Printing 9,000 Question Papers: Cost Against a Hard Deadline
A school must print 9,000 question papers for a scholarship test. Three machines are available, each with a fixed setup charge payable once and an hourly running charge. Machine A prints 750 sheets an hour, costs ₹600 to set up and ₹400 for every hour it runs. Machine B prints 500 sheets an hour, costs ₹500 to set up and ₹300 an hour. Machine C prints 250 sheets an hour, costs ₹400 to set up and ₹200 an hour. Any machine that is used must run continuously from the moment it is switched on, and when several machines run together their speeds simply add. The examination is at 3 p.m.; the papers must be printed and stacked by 8 a.m. the same morning, which leaves a working window of exactly eight hours once the building opens. The head of the printing unit says the obvious answer is to switch on all three machines, because that finishes the job fastest. The principal wants the least expensive plan that still meets the eight-hour window.
1. If all three machines run together, how long does the job take and what does it cost? Show the cost build-up per machine.
Read the worked answer
Time. Combined speed = 750 + 500 + 250 = 1,500 sheets/hour. Time = 9,000 ÷ 1,500 = 6 hours.
Cost per machine (setup charge paid once, running charge for the 6 hours each machine actually runs):
- A: ₹600 + 6 × ₹400 = 600 + 2,400 = ₹3,000
- B: ₹500 + 6 × ₹300 = 500 + 1,800 = ₹2,300
- C: ₹400 + 6 × ₹200 = 400 + 1,200 = ₹1,600
Total = ₹6,900.
This is indeed the fastest possible plan - 6 hours is the floor set by the combined speed of all three machines - but as the next task shows, fastest and cheapest are not the same thing here.
2. Find the least expensive combination of machines that still completes the job inside the eight-hour window. Compare every viable combination.
Read the worked answer
For each combination, time = 9,000 ÷ combined speed, and cost = Σ(setup + time × hourly rate).
| Machines | Speed (sheets/h) | Time | Cost | Within 8 h? |
|---|---|---|---|---|
| A only | 750 | 12.0 h | 600 + 12×400 = ₹5,400 | No |
| B only | 500 | 18.0 h | 500 + 18×300 = ₹5,900 | No |
| C only | 250 | 36.0 h | 400 + 36×200 = ₹7,600 | No |
| A + B | 1,250 | 7.2 h | (600+2,880) + (500+2,160) = ₹6,140 | Yes |
| A + C | 1,000 | 9.0 h | (600+3,600) + (400+1,800) = ₹6,400 | No |
| B + C | 750 | 12.0 h | (500+3,600) + (400+2,400) = ₹6,900 | No |
| A + B + C | 1,500 | 6.0 h | (600+2,400) + (500+1,800) + (400+1,200) = ₹6,900 | Yes |
Only A + B and A + B + C fit inside eight hours, and A + B is ₹760 cheaper.
Answer: run A and B together for 7.2 hours, costing ₹6,140. The two pairings that would have been cheaper on paper - A alone, and A + C - are excluded purely on time, and it is the deadline, not the price list, that decides the answer.
3. The head of the printing unit objects: "Machine A on its own costs only ₹5,400, so your ₹6,140 plan is not the cheapest." Respond to the objection.
Read the worked answer
The ₹5,400 figure is arithmetically correct - A alone is genuinely the cheapest way to print 9,000 sheets - but it is not a feasible answer to the question asked. Machine A needs 9,000 ÷ 750 = 12 hours, and the papers must be stacked by 8 a.m. within an eight-hour window. The plan therefore fails a hard constraint, and a plan that misses the deadline is not a cheaper plan; it is a plan that does not deliver the papers.
The objection is a classic conflation of unconstrained minimum cost with minimum cost subject to a constraint. Once the eight-hour cap is imposed, the feasible set shrinks to A + B and A + B + C, and the answer follows from that set alone.
A fair follow-up, though: how firm is the window? A + B buys the job four extra hours at ₹740 more than A alone, an effective ₹185 per hour. If overtime, security or supervision for a 12-hour overnight run costs less than ₹185 an hour, the objection becomes a genuine challenge to the constraint rather than an arithmetic error - and that is a decision for the principal, not the printing unit.