TASC Mathematics Methods – Foundation Level 3 study materials
TASC Mathematics Methods – Foundation Level 3 study resources

Modelling a Rainwater Tank for a School Garden

TASC Mathematics Methods – Foundation Level 3 · Put your knowledge to work in context.

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Modelling a Rainwater Tank for a School Garden

A school installs a 2000-litre rainwater tank feeding a garden. Daily inflow depends on rainfall r (mm/day) and roof area 60 m², giving inflow ≈ 60r litres per day (ignoring losses). Daily garden demand is 120 litres in term time but 260 litres during the six-week summer holiday. The tank starts the holiday full at 2000 litres. Historical summer rainfall averages 1.2 mm/day but with high variability; a dry spell of 20 consecutive days with r = 0.2 mm/day is plausible. The gardener wants to know whether the tank will run dry, and what safety margin to quote to the head teacher.

1. Model the tank volume during the holiday and determine whether the tank runs dry in the plausible dry spell.

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Let V(n) be litres after n days. V(n) = 2000 + n(60r − 260). Dry spell, r = 0.2: inflow = 12 L/day, net change = 12 − 260 = −248 L/day. V(n) = 2000 − 248n. Set V = 0: n = 2000/248 ≈ 8.06 days. So the tank empties in about 8 days of the dry spell, well before the 20-day spell ends. The garden would be without stored water for roughly 12 days unless supplemented. Average case, r = 1.2: inflow = 72 L/day, net = 72 − 260 = −188 L/day, emptying in 2000/188 ≈ 10.6 days. Even the average case does not survive the full holiday, so demand reduction or a second tank is needed.

2. The head teacher asks for a single 'days of supply' figure with a stated assumption. Propose a defensible figure and justify the modelling choice.

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A single figure should be tied to an explicit rainfall assumption. Using the historical average r = 1.2 mm/day gives about 10.6 days, but this hides the risk that half of summers are drier than average. A more defensible approach: quote a range. Worst plausible case (r = 0.2): ≈ 8 days. Average case (r = 1.2): ≈ 11 days. Wet case (r = 3.0, inflow 180 L/day, net −80): 25 days. Recommend quoting 'about 8 to 11 days under typical-to-dry conditions' rather than a single number, and note that the model assumes constant demand, no evaporation and no overflow losses, all of which would shorten supply. This is a modelling judgement, not a calculation: stating assumptions and a range is more honest than a falsely precise figure.