A can complete a piece of work in 25 days while B can complete the same work in 30 days. They work on alternate basis, starting with A. Both A and B follow this pattern for 5 days and then A leaves the work. In how many days will B finish the remaining work?

The correct answer is $24\frac{2}{5}$.

Step 1: Find the 1-day work of each person.

A completes the work in 25 days, so A's work per day:

$\text{A's 1-day work}=\frac{1}{25}$

B completes the work in 30 days, so B's work per day:

$\text{B's 1-day work}=\frac{1}{30}$

Step 2: Determine what happens over the 5 alternating days (starting with A).

Since they work on alternate days starting with A, the schedule is:

Day 1 → A works
Day 2 → B works
Day 3 → A works
Day 4 → B works
Day 5 → A works

So in 5 days: A works 3 days and B works 2 days.

Step 3: Calculate total work done in 5 days.

$\text{Work done}=3\times \frac{1}{25}+2\times \frac{1}{30}$

$=\frac{3}{25}+\frac{2}{30}=\frac{3}{25}+\frac{1}{15}$

Finding LCM of 25 and 15, which is 75:

$=\frac{9}{75}+\frac{5}{75}=\frac{14}{75}$

Step 4: Find the remaining work after A leaves.

$\text{Remaining work}=1-\frac{14}{75}=\frac{75}{75}-\frac{14}{75}=\frac{61}{75}$

Step 5: Find how many days B alone takes to finish the remaining work.

B's 1-day work is $\frac{1}{30}$, so the number of days B needs:

$\text{Days for B}=\frac{61}{75}÷\frac{1}{30}=\frac{61}{75}\times 30=\frac{61}{\times }=\frac{1830}{75}=\frac{122}{5}$

$=24\frac{2}{5}\text{days}$

Therefore, B will finish the remaining work in $24\frac{2}{5}$ days.