Planning a school field trip: choosing a route on a 1:50 000 map
A Grade 7 class is planning a trip to a farm outside their growth point. On the 1:50 000 map the school gate is at grid reference 452318 and the farm gate at 489296. The straight-line distance between the two points measures 6.4 cm on the map. A footpath that follows the river measures 9.1 cm. The tarred road, the only route the bus can use, measures 12.3 cm. The class must be back at school by 15:30. Walking speed on the footpath is about 4 km/h. The bus averages 40 km/h but must leave 20 minutes early to collect parents on the way. The deputy head says the walk is 'too far for Grade 7s'.
1. Convert the three map measurements into real distances in kilometres. Show your working.
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Step 1 – Read the scale. 1:50 000 means 1 cm on the map represents 50 000 cm on the ground.
Step 2 – Convert to a usable unit. 50 000 cm = 500 m = 0.5 km. So every 1 cm on the map is 0.5 km in reality.
Step 3 – Multiply each measurement by 0.5:
| Route | Map distance | Real distance |
|---|---|---|
| Straight line (school to farm) | 6.4 cm | 6.4 × 0.5 = 3.2 km |
| River footpath | 9.1 cm | 9.1 × 0.5 = 4.55 km |
| Tarred road (bus route) | 12.3 cm | 12.3 × 0.5 = 6.15 km |
Reasoning note: the straight-line distance is the shortest possible distance and is never the distance anyone actually travels. The road is nearly double the straight line because it bends around a hill or follows property boundaries. Always state which distance you are quoting.
2. State the direction from the school gate to the farm gate and explain how the grid references tell you this.
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Step 1 – Split the six-figure references. 452318 means easting 452, northing 318. 489296 means easting 489, northing 296.
Step 2 – Compare eastings. 489 − 452 = 37 squares further east. Northing tells us the farm is east of the school.
Step 3 – Compare northings. 296 − 318 = −22 squares, i.e. 22 squares further south. Northing decreases southwards, so the farm is south of the school.
Step 4 – Combine. The farm lies east and south, with more easting than northing (37 against 22), so the direction is south-east, closer to east than to south (about east-south-east).
Reasoning note: grid references are read 'along the corridor, then up the stairs' - eastings first, northings second. A learner who reads 452318 as 'north 452, east 318' will get the direction roughly reversed, so the order matters.
3. Using the times given, recommend which route the class should use. Answer the deputy head's objection directly.
Read the worked answer
Step 1 – Time the footpath. 4.55 km at 4 km/h = 4.55 ÷ 4 = 1.14 hours ≈ 68 minutes each way. Return trip = 136 minutes ≈ 2 hours 16 minutes, before any time spent at the farm. Leaving at 09:00, they would be back around 11:50 only if they stayed under 10 minutes - unrealistic, and any delay pushes them past 15:30.
Step 2 – Time the bus. 6.15 km at 40 km/h = 6.15 ÷ 40 = 0.154 hours ≈ 9 minutes each way. Add 20 minutes for collecting parents = about 38 minutes of travelling in total.
Step 3 – Weigh distance against time. The bus route is almost twice as long in distance (6.15 km against 4.55 km) yet takes roughly a quarter of the time. Distance alone does not decide the answer - speed and time budget do.
Step 4 – Answer the deputy head. He is half right: 4.55 km each way is a long walk for Grade 7 learners and 9 km in a day is tiring. But the real objection is not distance, it is time - the walk cannot fit inside the 15:30 deadline once farm activities are included.
Recommendation: use the bus, leave by 08:30 so the 20-minute collection does not eat into farm time, and keep a 30-minute buffer for delays. The footpath could be used later for a short, separate map-work exercise near the school.