Question

Arun, Varun and Tarun, if working alone, can complete a task in 24, 21, and 15 days, respectively. They charge Rs 2160, Rs 2400, and Rs 2160 per day, respectively, even if they are employed for a partial day. On any given day, any of the workers may or may not be employed to work. If the task needs to be completed in 10 days or less, then the minimum possible amount, in rupees, required to be paid for the entire task is?

• A. \(34400\)
• B. \(38400\)
• C. \(47040\)
• D. \(38880\)

Hint:

In mixed worker problems with a time limit and different wages, first compute \emph{cost per unit of work}. Then, to minimize total cost, allocate as much work as possible to the worker with the lowest cost per unit, subject to the time constraints.

Answer 1

The Correct Option is B

Approach: Cost per day is fixed, but a partial day still costs a full day's wage, so the only thing that matters is how many days each man works. Compare cost-per-unit-of-work, load the cheapest worker fully, and let the deadline (10 days) decide the rest.

Step 1: Fix total work and find daily rates. Take total work \(= \text{LCM}(24,21,15) = 840\) units.
Arun \(= 840/24 = 35\) units/day, Varun \(= 840/21 = 40\) units/day, Tarun \(= 840/15 = 56\) units/day.

Step 2: Compare cost per unit.
Tarun \(= 2160/56 \approx 38.6\), Varun \(= 2400/40 = 60\), Arun \(= 2160/35 \approx 61.7\) Rs/unit.
So Tarun is by far the cheapest, then Varun, then Arun.

Step 3: Use Tarun to the limit. In the allowed 10 days, Tarun can do at most \(56 \times 10 = 560\) units at a cost of \(10 \times 2160 = \text{Rs }21600\). That still leaves \(840 - 560 = 280\) units.

Step 4: Clear the remaining 280 units with the next-cheapest, Varun. Varun needs \(280/40 = 7\) days (\(\le 10\), so deadline is safe), costing \(7 \times 2400 = \text{Rs }16800\). Arun (most expensive) is never needed.

Intuition: Because a touched day always costs a full wage, you never want a costly worker doing odd leftover units — you push the cheapest worker to his 10-day ceiling first.

Step 5: Total. \[ 21600 + 16800 = \text{Rs }38400. \] The minimum amount is \(\boxed{38400}\), matching option (b).

Answer 2

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:

• Use Tarun as much as possible (most cost-efficient),
• Then use Varun if needed,
• Use Arun only if absolutely necessary.

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 \(38400\).