Question: A and B together can complete a piece of work in 12 days. A alone can do it in 20 days. In how many days B alone complete the work?

We can solve this problem step by step:

Given Data:

• A and B together can complete the work in 12 days.
• A alone can complete the work in 20 days.
• Find the number of days B alone would take to complete the work.

Step 1: Calculate Total Work

Assume the total work is the LCM of 12 and 20.

Total Work = LCM(12,20) = 60 units.

Step 2: Calculate Efficiencies

• Efficiency of A and B together = \frac{\text{Total Work}}{\text{Days}} = \frac{60}{12} = 5 \, \text{units/day}.
• Efficiency of A alone = \frac{\text{Total Work}}{\text{Days}} = \frac{60}{20} = 3 \, \text{units/day}.

Step 3: Find B’s Efficiency

Since A and B together complete 5 units/day, and A alone completes 3 units/day, B’s efficiency:

B’s efficiency = 5 − 3 = 2 units/day.

Step 4: Find Time Taken by B Alone

Time = \frac{\text{Total Work}}{\text{Efficiency of B}}

\text{Time} = \frac{60}{2} = 30 \, \text{days}.

Final Answer:

B alone would take 30 days to complete the work.

Similar Problem

1. A and B both together can do a work in 20 days. B alone can do it in 30 days. In how many days A alone can do it? [Ans: 60 days]
2. A and B together can do a piece of work in 7\displaystyle \frac{1}{2} days and B alone can do it in 25 days. In how many days can A alone finish the work? [Ans: 10\displaystyle \frac{5}{7} days]