Three bells A, B and C toll every 15 seconds, 24 seconds, and 42 seconds, respectively. If all of them toll together at 8 a.m., then between 8 a.m. and 10 a.m. on the same day, how many times will they all toll together, including the one at 8 a.m.?

Correct option is C

Given:

Toll time for bells A,B and C = 15 seconds, 24 seconds and 42 seconds

Toll together first at 8:00 am

Formula Used:

LCM is the smallest common number which is divisible by the given numbers without leaving any remainders

Solution:

LCM of 15,24 and 42:

Prime factorization of 15,24 and 42

15 = $$3 \times 5$$​​

24 = $$2^3 \times 3$$​​

42 =$$2 \times 3 \times 7$$​​

LCM = $$2^3 \times 3 \times 5 \times 7$$​=840

Hence all bells will toll together every 840 seconds =$$\frac{840}{60}$$​ =14 minutes

Total time from 8:00am to 10:00 am = 2 hours

Total minutes =$$2 \times 60$$​ = 120minutes

Total time bells will toll together =$$\frac{120}{14}= 8\frac{8}{14}$$​​

Hence total time bells will toll together including the 8:00am = 8+1 = 9