4 men and 4 boys can do a piece of work in 3 days, while 2 men and 5 boys canfinish it in 4 days. How long would it take 1 man alone to do it?

Suppose 1 man alone can finish it in x days and 1 boy alone can finish it in y days. Then,
1 man's 1 day's work = 1/x.
1 boy's 1 day's work = 1/y.
Therefore 4 × (1/x) + 4 × (1/y) = 1/3 ⇒ … (i)
and 2 × (1/x) + 5 × (1/y) = 1/4 ⇒ … (ii)
Putting (1/x) = u and (1/y) = v, these equations become:
u + v = (1/12) … (iii)
2u + 5v = ¼ … (iv)
Multiplying (iii) by 2, we get:
2u + 2v = (1/6) … (v)
Subtracting (v) from (iv), we get :
3v = 1/12

⇒ v = 1/36
Substituting v = 1/36 in (iii), we get:
u + (1/36) = (1/12) ⇒ u = .
Now, u = (1/18)

⇒ 1/x = 1/18

⇒ x = 18.
v = 1/36

⇒ 1/y = 1/36

⇒ y = 36.
Therefore 1 man alone can finish the work in 18 days.