Correct Option: D
| (A + B)’s 1 day’s work = | 1 | |
12 |
| (B + C)’s 1 day’s work = | 1 | |
15 |
| (C + A)’s 1 day’s work = | 1 | |
20 |
On adding ,
| 2 (A + B + C)’s 1 day's work = | 1 | + | 1 | + | 1 |
12 | 15 | 20 |
| 2 (A + B + C)’s 1 day's work = | 5 + 4 + 3 | = | 1 | |
60 | 5 |
| ∴ (A+B+C)’s 1 day’s work = | 1 | |
10 |
| ∴ B’s 1 day’s work = | 1 | - | 1 | |
10 | 20 |
| B’s 1 day’s work = | 2 -1 | = | 1 | |
20 | 20 |
∴ B alone can do the work in 20 days.
Second method to solve this question ,Here , x = 12 , y = 15 , z = 20
| Time taken = | 2xyz | |
-xy + yz + zx |
| B alone can do in = | 2 × 12 × 15 × 20 | |
- 12 × 15 + 15 × 20 + 20 × 12 |
| B alone can do in = | 24 × 300 | |
- 180 + 300 + 240 |
| B alone can do in = | 24 × 300 | = 20 days. |
360 |
