Let the number of days required to complete the job be n.
On day 1, one person works, on day 2, two people work, on day 3, three people work, and so on, up to n people on day n.
Each person has the same efficiency.
Work \(=1\left(\frac{1}{120}\right)+2\left(\frac{1}{120}\right)+3\left(\frac{1}{120}\right)........+n\left(\frac{1}{120}\right)\)
This is equivalent to 1.
\(⇒\frac{1}{120}+\frac{2}{120}+\frac{3}{120}.........+\frac{n}{120}=1\)
\(\Sigma\ n=120\)
\(⇒ n=15\)