Corey can finish a job in $4$ hours alone. When Corey and Casey work together, they finish the same job in $2$ hours. How long would Casey need to finish the job alone?

Let $t$ be Casey's solo time. Their hourly rates add to the combined rate: $\tfrac{1}{4} + \tfrac{1}{t} = \tfrac{1}{2}$. Isolate $\tfrac{1}{t}$: $\tfrac{1}{t} = \tfrac{1}{2} - \tfrac{1}{4} = \tfrac{2}{4} - \tfrac{1}{4} = \tfrac{1}{4}$. So $t = 4$ hours. Check: with both contributing $\tfrac{1}{4}$ per hour, together they do $\tfrac{2}{4} = \tfrac{1}{2}$ per hour, finishing in $2$ hours ✓. - Trap B ($2$) confuses Casey's solo time with the joint time. - Trap A ($1$ hour) would make Casey faster than the **combined** speed — impossible. - Trap D ($8$) gives a combined rate $1/4 + 1/8 = 3/8$ per hour, i.e. about $2.67$ hours together — wrong. Surprising result: Casey works at exactly the same pace as Corey ($4$ hours each). When two workers of equal speed pair up, the joint time is half — which is exactly what the problem describes.