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?
- A3
40
days - B3
28
days - C4
37
days - D4
53
days
Solution & Step-by-step Explanation
Let the total work be the LCM of 20,30, and 60, which is 60 units.
Efficiencies of A, B, and C:
Efficiency of A =
20
60
=3 units/day
Efficiency of B =
30
60
=2 units/day
Efficiency of C =
60
60
=1 units/day
According to the alternate day cycle pattern:
Day 1: A is assisted by B ⟹Work done=A+B=3+2=5 units
Day 2: A is assisted by C ⟹Work done=A+C=3+1=4 units
In a 2-day cycle, the amount of work completed is:
Work in 2 days=5+4=9 units
To get close to the total work of 60 units, let's multiply by 6 cycles:
Work in 6×2=12 days⟹6×9=54 units
Remaining work = 60−54=6 units.
On Day 13, it is the turn of (A + B), who can complete 5 units of work.
Remaining work after Day 13=6−5=1 unit
Total time passed=13 days
On Day 14, it is the turn of (A + C), who can do 4 units of work per day.
Time required to finish the final 1 unit =
4
1
days.
Total time taken = 13+
4
1
=
4
53
days.
Efficiencies of A, B, and C:
Efficiency of A =
20
60
=3 units/day
Efficiency of B =
30
60
=2 units/day
Efficiency of C =
60
60
=1 units/day
According to the alternate day cycle pattern:
Day 1: A is assisted by B ⟹Work done=A+B=3+2=5 units
Day 2: A is assisted by C ⟹Work done=A+C=3+1=4 units
In a 2-day cycle, the amount of work completed is:
Work in 2 days=5+4=9 units
To get close to the total work of 60 units, let's multiply by 6 cycles:
Work in 6×2=12 days⟹6×9=54 units
Remaining work = 60−54=6 units.
On Day 13, it is the turn of (A + B), who can complete 5 units of work.
Remaining work after Day 13=6−5=1 unit
Total time passed=13 days
On Day 14, it is the turn of (A + C), who can do 4 units of work per day.
Time required to finish the final 1 unit =
4
1
days.
Total time taken = 13+
4
1
=
4
53
days.