Question

Twenty two person working together can do a certain job in 70 days. If on the first day, one man starts the work; on the second day, two men join him; on the third day, three men join and so on, with exactly N men joining the work on the N^th day. Then, find the number of days in which the work would be completed?
1. 20
2. 22
3. 24
4. 25
5. None of these

Answer 1

Correct Answer - Option 1 : 20

The total work (in man days) is (22 × 70) = 1540 man days.

Days	Number of persons	Work done in man days
1	1	1
2	2	1 + 2
3	3	1 + 2 + 3
4	4	1 + 2 + 3 + 4
----	----	-----
N	N	(1 + 2 + 3 + 4 + ----- + N)

 

Therefore, on the Nth day the work done is (1 + 2 + 3 + ---- + N) = N(N + 1)/2

If we denote the work done on the N^th day by tn.

⇒ tn = N(N + 1)/2

∴ Total work done in N days = t₁ + t₂ + t₃ + ----- t_N

⇒ ∑ N(N + 1)/2

⇒ ∑ N²/2 + ∑N/2

⇒ N (N + 1) (2N + 1)/12 + N (N + 1)/4

⇒ N (N + 1) [(2N + 1 + 3)/12

⇒ N (N + 1) (N + 2)/6

We need to find a value of N, such that N (N + 1) (N + 1)/6 ≥ 1540

⇒ N (N + 1) (N + 2) ≥ (6 × 1540)

⇒ N (N + 1) (N + 2) ≥ 6 × 10 × 7 × 22

⇒ N (N + 1) (N + 2) ≥ (20 × 21 × 22)

∴ N = 20

Therefore, the work gets completed in 20 days.