A can complete a task in 16 days, and B can complete the same task in 24 days. If A and B work on alternate days starting with A, in how many days will the task be completed?

Correct option is D

Given:

A can complete the task in 16 days 

B can complete the task in 24 days

A and B work on alternate days, starting with A

Solution: 

A's 1-day work = $$\frac{1}{16}$$​​

B's 1-day work =  $$\frac{1}{24}$$​​

Work in 2 days = $$\frac{1}{16} + \frac{1}{24} = \frac{3 + 2}{48} = \frac{5}{48}$$​​

After 9 cycles (18 days): $$9 \times \frac{5}{48} = \frac{45}{48}$$​​

Remaining work = $$1 - \frac{45}{48} = \frac{3}{48} = \frac{1}{16}$$​​

On the 19th day, it’s A’s turn. A does $$\frac{1}{16}$$​ work/day

So, task will be completed on day 19

Alternate Solution:

Total work (LCM of 16 and 24) = 48 units

A’s rate = $$\frac{48}{16}$$​ = 3 units/day

B’s rate =$$\frac{48}{24}$$​ = 2 units/day

Total in 2 days = 3 + 2 = 5 units

After 9 full cycles (18 days):

Total work done = 9 × 5 = 45 units

Remaining work = 48 - 45 = 3 units

Day 19: A works and completes 3 units (exactly A’s 1-day work)

Total time taken: 19 days