A delivery driver leaves the depot at 08:15 and drives 96 km to a warehouse, arriving at 09:45. After unloading for 20 minutes, she drives back along a diff...
Question 1 Report
A delivery driver leaves the depot at 08:15 and drives 96 km to a warehouse, arriving at 09:45. After unloading for 20 minutes, she drives back along a different route of 108 km, arriving at the depot at 11:35.
- Work out the time taken for the outward journey. (1)
- Work out the average speed for the outward journey, in km/h. (2)
- Work out the time taken for the return journey. (2)
- Work out the average speed for the whole round trip, excluding the unloading stop, giving your answer to 1 decimal place. (2)
- The driver must average at least 55 km/h overall to meet her delivery schedule. Determine, with a reason, whether she met this target. (1)
Answer Details
This question tests calculating times and average speeds for the two stages of a round trip, combining them for an overall average speed, and comparing that result with a target.
- (a) The outward journey runs from 08:15 to 09:45: \[09{:}45 - 08{:}15 = 1 \text{ hour } 30 \text{ minutes}\] [1]
- (b) Writing this as \(1.5\) hours and using speed = distance ÷ time: \[96 \div 1.5 = 64 \text{ km/h}\] [2]
- (c) Unloading for 20 minutes after arriving at 09:45 means driving resumes at 10:05. The return journey runs to 11:35: \[11{:}35 - 10{:}05 = 1 \text{ hour } 30 \text{ minutes}\] [2]
- (d) The overall average speed uses the total distance driven divided by the total time spent driving, excluding the unloading stop: total distance \(= 96 + 108 = 204\) km, total driving time \(= 1.5 + 1.5 = 3\) hours, so \[204 \div 3 = 68.0 \text{ km/h}\] (1 d.p.) [2]
- (e) Since \(68.0\) km/h is greater than the required \(55\) km/h, the driver did meet her delivery schedule. [1]
The unloading time is deliberately excluded from part (d); an average driving speed is calculated only from time spent moving, not from the total time the driver was away from the depot.
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