A and B can do a piece of work separately in 10 days and 8 days, respectively. If they work alternately and A begins the work, in how many days will the work be completed?
- A10
- B8
- C9
- D12
Solution & Step-by-step Explanation
Let's find the total work by taking the LCM of the time taken by A and B.
Total Work=LCM(10,8)=40 units
Now, determine their individual daily efficiencies:
Efficiency of A=
10
40
=4 units/day
Efficiency of B=
8
40
=5 units/day
They work on alternate days starting with A:
Day 1 (A): 4 units
Day 2 (B): 5 units
In a 2-day cycle, the total work completed is:
4+5=9 units
To get close to 40 units, multiply by 4 full cycles:
Work done in 4×2=8 days=4×9=36 units
Remaining work after 8 days:
40−36=4 units
On the 9
th
day, it is A's turn to work. A can do 4 units of work in a full day.
Since the remaining work is exactly 4 units, A will complete it in exactly 1 day.
Total time taken = 8+1=9 days.
Total Work=LCM(10,8)=40 units
Now, determine their individual daily efficiencies:
Efficiency of A=
10
40
=4 units/day
Efficiency of B=
8
40
=5 units/day
They work on alternate days starting with A:
Day 1 (A): 4 units
Day 2 (B): 5 units
In a 2-day cycle, the total work completed is:
4+5=9 units
To get close to 40 units, multiply by 4 full cycles:
Work done in 4×2=8 days=4×9=36 units
Remaining work after 8 days:
40−36=4 units
On the 9
th
day, it is A's turn to work. A can do 4 units of work in a full day.
Since the remaining work is exactly 4 units, A will complete it in exactly 1 day.
Total time taken = 8+1=9 days.