Step 1: Determine work rates and cost efficiency.
Assume the total work is the LCM of their individual times: \[ \text{LCM}(24, 21, 15) = 840 \text{ units.} \] Then their individual rates (in units/day) are:
\[ \text{Arun's rate} = \frac{840}{24} = 35 \text{ units/day}, \] \[ \text{Varun's rate} = \frac{840}{21} = 40 \text{ units/day}, \] \[ \text{Tarun's rate} = \frac{840}{15} = 56 \text{ units/day}. \] Their daily charges are:
\[ \text{Arun: } \text{Rs } 2160 \text{ per day}, \] \[ \text{Varun: } \text{Rs } 2400 \text{ per day}, \] \[ \text{Tarun: } \text{Rs } 2160 \text{ per day}. \] Cost per unit of work:
\[ \text{Arun: } \frac{2160}{35} \approx 61.7 \text{ Rs/unit}, \] \[ \text{Varun: } \frac{2400}{40} = 60 \text{ Rs/unit}, \] \[ \text{Tarun: } \frac{2160}{56} \approx 38.6 \text{ Rs/unit}. \] Thus, Tarun is the cheapest per unit, then Varun, then Arun.
Step 2: Strategy to minimize cost.
We must finish 840 units of work in at most 10 days. To minimize cost:
Step 3: Assign work to Tarun.
Tarun can work at most 10 days (due to the time limit). Work done by Tarun in 10 days: \[ 56 \times 10 = 560 \text{ units.} \] Cost for Tarun: \[ 10 \times 2160 = \text{Rs } 21600. \] Remaining work: \[ 840 - 560 = 280 \text{ units.} \]
Step 4: Assign remaining work to Varun.
Varun's rate is 40 units/day. Days required by Varun to complete 280 units: \[ \frac{280}{40} = 7 \text{ days.} \] This is within the 10-day limit (they can work in parallel or sequentially within 10 days, since the constraint is on total project duration, not individual employment days). Cost for Varun: \[ 7 \times 2400 = \text{Rs } 16800. \] Remaining work: \(0\). Arun is not needed.
Step 5: Total minimum cost. \[ \text{Total Cost} = 21600 + 16800 = \text{Rs } 38400. \] Hence, the minimum possible amount required is Rs \(38400\).