Interpret timetables for buses, trains and ferries, including calculating the duration of a journey and the average speed of a trip

Part (a): one mark for the duration $1$ h $30$ min ($6\!:\!42$ to $7\!:\!42$ is one hour, then $30$ minutes to $8\!:\!12$). Part (b): one mark for converting the time to hours ($1$ h $30$ min $= 1.5$ h) - markers award this conversion separately, since it is the step most often skipped - one mark for the correct setup $135 \div 1.5$, and one mark for the answer $90$ km/h with the unit. A bald answer with no time-in-hours conversion shown typically caps at the answer mark and loses the method marks.

Part (a): one mark for the wait $7\!:\!50 - 7\!:\!38 = 12$ minutes. Part (b): one mark for recognising the total runs from first departure $7\!:\!05$ to last arrival $8\!:\!35$, and one mark for the duration $1$ h $30$ min ($90$ minutes); a further mark is available for verifying with the parts ($33 + 12 + 45 = 90$). The most common error is adding only the two travelling legs ($33 + 45 = 78$ min) and omitting the wait - this loses the total-time marks.

Part (a): one mark for the duration $1$ h $15$ min, one mark for the conversion to $1.25$ h, and one mark for $50 \div 1.25 = 40$ km/h with the unit. Part (b): one mark for comparing the arrival $11\!:\!40$ with the meeting time $11\!:\!45$, and one mark for the conclusion that the passenger arrives $5$ minutes early. Markers reward a clear statement of the comparison, not just the number $5$.

Subtract the departure from the arrival. Both times are in the same hour block once you split it at the hour. From $10\!:\!35$ to $11\!:\!00$ is $25$ minutes, and from $11\!:\!00$ to $11\!:\!20$ is a further $20$ minutes:

So the journey takes $45$ minutes. (Splitting the subtraction at the whole hour avoids the slip of treating times like ordinary decimals.)

Count the whole hours first, then the extra minutes. From $14\!:\!50$ to $15\!:\!50$ is $1$ hour, and from $15\!:\!50$ to $16\!:\!25$ is a further $35$ minutes:

Check by working in minutes. Departure is $14 \times 60 + 50 = 890$ minutes after midnight and arrival is $16 \times 60 + 25 = 985$ minutes, so the gap is $985 - 890 = 95$ minutes $= 1$ h $35$ min, which agrees.

Convert the arrival to 24-hour time so the times are comparable. $1\!:\!05$ pm is $13\!:\!05$. The trip crosses noon, which is exactly why writing both times the same way prevents an error.

Subtract, splitting at the hour. From $11\!:\!40$ to $12\!:\!40$ is $1$ hour, and from $12\!:\!40$ to $13\!:\!05$ is a further $25$ minutes:

So the journey takes $1$ h $25$ min ($85$ minutes).

Part (a) - duration. From $8\!:\!10$ to $9\!:\!02$, count $1$ hour to $9\!:\!10$ would overshoot, so count to $9\!:\!00$ first: $8\!:\!10$ to $9\!:\!00$ is $50$ minutes, then $9\!:\!00$ to $9\!:\!02$ is $2$ minutes:

Part (b) - average speed. Average speed is distance divided by time, and the time must be in hours to get km/h. Convert $52$ minutes to hours:

Then

So the train averages about $60$ km/h. (Neat check: $52$ km in $52$ minutes is exactly $1$ km per minute, and $1$ km/min $\times 60 = 60$ km/h.)

Part (a) - ferry leg. From $9\!:\!20$ to $9\!:\!48$:

Part (b) - waiting time at the interchange. The ferry arrives at $9\!:\!48$ and the bus leaves at $10\!:\!05$. From $9\!:\!48$ to $10\!:\!00$ is $12$ minutes, then $10\!:\!00$ to $10\!:\!05$ is $5$ minutes:

Part (c) - total travel time. The whole journey runs from the first departure $9\!:\!20$ to the final arrival $10\!:\!53$. From $9\!:\!20$ to $10\!:\!20$ is $1$ hour, then $10\!:\!20$ to $10\!:\!53$ is $33$ minutes:

Cross-check by adding the parts. Ferry $28$ min, wait $17$ min, and the bus leg ($10\!:\!05$ to $10\!:\!53 = 48$ min): $28 + 17 + 48 = 93$ minutes, which agrees. Always measure the total from first departure to last arrival, because that single span automatically includes the wait.

Part (a) - duration. From $6\!:\!48$ to $7\!:\!48$ is $1$ hour, then $7\!:\!48$ to $8\!:\!03$ is a further $15$ minutes:

Part (b) - average speed. First write the time in hours: $1$ h $15$ min $= 1 + \tfrac{15}{60} = 1.25$ h. Then

So the train averages $120$ km/h.

Part (c) - interpretation. Average speed is the total distance divided by the total time, so it already absorbs the slow parts. While stopped, the train's actual speed was $0$ km/h, so to still cover $150$ km in $1.25$ h it must have travelled faster than $120$ km/h between stations. The figure of $120$ km/h is the single steady speed that would have produced the same trip in the same time, not the speed at any one instant. (A marker awards full marks for stating that average speed uses total distance over total time and that the running speed therefore exceeds the average wherever the train was stopped.)

Part (a) - bus leg and wait. Bus leg, $16\!:\!38$ to $17\!:\!11$: from $16\!:\!38$ to $17\!:\!00$ is $22$ minutes, plus $11$ minutes, so $22 + 11 = 33$ minutes. Wait at the interchange, $17\!:\!11$ to $17\!:\!25$: $25 - 11 = 14$ minutes.

Part (b) - total travel time. Measure from the first departure $16\!:\!38$ to the final arrival $18\!:\!40$. From $16\!:\!38$ to $18\!:\!38$ is $2$ hours, then $18\!:\!38$ to $18\!:\!40$ is $2$ minutes:

Check against the parts: bus $33$ min, wait $14$ min, train leg ($17\!:\!25$ to $18\!:\!40 = 75$ min): $33 + 14 + 75 = 122$ minutes. Agrees.

Part (c) - average speed of the train leg. The train leg lasts $75$ minutes $= \tfrac{75}{60} = 1.25$ h. Then

Part (d) - arrival check. The student arrives at $18\!:\!40$ and the exam starts at $18\!:\!45$, so they arrive

early. The student makes it with $5$ minutes to spare.