Diptesh and Tushar working separately can do a piece of work in 9 and 12 days respectively. If they work for a day alternately, Diptesh beginning, in how many days, the work will be completed?

Given:

Diptesh's rate = 1/9 work/day

Tushar's rate = 1/12 work/day

Formula:

Combined work in two days = Diptesh's 1 day work + Tushar's 1 day work

Calculation:

⇒ The combined work in two days: 1/9 + 1/12

⇒ LCM of 9 and 12 = 36

⇒ (4/36 + 3/36) = 7/36 work in 2 days

⇒ Full cycles to finish the work = Total work / Work done in two days = 1 / (7/36) = 36/7

⇒ Days taken for full cycles = floor(36/7) × 2 = 5 × 2 = 10 days

⇒ Remaining work after 5 cycles: 1 - (35/36) = 1/36

⇒ On the 11th day, Diptesh works, completing more than the remaining work as he can do 1/9 per day.

So, time taken by Diptesh to complete the remaining work = (1/36)/(1/9) = 1/4 days

Thus, total days required = 10 + 1/4 = 41/4 days

Hence, option 4 is the correct answer.

Alternate Metho

Total amount of work = LCM of 9 and 12

⇒ Total amount of work = 3 × 3 × 4 = 36 units

⇒ Work done by Diptesh per day = \(\frac{36}{9}\) = 4 units

⇒ Work done by Tushar per day = \(\frac{36}{12}\) = 3 units

Starting the work alternatively with Diptesh,

First day work = 4 units

Second day work = 3 units and so on...

After 10 days, Total amount of work done = 4 × 5 + 3 × 5 = 35 units

Unit of work left = 1 unit

On the 11th day, Diptesh will work,

So, Time taken by Diptesh to complete 1 unit = \(\frac{1}{4}\)

Total work finished in working alternatively = 10\(\frac{1}{4}\) days

Therefore, working alternately starting with Diptesh, the work will be finished in 10\(\frac{1}{4}\) days.