​The team of A, B, and C was supposed to complete a piece of work. Working alone, A could have completed the same work in 190 hours, B could have completed the same work in 285 hours, while C could have completed it in 342 hours. The trio decided that only one of the three would work at any point of time. In the first cycle of three days A worked for two hours on Day 1, B worked for two hours on Day 2, and C worked for two hours on Day 3. In the second cycle of three days A worked for four hours on Day 1, B worked for four hours on Day 2, and C worked for four hours on Day 3. In the third cycle of three days A worked for six hours on Day 1, B worked for six hours on Day 2, and C worked for six hours on Day 3. This continued till the piece of work was completed, with each working for 2n hours in Cycle n, the sole exception being made, if necessary, in the last cycle, as the work may get completed mid-cycle. What was the total number of days needed for the work to be completed?​

Correct option is A

Given:

A completes the work in 190 hours.

B completes the work in 285 hours.

C completes the work in 342 hours.

Work schedule:

Cycle 1 (Days 1-3): A works 2 hours, B works 2 hours, C works 2 hours.

Cycle 2 (Days 4-6): A works 4 hours, B works 4 hours, C works 4 hours.

Cycle n (Days 3n-2 to 3n): Each works 2n2n hours in their respective days.

The work may complete mid-cycle.

Formula Used:

Total work = Efficiency × time

Solution:

Total work = LCM of 190, 285, and 342 = 1710

A's efficiency =$$\frac{1710}{190}$$​ = 9 units/hour

B's efficiency = $$\frac{1710}{285}$$​ = 6 units/hour

C's efficiency = $$\frac{1710}{342}$$​ = 5 units/hour

In Cycle n, each works 2n hours on their day:

Day 1 (A) = $$2n \times 9$$​ = 18n units

Day 2 (B) = $$2n \times 6$$​ = 12n units

Day 3 (C) = $$2n \times 5$$​= 10n units

Total per cycle = 18n + 12n + 10n = 40n units

Total number of cycles:

Sum of work over cycles until it reaches or exceeds 1710 units:

​$$\sum_{n=1}^{k} 40n = 40 \times \frac{k(k+1)}{2} \geq 1710$$​​

​$$20k(k+1) \geq 1710 \implies k(k+1) \geq 85.5$$​​

Solving $$k^2 + k - 85.5 = 0$$​ :

​$$k = \frac{-1 \pm \sqrt{1 + 342}}{2} = \frac{-1 \pm \sqrt{343}}{2} \approx \frac{-1 \pm 18.52}{2}$$​​

Positive root: $$k \approx 8.76$$​ 

Thus, 8 full cycles complete.

Work after 8 cycles (24 days):

$$20 \times 8 \times 9$$​ = 1440  units

Remaining work = 1710 - 1440 = 270 units.

Work in 9th Cycle :

In Cycle 9, each works $$2 \times 9$$​ = 18 hours/day:

Day 25 (A) = $$18 \times 9$$​ = 162 units

Remaining after Day 25: 270 - 162 = 108 units.

Day 26 (B) = $$18 \times 6$$​ = 108 units

Thus, The work is completed in 26 days.