Charge-Sheet Typing Deadline: Breaks, Effective Speed and Rework Allowance
The record room of a Delhi Police district lines must compile a 15,000-word charge-sheet bundle before 4:00 p.m., with typing starting at 12:00 noon. Three typists are available. Typist P types 30 words per minute but, because of an existing roster, is released from other duty for only 3 hours and is allowed a 10-minute break in every 60 minutes. Typist Q types 24 words per minute, is available for the full 4 hours, but is entitled to 15 minutes of break in every hour. Typist R types 20 words per minute, is available for the full 4 hours and takes no breaks. Output is measured on finished text; none of the typists can work on another's portion and the bundle cannot be split unevenly between them. The inspector also insists that 1 word in every 25 will have to be re-typed during proof-reading, and that this rework allowance must be added to the target before anyone declares the deadline safe.
1. Using effective typing rates, calculate how many words the three typists will actually produce by 4:00 p.m. and decide whether the 15,000-word target is met.
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Convert nominal speed into effective speed, because break minutes produce no words:
- P: 30 × 50/60 = 25 wpm. Time available 3 h = 180 min → 25 × 180 = 4,500 words.
- Q: 24 × 45/60 = 18 wpm. Time available 240 min → 18 × 240 = 4,320 words.
- R: 20 wpm with no break → 20 × 240 = 4,800 words.
Total = 4,500 + 4,320 + 4,800 = 13,620 words.
Target 15,000 → shortfall 1,380 words (9.2%). The deadline is missed as the roster stands.
The trap is the shortcut 30×180 + 24×240 + 20×240 = 15,960, which 'clears' the target by pretending the 10 and 15 break minutes per hour are productive. That single error flips the decision, so the arithmetic must be done on effective minutes, not clock minutes.
2. If P, Q and R all worked without breaks (P still available for only 3 hours), what would total output be, and how many words are lost to breaks?
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Without breaks:
- P: 30 × 180 = 5,400
- Q: 24 × 240 = 5,760
- R: 20 × 240 = 4,800
- Total = 15,960 words, i.e. 960 above target.
Loss caused by breaks, measured at each typist's own speed over the window actually available:
- P: 5 wpm lost × 180 min = 900 words
- Q: 6 wpm lost × 240 min = 1,440 words
- R: 0
- Total loss = 2,340 words.
Check: 15,960 − 2,340 = 13,620, which matches the earlier figure, so the two methods are consistent.
The combined continuous rate is 30 + 24 + 20 = 74 wpm, so 15,000 words would take 15,000/74 = 202.7 min ≈ 3 h 23 min if all three were free throughout. This shows the shortfall is not caused by slow typing but by break entitlement plus P's restricted availability - a scheduling problem, not a skill problem.
3. The inspector can get only one of two changes approved: (i) Q works without breaks, or (ii) P is released from other duty for the fourth hour. Which single change is better, and is either of them sufficient once the 4% rework allowance is added?
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Rework allowance: 1 word in 25 = 4% of 15,000 = 600 words, so the working target is 15,600 words. Present capacity 13,620 → gap 1,980 words.
- Change (i), Q without breaks: Q becomes 24 wpm × 240 = 5,760, a gain of 1,440. New total 15,060. Still 540 short.
- Change (ii), P for the full 4 hours at 25 wpm = 6,000, a gain of 1,500. New total 15,120. Still 480 short.
Conclusion: neither change alone is sufficient, so approving only one still misses the deadline. Change (ii) is the marginally better of the two (+1,500 against +1,440) and, more importantly, it is the more reliable one: P's extra hour is a rostering decision that cannot be interrupted, whereas Q's 'no break' depends on Q actually forgoing rest, which no supervisor should bank on.
With both changes: 13,620 + 1,440 + 1,500 = 16,560 words, which clears 15,600 with 960 words of margin (about 13 minutes at the combined 74 wpm rate). If only one is approved, the residual gap of roughly 500 words must be covered by R doing about 25 minutes of overtime at 20 wpm, or by cutting the rework allowance through a second proof-reader.