Discussion

Explanation:

1 day’s work of A = 1/10 and that of B = 1/20.

Since they work on alternate days, they would work in cycles of 2 days.

Work done by (A + B) in 1 cycle of two days = 1/10 + 1/20 = 3/20.

Hence, they take $\frac{1}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}$ = $\frac{20}{3}$ = $6\frac{2}{3}$ cycles to complete the work.

Since, the no. of cycles is a fraction, we will have to analyze the last incomplete cycle.

Work done in 6 complete cycles = 6 × (3/20) = 18/20 = 9/10

Time that has passed till 6 cycles = 6 × 2 = 12 days

Hence, work left for the 7th incomplete cycle = 1 – 9/10 = 1/10

Case I: If A starts the work, he would work on 13th day and take $\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}$ = 1 day to complete the work. Hence, the work gets completed in 13 days.

Case II: If B starts the work, he would work on 13th day and would complete 1/20 of the work. Work still left = 1/10 – 1/20 = 1/20. This work would finally be completed by A on 14th day. Time taken by A = $\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$20$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}$ = 1/2 day. Hence the work gets completed in 13.5 days.

The work, at most can be completed in 13.5 days.

Hence, 13.5.

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