Question

A and B can do a work in 20 days. When A works at 60% capacity, B has to work at 150% capacity to finish the work. Find in how many days the faster one will finish the work alone.

Answer 1

Approach: Both scenarios finish the same job in the same 20 days, so the team's one-day work is identical in both — equate them to get the ratio of A's and B's speeds.

Step 1: Let A and B alone finish in $a$ and $b$ days, so their one-day works are $\frac1a$ and $\frac1b$ with $\frac1a+\frac1b=\frac1{20}$.

Step 2: In the second case the team still finishes in 20 days, so $0.6\cdot\frac1a+1.5\cdot\frac1b=\frac1{20}$.

Step 3: Equating: $\frac1a+\frac1b=0.6\cdot\frac1a+1.5\cdot\frac1b\Rightarrow 0.4\cdot\frac1a=0.5\cdot\frac1b\Rightarrow \frac1a:\frac1b=5:4$.

Step 4: So the one-day works are $5t$ and $4t$ with $9t=\frac1{20}\Rightarrow t=\frac1{180}$, giving A's rate $\frac{5}{180}=\frac1{36}$.

Answer: The faster worker, A, finishes the work alone in $36$ days.