त्योहार ड्यूटी: रिज़र्व, गेट-वार तैनाती और समय-गणना
For a festival, 240 constables are available for three gates, A, B and C. The departmental norm is that 10 per cent of the total strength must first be kept as a mobile reserve, and only the remaining force is divided among the gates in the ratio 5 : 4 : 3. Gate A additionally gets 12 home guards, who are not part of the 240. A patrol pair must cover the 3 km road between gates B and C. One constable starts from B at 4 km/h and another from C at 5 km/h, both at the same time, walking towards each other. Over the first five days, the beat issued 42, 38, 50, 46 and 54 challans against a daily target of 50. The inspector wants the deployment figures, the meeting time of the patrol pair and the challan position before the next shift briefing.
1. Calculate the reserve, the gate-wise deployment and Gate A's actual total. Does the 5 : 4 : 3 ratio survive the reserve deduction?
Read the worked answer
Reserve = 10 per cent of 240 = 24 constables. Force left for the gates = 240 minus 24 = 216. Ratio 5 : 4 : 3, so total parts = 5 + 4 + 3 = 12. One part = 216 divided by 12 = 18 constables. Gate A = 5 times 18 = 90. Gate B = 4 times 18 = 72. Gate C = 3 times 18 = 54. Check: 90 + 72 + 54 = 216, and 216 + 24 = 240. Gate A's actual strength on the ground = 90 + 12 home guards = 102.
Yes, the ratio among the three gates is still 5 : 4 : 3, because the reserve was deducted before the division. The home guards must not be counted inside the 240, otherwise the reserve falls below 10 per cent. Reporting 'A has 102' without saying that 12 are home guards would make the reserve and gate figures look inconsistent.
2. When and where will the two constables meet, and how far is that point from gate B?
Read the worked answer
They walk towards each other, so their relative speed is 4 + 5 = 9 km/h. The gap is 3 km. Time = distance divided by relative speed = 3 divided by 9 = 1/3 hour = 20 minutes. Distance covered from gate B = 4 km/h times 1/3 hour = 4/3 km, that is 1 km 333 m. Check from C: 5 times 1/3 = 5/3 km, about 1 km 667 m, and 1.333 + 1.667 = 3 km.
So the pair meets 20 minutes after starting, about 1.33 km from B and 1.67 km from C. A common mistake is to add the times or to use the average speed 4.5 km/h; the correct method is relative speed for opposite directions. The same calculation tells the control room exactly which stretch to keep clear.
3. What is the average daily performance over the five days, and what daily figure is needed over the next five days to reach an overall average of 50?
Read the worked answer
Total challans = 42 + 38 + 50 + 46 + 54 = 230. Average = 230 divided by 5 = 46 per day. Against the target of 50, the shortfall is 4 per day, that is 4/50 = 8 per cent below target.
For an overall average of 50 over 10 days, the total must be 500. Already achieved = 230, so the next five days must yield 500 minus 230 = 270. Required daily average = 270 divided by 5 = 54.
So the beat must improve from 46 to 54 per day, not merely to 50, because the earlier shortfall has to be recovered. Reporting only 'we are 8 per cent short' hides the size of the correction needed; the inspector needs the figure 54.