One contractor had undertaken a construction work. To complete the given job, he sought labour comprising 24 men for 12 days for working 10 hours a day. But due to festival season, as the labour had gone to their respective hometowns, he could get only 15 men and they too were willing to work 8 hours a day. With this much labour, in how many days would he be able to complete the same job?

This is a work and time problem. The amount of work done is proportional to the number of men, the number of days they work, and the number of hours they work per day. Let $M$ be the number of men, $D$ be the number of days, and $H$ be the number of hours per day. The total work can be represented as $W = M \times D \times H \times k$, where $k$ is a constant representing the work rate per man per hour. Since the same job is to be completed in both scenarios, the total work done is the same. Scenario 1: $M_1 = 24$, $D_1 = 12$, $H_1 = 10$. Work$_1 = 24 \times 12 \times 10 \times k$. Scenario 2: $M_2 = 15$, $H_2 = 8$, $D_2 = ?$ Work$_2 = 15 \times D_2 \times 8 \times k$. Equating the work done in both scenarios: $24 \times 12 \times 10 \times k = 15 \times D_2 \times 8 \times k$. Assuming $k \neq 0$, we can cancel $k$ from both sides: $24 \times 12 \times 10 = 15 \times D_2 \times 8$. $2880 = 120 \times D_2$. $D_2 = 2880 / 120$. $D_2 = 288 / 12$. $D_2 = 24$.