Question

At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?

• A. 24
• B. 18
• C. 12
• D. 36

Answer 1

The Correct Option is B

Step 1: Assume total work = LCM of 9 and 12 = 36 units. 

Let the daily work rate of A be \(x\) units and that of B be \(y\) units.

From the first condition: A and B together finish the work in 12 days.

\[ (x + y) \times 12 = 36 \Rightarrow x + y = 3 \quad \text{... (1)} \]

From the second condition: A works at half efficiency, B works at thrice efficiency, and they finish the work in 9 days.

\[ \left(\frac{x}{2} + 3y\right) \times 9 = 36 \Rightarrow \frac{x}{2} + 3y = 4 \quad \text{... (2)} \]

Step 2: Solve equations (1) and (2):

From equation (1): \(x = 3 - y\)

Substitute into equation (2):

\[ \frac{3 - y}{2} + 3y = 4 \] \[ \Rightarrow \frac{3}{2} - \frac{y}{2} + 3y = 4 \Rightarrow \frac{3}{2} + \frac{5y}{2} = 4 \Rightarrow \frac{5y}{2} = \frac{5}{2} \Rightarrow y = 1 \]

Then from equation (1): \(x = 3 - y = 2\)

Step 3: Calculate time taken by A alone:

\[ \text{Time} = \frac{\text{Total Work}}{\text{Work per day by A}} = \frac{36}{2} = \boxed{18 \text{ days}} \]