1. **Problem:** A farmer planned to purchase 350 kg of CAN (Calcium Ammonium Nitrate). CAN was not available, so the farmer wants to buy Urea instead. Find how much Urea is needed. 2. **Formula and Explanation:** CAN contains 27% N (Nitrogen), and Urea contains 46% N. To supply the same amount of N, the quantity of Urea needed is calculated by: $$\text{Urea required} = \frac{\text{CAN quantity} \times \text{N in CAN}}{\text{N in Urea}}$$ 3. **Calculation:** $$\text{Urea required} = \frac{350 \times 0.27}{0.46} = 205.43 \text{ kg}$$ 4. **Answer:** The farmer needs to purchase approximately 205.43 kg of Urea instead of 350 kg of CAN. 5. **Problem:** Fertilize an area of 80 m × 50 m with N-P2O5-K2O @ 80-40-40 kg/ha using Urea, SSP, and MOP. Find the quantity of each fertilizer. 6. **Area Conversion:** $$80 \times 50 = 4000 \text{ m}^2 = 0.4 \text{ ha}$$ 7. **Nutrient requirement for 0.4 ha:** $$N = 80 \times 0.4 = 32 \text{ kg}$$ $$P_2O_5 = 40 \times 0.4 = 16 \text{ kg}$$ $$K_2O = 40 \times 0.4 = 16 \text{ kg}$$ 8. **Fertilizer nutrient contents:** - Urea: 46% N - SSP: 16% P2O5 - MOP: 60% K2O 9. **Calculate quantities:** $$\text{Urea} = \frac{32}{0.46} = 69.57 \text{ kg}$$ $$\text{SSP} = \frac{16}{0.16} = 100 \text{ kg}$$ $$\text{MOP} = \frac{16}{0.60} = 26.67 \text{ kg}$$ 10. **Answer:** Urea = 69.57 kg, SSP = 100 kg, MOP = 26.67 kg 11. **Problem:** A farmer has DAP instead of SSP for fertilizing 80 m × 50 m area with 80-40-40 kg/ha. 12. **DAP nutrient content:** 18% N, 46% P2O5 13. **Calculate quantities:** $$N = 32 \text{ kg}, P_2O_5 = 16 \text{ kg}, K_2O = 16 \text{ kg}$$ $$\text{DAP} = \frac{16}{0.46} = 34.78 \text{ kg}$$ $$\text{N from DAP} = 34.78 \times 0.18 = 6.26 \text{ kg}$$ $$\text{Remaining N} = 32 - 6.26 = 25.74 \text{ kg}$$ $$\text{Urea} = \frac{25.74}{0.46} = 55.96 \text{ kg}$$ $$\text{MOP} = 26.67 \text{ kg}$$ 14. **Answer:** Urea = 55.96 kg, DAP = 34.78 kg, MOP = 26.67 kg 15. **Problem:** Calculate Urea, DAP, MOP to supply 40-40-40 kg/ha in 2 acres. 16. **Area conversion:** $$2 \text{ acres} = 0.8094 \text{ ha}$$ 17. **Nutrient requirement:** $$N = 40 \times 0.8094 = 32.38 \text{ kg}$$ $$P_2O_5 = 40 \times 0.8094 = 32.38 \text{ kg}$$ $$K_2O = 40 \times 0.8094 = 32.38 \text{ kg}$$ 18. **Calculate quantities:** $$\text{DAP} = \frac{32.38}{0.46} = 70.39 \text{ kg}$$ $$\text{N from DAP} = 70.39 \times 0.18 = 12.67 \text{ kg}$$ $$\text{Remaining N} = 32.38 - 12.67 = 19.71 \text{ kg}$$ $$\text{Urea} = \frac{19.71}{0.46} = 42.85 \text{ kg}$$ $$\text{MOP} = \frac{32.38}{0.60} = 53.97 \text{ kg}$$ 19. **Answer:** Urea = 42.85 kg, DAP = 70.39 kg, MOP = 53.97 kg 20. **Problem:** Fertilizer N-P2O5-K2O @ 60-20-40 kg/ha on 10 ha rice field with basal and top dress application. 21. **Nutrient requirement:** $$N = 60 \times 10 = 600 \text{ kg}$$ $$P_2O_5 = 20 \times 10 = 200 \text{ kg}$$ $$K_2O = 40 \times 10 = 400 \text{ kg}$$ 22. **Fertilizer quantities:** $$\text{Urea} = \frac{600}{0.46} = 1304.35 \text{ kg}$$ $$\text{SSP} = \frac{200}{0.16} = 1250 \text{ kg}$$ $$\text{MOP} = \frac{400}{0.60} = 666.67 \text{ kg}$$ 23. **Application:** Half Urea basal = 652.17 kg, half Urea top dressed in two equal parts = 326.08 kg each 24. **Answer:** Basal: Urea = 652.17 kg, SSP = 1250 kg, MOP = 666.67 kg; Top dressing: Urea = 326.08 kg at maximum tillering and 326.08 kg at panicle initiation 25. **Problem:** SSP not available, use DAP instead for above case. 26. **DAP for P2O5:** $$\text{DAP} = \frac{200}{0.46} = 434.78 \text{ kg}$$ 27. **N from DAP:** $$434.78 \times 0.18 = 78.26 \text{ kg}$$ 28. **Remaining N:** $$600 - 78.26 = 521.74 \text{ kg}$$ 29. **Urea:** $$\frac{521.74}{0.46} = 1134.22 \text{ kg}$$ 30. **MOP:** $$666.67 \text{ kg}$$ 31. **Answer:** Urea = 1134.22 kg, DAP = 434.78 kg, MOP = 666.67 kg 32. **Problem:** Fertilizer @ 100-50-50 kg/ha on 1.5 ha maize crop. 33. **Nutrient requirement:** $$N = 100 \times 1.5 = 150 \text{ kg}$$ $$P_2O_5 = 50 \times 1.5 = 75 \text{ kg}$$ $$K_2O = 50 \times 1.5 = 75 \text{ kg}$$ 34. **Fertilizer quantities:** $$\text{Urea} = \frac{150}{0.46} = 326.09 \text{ kg}$$ $$\text{DAP} = \frac{75}{0.46} = 163.04 \text{ kg}$$ $$\text{MOP} = \frac{75}{0.60} = 125 \text{ kg}$$ 35. **Answer:** Urea = 326.09 kg, DAP = 163.04 kg, MOP = 125 kg 36. **Problem:** Fertilizer @ 60-30-30 kg/ha on 3 ha with DAP, CAN, MOP. 37. **Nutrient requirement:** $$N = 60 \times 3 = 180 \text{ kg}$$ $$P_2O_5 = 30 \times 3 = 90 \text{ kg}$$ $$K_2O = 30 \times 3 = 90 \text{ kg}$$ 38. **Fertilizer contents:** CAN = 27% N 39. **Quantities:** $$\text{DAP} = \frac{90}{0.46} = 195.65 \text{ kg}$$ $$\text{N from DAP} = 195.65 \times 0.18 = 35.22 \text{ kg}$$ $$\text{Remaining N} = 180 - 35.22 = 144.78 \text{ kg}$$ $$\text{CAN} = \frac{144.78}{0.27} = 536.22 \text{ kg}$$ $$\text{MOP} = \frac{90}{0.60} = 150 \text{ kg}$$ 40. **Answer:** DAP = 195.65 kg, CAN = 536.22 kg, MOP = 150 kg 41. **Problem:** Make 1 ton fertilizer mixture grade 4-8-10 using nitrate of soda (16% N), ammonium sulphate (20% N) equally for N, SSP, and MOP. 42. **N required:** 4% of 1000 kg = 40 kg N 43. **N from nitrate of soda and ammonium sulphate:** each provides 20 kg N 44. **Quantities:** $$\text{Nitrate of soda} = \frac{20}{0.16} = 125 \text{ kg}$$ $$\text{Ammonium sulphate} = \frac{20}{0.20} = 100 \text{ kg}$$ 45. **P2O5 from SSP:** 8% of 1000 kg = 80 kg $$\text{SSP} = \frac{80}{0.16} = 500 \text{ kg}$$ 46. **K2O from MOP:** 10% of 1000 kg = 100 kg $$\text{MOP} = \frac{100}{0.60} = 166.67 \text{ kg}$$ 47. **Answer:** Nitrate of soda = 125 kg, Ammonium sulphate = 100 kg, SSP = 500 kg, MOP = 166.67 kg 48. **Problem:** Make 1 ton of 8-10-10 fertilizer using Urea, SSP, and MOP. 49. **N required:** 8% of 1000 = 80 kg 50. **P2O5 required:** 10% = 100 kg 51. **K2O required:** 10% = 100 kg 52. **Quantities:** $$\text{Urea} = \frac{80}{0.46} = 173.91 \text{ kg}$$ $$\text{SSP} = \frac{100}{0.16} = 625 \text{ kg}$$ $$\text{MOP} = \frac{100}{0.60} = 166.67 \text{ kg}$$ 53. **Answer:** Urea = 173.91 kg, SSP = 625 kg, MOP = 166.67 kg 54. **Problem:** Find percentages of nutrients in fertilizer mixture ratio 1:2:1 with 10% N. 55. **Given:** N = 10%, ratio N:P2O5:K2O = 1:2:1 56. **Calculate P2O5 and K2O:** $$P_2O_5 = 2 \times 10 = 20\%$$ $$K_2O = 1 \times 10 = 10\%$$ 57. **Answer:** Nutrient percentages are N=10%, P2O5=20%, K2O=10% 58. **Problem:** Cost of 1 kg N, P2O5, K2O if costs of Urea, SSP, MOP are 12, 10, 15 respectively. 59. **Cost per kg nutrient:** $$\text{Cost of N} = \frac{12}{0.46} = 26.09$$ $$\text{Cost of P2O5} = \frac{10}{0.16} = 62.5$$ $$\text{Cost of K2O} = \frac{15}{0.60} = 25$$ 60. **Answer:** Cost per kg N = 26.09, P2O5 = 62.5, K2O = 25 61. **Problem:** Cost of 1 kg N, P, K if costs of Urea, SSP, MOP are 12, 10, 15. 62. **Convert P2O5 to P:** $$P = P_2O_5 \times \frac{30}{44} = 0.6818 \times P_2O_5$$ 63. **Cost per kg P:** $$\frac{62.5}{0.6818} = 91.68$$ 64. **Cost per kg K:** K2O to K factor is 0.83 $$\text{Cost per kg K} = \frac{25}{0.83} = 30.12$$ 65. **Answer:** Cost per kg N = 26.09, P = 91.68, K = 30.12