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Soil Fertility — Soil Science Reviewer Questions

17 board-style Soil Fertility items for the Agriculturist Licensure Examination. Try 40 questions free; lifetime access is ₱49. Texture drives water, water drives aeration, and aeration drives nutrient availability. Reason down that chain and most items unlock.

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Sample Soil Fertility questions with answers and explanations

Board-style items taken from the Soil Science bank. Every answer is explained, which is the part that makes a review question worth doing twice.

  1. A recommendation calls for 90 kg N/ha and the only source is urea (46-0-0). How much urea is needed?

    • A. 414 kg/ha
    • B. 196 kg/ha correct
    • C. 41 kg/ha
    • D. 90 kg/ha

    Why: Divide the nutrient requirement by the grade expressed as a decimal: 90 / 0.46 = 195.7, rounded to 196 kg/ha. Multiplying by 0.46 instead gives 41 kg and would supply less than a quarter of the nitrogen the crop was recommended.

  2. A farmer applies 300 kg/ha of complete fertiliser (14-14-14). How much actual nitrogen is supplied?

    • A. 300 kg N/ha
    • B. 14 kg N/ha
    • C. 42 kg N/ha correct
    • D. 21 kg N/ha

    Why: 0.14 x 300 = 42 kg N/ha, and the same 42 kg each of P2O5 and K2O. The grade is a percentage of the BAG, so the nutrient delivered always depends on how much material is spread, not on the analysis alone.

  3. A recommendation is 60-40-40 kg NPK/ha. If 286 kg/ha of 14-14-14 is applied as basal, how much additional N must come from urea (46-0-0)?

    • A. 20 kg/ha of urea
    • B. 96 kg/ha of urea
    • C. 130 kg/ha of urea
    • D. 44 kg/ha of urea correct

    Why: The complete supplies 0.14 x 286 = 40 kg each of N, P2O5 and K2O, meeting P and K exactly. The nitrogen shortfall is 60 - 40 = 20 kg N, and 20 / 0.46 = 43.5, rounded to 44 kg of urea. Answering 20 kg confuses the NUTRIENT needed with the MATERIAL that carries it.

  4. A soil test recommends 100 kg P2O5/ha. Expressed as elemental phosphorus, how much is that?

    • A. 43.6 kg P/ha correct
    • B. 229 kg P/ha
    • C. 83.0 kg P/ha
    • D. 100 kg P/ha

    Why: P = P2O5 x 0.436 = 43.6 kg/ha. The factor exists because fertiliser grades are stated as oxides by long convention while plant uptake and soil test reports are often elemental -- mixing the two conventions is a frequent and expensive error.

  5. A farmer needs 90 kg K2O/ha and uses muriate of potash (0-0-60). How much material is required?

    • A. 54 kg/ha
    • B. 150 kg/ha correct
    • C. 90 kg/ha
    • D. 180 kg/ha

    Why: 90 / 0.60 = 150 kg/ha. That also delivers roughly 75 kg of chloride, which matters for chloride-sensitive crops such as tobacco -- a reason sulphate of potash is sometimes chosen despite its higher cost.

  6. Ammonium sulfate is 21-0-0 and urea is 46-0-0. To supply 84 kg N/ha, how much MORE material must be spread if ammonium sulfate is used instead of urea?

    • A. 400 kg/ha more
    • B. 183 kg/ha more
    • C. 217 kg/ha more correct
    • D. 62 kg/ha more

    Why: Ammonium sulfate: 84 / 0.21 = 400 kg; urea: 84 / 0.46 = 183 kg. The difference is 217 kg/ha of extra material to buy, haul and spread. Ammonium sulfate can still be preferred where its sulfur or its acidifying effect is wanted, which is a judgement the arithmetic informs rather than settles.

  7. A recommendation is 120 kg N/ha for one hectare. A farmer's plot measures 40 m x 50 m. How much urea (46-0-0) should be applied to the plot?

    • A. 261 kg
    • B. 24 kg
    • C. 120 kg
    • D. 52 kg correct

    Why: The plot is 2,000 m2, or 0.2 ha, so it needs 24 kg N. As urea: 24 / 0.46 = 52 kg. Two conversions are required -- area to hectares and nutrient to material -- and skipping either produces a rate that is wrong by a factor of five or of two.

  8. Rice yielding 5 t/ha removes about 15 kg N per tonne of grain. If fertiliser nitrogen is only 40% recovered, how much N must be applied to replace the removal?

    • A. 188 kg N/ha correct
    • B. 75 kg N/ha
    • C. 30 kg N/ha
    • D. 125 kg N/ha

    Why: Removal = 5 x 15 = 75 kg N/ha. Because only 40% of applied N reaches the crop, the rate needed is 75 / 0.40 = 187.5, rounded to 188 kg N/ha. Recovery efficiency is the step that separates crop REMOVAL from a fertiliser RECOMMENDATION, and it is why split application, which raises recovery, lowers the rate needed.

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