Fog Delay on the Moga–Chandigarh Bus: A Quadratic Model
A private bus runs daily between Moga and Chandigarh, a distance of 240 km along the highway. On a clear day the bus keeps a steady average speed and completes the run in a whole number of hours. During the winter fog the driver must slow down by 8 km/h, and on one such day the same 240 km run took exactly one hour longer than usual. The bus is scheduled to leave at 6:00 a.m., and the operator promises passengers an 11:00 a.m. arrival. Diesel costs ₹89.50 per litre, the bus covers 4 km per litre, and the operator has budgeted ₹5,000 for fuel on each single trip. Before reprinting the timetable, the manager wants two claims checked: the driver says that leaving 45 minutes earlier will be enough to beat the fog delay, and the accountant says the fog raises the fuel bill.
1. Form a quadratic equation and find the usual speed, the fog-day speed, and the two journey times.
Read the worked answer
Let the usual average speed be x km/h. Then the fog-day speed is (x − 8) km/h.
Usual time = 240/x hours; fog-day time = 240/(x − 8) hours.
Fog day takes 1 hour more: 240/(x − 8) − 240/x = 1
Multiply both sides by x(x − 8): 240x − 240(x − 8) = x(x − 8) 1920 = x² − 8x x² − 8x − 1920 = 0
Solve: x = [8 ± √(64 + 4 × 1920)]/2 = [8 ± √(64 + 7680)]/2 = [8 ± √7744]/2 = (8 ± 88)/2
x = 48 or x = −40.
The root x = −40 satisfies the algebra but is rejected: a speed must be positive, and the problem also requires x − 8 > 0 because the bus slows down.
So usual speed = 48 km/h, usual time = 240/48 = 5 hours. Fog-day speed = 40 km/h, fog-day time = 240/40 = 6 hours.
Check: 6 − 5 = 1 hour ✓ (factorising gives (x − 48)(x + 40) = x² − 8x − 1920 ✓).
2. Is the driver right that leaving 45 minutes earlier will beat the fog delay? Give the actual arrival time under that plan.
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The fog costs a full extra 60 minutes (6 h instead of 5 h). Leaving 45 minutes earlier recovers only 45 of those 60 minutes.
Departure would be 6:00 a.m. − 45 min = 5:15 a.m. Arrival = 5:15 a.m. + 6 h = 11:15 a.m.
The promised arrival is 11:00 a.m., so the bus is still 15 minutes late. The driver's claim is nearly right but not right: he has under-estimated the delay by a quarter of an hour.
To keep the 11:00 a.m. promise on fog days, the bus must leave at 11:00 a.m. − 6 h = 5:00 a.m., a full hour earlier, or the timetable must be reprinted with an 11:00 a.m.-plus-1-hour winter arrival.
3. Compute the fuel cost of one trip and judge both the ₹5,000 budget and the accountant's claim that fog raises the fuel bill.
Read the worked answer
Distance = 240 km; mileage = 4 km per litre. Diesel used = 240 ÷ 4 = 60 litres Cost = 60 × 89.50 = ₹5,370
Budget check: ₹5,370 − ₹5,000 = ₹370 over budget (about 7.4% short). The ₹5,000 figure is unrealistic; at ₹89.50 per litre it would buy only 5000 ÷ 89.50 ≈ 55.9 litres, i.e. about 223 km at 4 km/L. The operator must budget ₹5,370 per trip (or find a mileage of 240 ÷ 55.9 ≈ 4.3 km/L).
Accountant's claim: on this model the bus burns fuel per kilometre, not per hour, and the route is still 240 km on a foggy day - so the fuel used and the fuel cost are the same ₹5,370 on both days. The fog costs time, not fuel. The claim would only become true if the fog forced a longer diversion route or extra idling, neither of which is stated.