General Foundation Science and Mathematics for Fisheries
What the exam tests here
Nearly every recommendation in inland fisheries reaches the student as a rate: lime at 300 kg/ha, urea at 100 kg/ha, KMnO4 at 2-4 mg/L, feed at 3% of body weight, salt at 3% for a dip. The paper checks whether you can turn a rate into a weight or a volume. The core conversions are: 1 ha = 10,000 m2, 1 m3 = 1000 L, 1 ppm = 1 mg/L = 1 g/m3, 1 quintal = 100 kg, 1 tonne = 1000 kg. Percentage, ratio, simple average and weighted average also appear, and chemistry reappears when hardness or alkalinity is expressed as CaCO3 equivalent.
The five-step method
- Convert all lengths to metres and compute water area x mean depth = volume in m3 (water volume, not the excavated volume of the pond).
- Convert the dose to g/m3; the mg/L figure is numerically the same as g/m3.
- Mass in grams = dose (g/m3) x volume (m3); divide by 1000 for kilograms.
- For a per-hectare dose, multiply by area in hectares, never by volume.
- Sanity-check: a 0.1 ha pond at 1.5 m depth holds 1500 m3, so a 1 mg/L treatment needs about 1.5 kg of chemical, never 1.5 g.
Worked example
A pond measures 20 m x 15 m with 1.2 m mean depth. Treat with potassium permanganate at 4 mg/L. Area = 300 m2. Volume = 300 x 1.2 = 360 m3. Since 4 mg/L = 4 g/m3, mass = 4 x 360 = 1440 g = 1.44 kg. If instead the instruction had been lime at 300 kg/ha, the answer would be 300 x 0.03 ha = 9 kg, because liming follows area, not water volume.
Traps that cost marks
- Using 1 ha = 1000 m2 or entering depth in centimetres.
- Applying a volume-based chemical on an area basis, or the reverse.
- Treating the pond's excavated depth as the water depth.
- Forgetting that ppm of a commercial product means ppm of product, not of active ingredient, unless purity is stated.