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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 Nth 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

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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 Nth day by tn.

⇒ tn = N(N + 1)/2

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

⇒ ∑ 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.

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