Question : A, B, and C can do a piece of work in 30 days, 40 days, and 50 days, respectively. Beginning with A, if A, B, and C do the work alternatively then in how many days will the work be finished?
Option 1: $38 \frac{1}{12}$
Option 2: $36 \frac{1}{2}$
Option 3: $36$
Option 4: $39 \frac{1}{12}$
Correct Answer: $38 \frac{1}{12}$
Solution :
A, B, and C can do a piece of work in 30 days, 40 days, and 50 days, respectively.
Beginning with A, they do the work alternatively.
Let the total work = LCM of 30, 40, and 50 = 600 units.
A's 1 day's work = $\frac{600}{30}$ = 20 units.
B's 1 day's work = $\frac{600}{40}$ = 15 units.
C's 1 day's work = $\frac{600}{50}$ = 12 units.
If they work on alternate days they will finish (20 + 15 + 12) = 47 units of work in 3 days,
In (12 × 3) = 36 days, they will finish (47 × 12) = 564 units of work,
In the next 2 days, A and B will finish (20 + 15) = 35 more units of work.
So, (564 + 35) = 599 units of work is done.
We have remaining (600 – 599) = 1 unit of work, but C can finish 12 units in a day.
So, C would be able to finish 1 unit in $\frac{1}{12}$ day itself.
Therefore, the work will be completed in $(36+2+\frac{1}{12})=38\frac{1}{12}$ days.
Hence, the correct answer is $38\frac{1}{12}$.
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