A, B, and C can do a piece of work in 20 days, 30 days, and 60 days, respectively,
working alone. How soon can the work be done if A is assisted by B and C, each on
every alternate day?
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. days.
LCM(20, 30, and 60) = 60
The efficiency of A = 60/20 = 3 units
The efficiency of B = 60/30 = 2 units
The efficiency of C = 60/60 = 1 unit
Day 1: A + B = 3 + 2 = 5 units
Day 2: A + C = 3 + 1 = 4 units
2 days = 9 units
↓×6 ↓× 6
12 days = 54 units
Remaining = 60 - 54 = 6 units
A + B will work 5 units, so the remaining = 1 unit
A + C will work 1 unit in = days
Total days = 12 + 1 + = 13 + = days