Let the total amount of work be $W$. Let the efficiency (work per day) of B be $b$. Since A is 4 times more efficient than B, the efficiency of A is $4b$.

When A and B work together, their combined efficiency is: $b + 4b = 5b$ It is given that together, A and B complete the work in 20 days. So, in 1 day, A and B together complete $\frac{1}{20}$ of the total work: $5b = \frac{1}{20}$ Solving for $b$: $b = \frac{1}{20 \times 5} = \frac{1}{100}$ Thus, B’s efficiency is $\frac{1}{100}$, meaning B completes $\frac{1}{100}$ of the work per day.

Therefore, B alone will complete the entire work in: $\text{Number of days} = \frac{1}{\frac{1}{100}} = 100 \text{ days}$

So, B will complete the work in 100 days.

Would you like more details on any step?

Here are 5 related questions for practice:

1. If A completes the work alone, how many days will it take?
2. If B’s efficiency were doubled, how many days would A and B together take to complete the work?
3. How many days will it take if A works for 10 days alone, and then B completes the rest of the work?
4. If A and B together complete 50% of the work in 10 days, how many days would it take them to complete the remaining 50%?
5. What fraction of the total work is done by A in one day?

Tip: When dealing with work-related problems, always consider defining the total work as 1 unit for simplicity.