Recruitment Drive: Seating, OMR Packets and the 1600 m Cut-off
A district police recruitment drive has 2,500 applications for 40 constable posts. Document scrutiny rejects 8% of applications. The written test is held in three halls with 100, 80 and 60 seats, in two shifts per day; no candidate may sit in a hall beyond its capacity. Optical mark recognition (OMR) sheets are printed with an extra 5% buffer over the number of eligible candidates and are supplied in sealed packets of 50; a packet opened in one shift cannot be used in another shift. Candidates who clear the written test must run 1,600 m on a standard 400 m track within 6 minutes 40 seconds. In practice, Candidate A covers the first three laps at 4.5 m/s and the final lap at 3 m/s, with no time lost at the start. Past attendance data shows that roughly 10% of eligible candidates stay absent on the day of the written test.
1. Calculate the number of eligible candidates, the minimum number of shifts required, the number of days needed at two shifts per day, and the total OMR sheets to be printed under the 5% rule.
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Eligible candidates 8% of 2,500 = 200 rejected. Eligible = 2,500 − 200 = 2,300.
Seats per shift 100 + 80 + 60 = 240 candidates per shift.
Shifts required 2,300 ÷ 240 = 9.58 → must round up, because a partially filled shift still occupies a hall and an invigilator team. Minimum shifts = 10.
Days required 10 shifts ÷ 2 shifts per day = 5 days.
OMR sheets under the 5% rule Buffer is computed on eligible candidates, not on hall capacity: 2,300 × 1.05 = 2,415. Sheets come in packets of 50, so round up to the next multiple of 50: 2,415 → 2,450 sheets = 49 packets.
Why rounding matters here Rounding down to 2,400 would breach the buffer; rounding up to 2,450 costs 50 extra sheets but keeps the audit clean. Note also that the last shift carries only 2,300 − (9 × 240) = 140 candidates, so the halls will be substantially empty on day 5 - a point to raise with the Superintendent, since one more hall of 100 seats would compress the schedule to 9 shifts.
2. How would you allot OMR packets shift by shift so the 5% buffer is honoured as far as possible without opening extra packets? State the packet count and the residual risk.
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The trap: applying 5% to each shift's capacity (240 × 1.05 = 252 → 6 packets = 300 sheets) gives 10 × 6 = 60 packets = 3,000 sheets, which is 600 sheets more than the sanctioned 2,450. That is not a buffer, it is waste.
Working allotment (49 packets = 2,450 sheets):
| Shift | Candidates | Packets issued | Sheets | Buffer on the shift |
|---|---|---|---|---|
| 1–9 | 240 each | 5 each = 45 | 2,250 | 10 sheets per shift (4.2%) |
| 10 | 140 | 3 | 150 | 10 sheets (7.1%) |
| Reserve | - | 1 | 50 | held centrally |
| Total | 2,300 | 49 | 2,450 | - |
Reasoning for issuing 5 packets, not 6: 250 sheets cover 240 candidates plus 10 spares. The 5% requirement is a document-level requirement (2,415 of 2,450), which this allotment satisfies overall; insisting on 5% inside every single shift would triple the surplus.
Residual risk and mitigation:
- Per-shift buffer is only 10 sheets. A torn or misprinted sheet in a shift of 240 leaves almost no slack.
- Mitigation: hold one unopened reserve packet with the Centre Observer, issue it only against a written log entry (shift, count, reason), and never open a packet in anticipation.
- The 10% absenteeism worsens the arithmetic: actual attendance ≈ 2,300 × 0.9 = 2,070, so about 430 printed sheets go unused. That is a planning cost, not an error - but it is the argument for printing shift-wise in future rather than one bulk lot.
3. Does Candidate A qualify in the 1,600 m run? Show the timing, state the margin, and explain what uniform speed is needed.
Read the worked answer
Convert the cut-off first. 6 minutes 40 seconds = (6 × 60) + 40 = 400 seconds.
Candidate A's split times
- Track = 400 m, so 1,600 m = 4 laps.
- First 3 laps = 1,200 m at 4.5 m/s: 1,200 ÷ 4.5 = 266.67 s.
- Final lap = 400 m at 3.0 m/s: 400 ÷ 3 = 133.33 s.
- Total = 266.67 + 133.33 = 400.00 s = 6:40 exactly.
Verdict: qualifies only on the boundary - margin = 0 seconds.
This is the judgement point the question is testing. A zero margin is not a comfortable pass:
- Timing is taken by hand at many centres, and reaction/rounding error of half a second flips the result.
- The slow final lap is the structural problem: dropping from 4.5 m/s to 3.0 m/s means the athlete is slowing when everyone else kicks.
Required uniform speed 1600 ÷ 400 = 4.0 m/s exactly. Any constant pace above 4.0 m/s clears the bar.
Practical recommendation: target 6:25–6:30 (385–390 s), i.e. about 4.10–4.16 m/s, which gives a 10–15 second cushion. A better race plan is an even or slightly negative split - for example 4.1 m/s for all four laps (≈ 390 s) rather than banking time early and losing it on the last lap.