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: Instead of cost-per-unit, reason directly in terms of "days of work" and check the few sensible combinations, since each man can give at most 10 days and the total job is 840 units.

Step 1: Set up. Total work $= \text{LCM}(24,21,15) = 840$ units; daily outputs are Arun $35$, Varun $40$, Tarun $56$. A day employed costs the full wage even for a partial day, so each man contributes (days worked) $\times$ (his rate) units and is paid (days worked) $\times$ (his wage).

Step 2: Tarun must be maxed out. Even at his ceiling Tarun gives only $56 \times 10 = 560 < 840$ units, so he should work all $10$ days — he is the cheapest per unit and we still fall short, so there is no reason to cut him. That fixes $560$ units done and $\text{Rs }21600$ spent, leaving $280$ units.

Step 3: Pick the helper for 280 units. Two clean choices:
Varun: $280/40 = 7$ days $\Rightarrow 7 \times 2400 = \text{Rs }16800$.
Arun: $280/35 = 8$ days $\Rightarrow 8 \times 2160 = \text{Rs }17280$.
Varun is cheaper, so use Varun.

Step 4: Add up. \[ 21600 + 16800 = \text{Rs }38400. \] Trying any split of the 280 units across both helpers only raises cost, so $\text{Rs }38400$ is the minimum. Answer: option (b).