Determining the minimum time required for alternating individual work schedules

Set the total work as the LCM of individual times: LCM(6, 8, 8) = 24 units. Calculate individual hourly rates: X = 24 / 6 = 4 units/hour. Y = 24 / 8 = 3 units/hour. Z = 24 / 8 = 3 units/hour.

To minimize total duration, we must maximize the use of our most efficient worker (X). However, X cannot work consecutive hours. The optimal alternation pattern to maximize X is: X, Y, X, Z, X, Y, X \dots

Let's compute work accumulated hour by hour: Hour 1 (X): 4 units (Total = 4) Hour 2 (Y): 3 units (Total = 7) Hour 3 (X): 4 units (Total = 11) Hour 4 (Z): 3 units (Total = 14) Hour 5 (X): 4 units (Total = 18) Hour 6 (Y): 3 units (Total = 21)

At the end of 6 hours, 21 units are completed. Remaining work = 24 - 21 = 3 units. In the 7th hour, it is X's turn again. X can do 4 units per hour. Time required for the final 3 units = 3/4 hour = 45 minutes. Total time = 6 hours 45 minutes.

Self-Correction Check: Wait, look at option choices. If we change order to alternate X and Y/Z more aggressively or choose a different scheduling block: can we finish faster? Let's check if Z can work instead of Y on hour 6: it doesn't change the units. The layout is solid.

Let's re-verify the pattern. Can we do X, Y, Z, X, Y, Z? That wastes X. The pattern X, Y, X, Z, X, Y, X uses X four times in 7 hours. Total work in 6 hours = 21. 7th hour needs 3 units. Since X works the 7th hour, time is 3/4 hr = 45 mins.

Let's double-check if we can start with Y, X, Z, X, Y, X, Z - that delays X. So 6 hours 45 mins is the calculated minimum.

Let's re-verify if we can pool inputs differently. Hour 1: X(4), Hour 2: Y(3), Hour 3: X(4), Hour 4: Z(3), Hour 5: X(4), Hour 6: Y(3). Sum = 21. Remaining = 3. Next is X. 3/4 of an hour = 45 mins. Total = 6 hr 45 min.

Answer: (c).