2024 CSAT — Q8

Use a common “work” unit of $24$ (LCM of $6,8,8$) so rates are whole numbers per hour:

total work $= 24$ units.

To finish fastest, use the strongest worker ($X$, $4$/hr) as often as possible — but $X$ cannot work two hours in a row. So alternate $X$ with a filler ($Y$ or $Z$, $3$/hr). Greedy schedule, hour by hour (cumulative units):

After $6$ hours: $21$ units done, $3$ units left. Hour $6$ was $Y$, so hour $7$ may be $X$. $X$ does $4$ units/hour, so the remaining $3$ units take $\tfrac{3}{4}$ hour $= 45$ minutes.

Total $= 6\text{ h} + 45\text{ min} = \boxed{6\text{ h }45\text{ min}}$.

Answer: (c) 6 hours 45 minutes.

Two moves crack this whole template. First, convert rates to integer units via the LCM ($24$) — fractions like $\tfrac16+\tfrac18$ are where people lose the thread; integers per hour make the schedule readable. Second, recognise it as a greedy scheduling problem: to minimise time, pack the fastest worker as densely as the constraint allows — here $X$ on every odd hour, fillers on the evens. The decisive insight the generalist misses is that the final block can be partial: once the leftover is below one worker’s hourly output, you prorate ($\tfrac{3}{4}$ h), you do not round up to a whole hour. That single point is the gap between the right answer (c) and the tempting (d). The “no two consecutive hours” clause is the entire difficulty — without it the answer would just be $24/4 = 6$ hours with $X$ only; the constraint forces one extra fractional turn.