Five men A, B, C, D and E can do a piece of work in 20, 10, 30, 40 and 50 days respectively. A, B, and C started the work and they work for 2 days after which A and B left the work and D joins the work. D and C work another 3 days together after which C leaves and E joins the work. Find in how many days E and D completed the rest of the work.
1). 10.2 days
2). 12.64 days
3). 20.5 days
4). 11.48 days
Let the total units of work be 600
|
DAYS TO COMPLETE THE WORK |
Per day capacity |
A |
20 |
30 units |
B |
10 |
60 units |
C |
30 |
20 units |
D |
40 |
15 units |
E |
50 |
12 units |
A, B, and C started the work and they work for 2 days
⇒ Number of units produced in two days by A, B and C = (30 + 60 + 20) × 2 = 220 units
Then A and B left the work and D and C work another 3 days together
⇒ Number of units produced in 3 days by D and C = (15 + 20) × 3 = 35 × 3 = 105 units
⇒ Work left for D and E = 600 – 220 – 105 = 275 units
∴ D and E will complete rest of the task in 275/(15 + 12) days = 10.2 days