Question

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

• A. 24 days
• B. 30 days
• C. 32 days
• D. 36 days

Hint:

Using the LCM method for total work is faster: LCM of 12 and 20 is 60 units. Combined efficiency is 5 units/day, A's efficiency is 3 units/day. Thus, B's efficiency is 2 units/day. Time taken by B = 60/2 = 30 days.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a time and work problem. We are given the time taken by two people working together and by one person working alone, and we need to find the time taken by the other person working alone.


Step 2: Key Formula or Approach:
If a person completes a task in \(x\) days, their 1 day's work rate is \(\frac{1}{x}\).
The combined work rate of A and B is the sum of their individual work rates: \(\text{Rate}(A+B) = \text{Rate}(A) + \text{Rate}(B)\).


Step 3: Detailed Explanation:
Given that A and B together can do the work in 12 days, their combined 1 day's work is:
\[ \text{Rate}(A+B) = \frac{1}{12} \] A alone can do the work in 20 days, so A's 1 day's work is:
\[ \text{Rate}(A) = \frac{1}{20} \] Let B's 1 day's work be \(\frac{1}{x}\), where \(x\) is the number of days B needs to complete the work alone.
\[ \text{Rate}(B) = \text{Rate}(A+B) - \text{Rate}(A) \] \[ \text{Rate}(B) = \frac{1}{12} - \frac{1}{20} \] To subtract the fractions, find the least common multiple (LCM) of 12 and 20, which is 60.
\[ \text{Rate}(B) = \frac{5}{60} - \frac{3}{60} = \frac{2}{60} = \frac{1}{30} \] Since B's 1 day's work is \(\frac{1}{30}\), B alone can complete the work in 30 days.


Step 4: Final Answer:
The correct choice is (B).