A, B, & C can do same job in 8, 24, & 48 hours respectively working

In one hour, A does 1/8 of work, B does 1/24 of work and C does 1/48 of work. Since the question states that the machines work on every alternative hour, eg A for the first hour, B for the second hour, and C for the third hour. For any multiples of 3, all 3 machines perform equal hours. In case the hours are 3n+1 ie 1 more than the multiple of 3, the one hour can be added to A, B or C. in case of 3n+2 hours, the additional 2 hours are added to pair of machines ie both to A and B or B and C etc.
For the minimum number of hours taken by all 3, most of the time should be assigned to the machine that does the most work. 15 is thus rejected, since for within 15 hours, all the 3 machines will have to work for 5 hours. Considering 15.5 hours, the number of hours can be distribution in the below fashion, 5 complete cycle of A-B-C and an extra 0.5 to A.
5.5/8 +5/24 + 5/48 = 48/48 hence this is the minimum number of hours.
For maximum number of hours, most of the time should be assigned to the machine that does the least amount of work. Because 17 is a whole number I will choose that. 5 cycles of C-B-A and additional 1 hour for machine C and 1 hour for machine B. 6/48+ 6/24 + 5/24= 48/48- this is the maximum number of hours­

Lets take the total work done as the LCM of 8,24,48 = 48
So the total work is 48 units
Individual work in units :
A = 48/8 = 6 units
B = 48/24 = 2 units
C = 48/48 = 1 unit
Total work in 3 hours is 9 units
Now 48/9 = 5.xx X 3 = 15.xx
remaining units = 3 units
if A + B work together in 2 hours we can complete the 3 units work.
therefore maximum will be 17 hours

if we want the minimum time to complete those 3 units, we can ask A
A will complete 3 units in half an hour
So, minimum will be 15.5 hours