The ratio of efficiencies of workers A and B is 4:3. If A can finish a task in 18 days, in how many days will B finish the same task alone?

Correct Option: 24 days

Step-by-Step Explanation:

1. Understand the given values:
Let the efficiency of worker A be $E_A$ and the efficiency of worker B be $E_B$.
We are given that the ratio of efficiencies of A and B is 4:3.

$\frac{E_A}{E_B}=\frac{4}{3}$

This means worker A completes 4 units of work per day, while worker B completes 3 units of work per day.

2. Calculate the Total Work:
We are given that A alone can finish the task in 18 days.
Total Work = (Efficiency of A) × (Time taken by A)

$\text{Total Work}=4\times 18=72\text{units}$

3. Calculate the time taken by B alone:
Now, we need to find how many days worker B will take to complete the total work of 72 units.
Time taken by B = (Total Work) / (Efficiency of B)

$\text{Time taken by B}=\frac{72}{3}=24\text{days}$

Thus, worker B alone will finish the same task in 24 days.