2*2*3*3*5
=360
1 Hour = 3600 sec
i.e no. of tolls= 3600/360
=10
Therefore, no of tolls = 10
10
To find out how many times the four bells toll together in one hour, we need to determine their common multiples up to 1 hour, excluding the first toll.
First, let's find the least common multiple (LCM) of the four intervals: 8, 12, 15, and 18 seconds.
Step 1: Find the prime factorization of each interval:
- 8 = 2^3
- 12 = 2^2 * 3
- 15 = 3 * 5
- 18 = 2 * 3^2
Step 2: Take the highest power of each prime factor present in the four intervals:
- 2^3 is the highest power of 2
- 3^2 is the highest power of 3
- 5 is the highest power of 5
Step 3: Multiply all the highest powers of the prime factors together to get the LCM:
LCM = 2^3 * 3^2 * 5 = 360
Now we know that the four bells will toll together every 360 seconds.
To calculate how many times they toll together in one hour (3600 seconds), we divide 3600 by 360:
3600 ÷ 360 = 10
Therefore, the four bells will toll together 10 times in one hour, excluding the one at the start.