ABE-Board-Exam-Reviewer-Part-2.pdf

ABE Board Exam Reviewer Part II Soil and Water Resources Development and Conservation, Irrigation and Drainage, and Allied Subjects Copyright © 2022 by Arthur It. Tambong, FPSAE SPECIFIC COVERAGE BASED ON PRC TABLE OF SPECIFICATIONS I. Hydrology (10%) II. Irrigation and Drainage Engineering...

ABE Board Exam Reviewer Part II Soil and Water Resources Development and Conservation, Irrigation and Drainage, and Allied Subjects Copyright © 2022 by Arthur It. Tambong, FPSAE SPECIFIC COVERAGE BASED ON PRC TABLE OF SPECIFICATIONS I. Hydrology (10%) II. Irrigation and Drainage Engineering (45%) III. Soil and Water Conservation Engineering (30%) IV. Mathematics, Agricultural Statistics and Operations Research (10%) V. Agricultural and Fishery Sciences (5%) I. Hydrology (10%) EASY QUESTIONS 1. Area which contributes runoff or drains water into the reservoir (PAES 609:2016). A. River basin B. Runoff reservoir C. Watershed D. Drainage divide Answer : C EASY QUESTIONS 2. A thermometer that has a constriction above the bulb that permits the mercury to rise in the capillary tube but does not allow it to descend the capillary tube unless the thermometer is reset (PAGASA, 2022). A. Minimum thermometer B. Maximum thermometer C. Air thermometer D. Constrithermometer Answer : B EASY QUESTIONS 3. The science that encompasses the occurrence, distribution, movement and properties of the waters of the earth and their relationship with the environment (USGS, 2022). A. Meteorology B. Hydrometeorology C. Hydrology D. Limnology Answer : C EASY QUESTIONS 4. The study of the biological, chemical, and physical features of lakes and other bodies of fresh water (Oxford Languages, 2022). A. Meteorology B. Hydrology C. Hydrometeorology D. Limnology Answer : D EASY QUESTIONS 5. A branch of meteorology and hydrology that studies the transfer of water and energy between the land surface and the lower atmosphere (Wikipedia, 2022). A. Meteorology B. Hydrometeorology C. Hydrology D. Limnology Answer : B EASY QUESTIONS 6. Heavy rainfall or rain that accumulates at a rate of 3 tenths of an inch (0.3 inch or about 7.5 mm), or more, per hour (US National Weather Service, 2022). A. Storm B. Typhoon C. Torrential rain D. Heavy rain Answer : C EASY QUESTIONS 7. Dominant form of precipitation in the Philippines. A. Flood B. Rainfall C. Runoff D. Evaporation Answer : B EASY QUESTIONS 8. Kind of thermometer with pin inside its tube which does not go with the expanding liquid when temperature increases. A. Maximum thermometer B. Minimum thermometer C. Air thermometer D. Soil thermometer Answer : B EASY QUESTIONS 9. Method of areal rainfall determination done by computing the weighted average. A. Averaging method B. Polygon Method C. Arithmetic method D. Isohyetal method Answer : C EASY QUESTIONS 10. An instrument to measure depth of rainfall. A. Rainfall dip stick B. Rain gage C. Bucket D. Rainfall meter Answer : B EASY QUESTIONS 11. Part of rainfall which runs off the soil as surface or subsurface flow. A. Flood B. Runoff C. Percolation D. Infiltration Answer : B EASY QUESTIONS 12. Ratio of runoff over rainfall. A. Runoff ratio B. Runoff/rainfall ratio C. Rainfall ratio D. Runoff coefficient Answer : D EASY QUESTIONS 13. The part of rainfall that is stored in the root zone and can be used by the plants (FAO, 2022). This excludes rainfall that does not reach the soil and percolation below the root zone. For grassed soil surface, it is estimated that this rainfall is greater than 0.5 mm/day. A. Atmospheric rainfall B. Effective rainfall C. Precipitation D. Torrential rainfall Answer : B EASY QUESTIONS 14. The inverse of the probability of exceedance of a certain hydrologic event. A. Hydro probability B. Probability of inceedance C. Return period D. Event probability Answer : C EASY QUESTIONS 15. Airmass lifting process which is mountain- facilitated. A. Natural B. Convective C. Mountain lifting D. Orographic Answer : D EASY QUESTIONS 16. Airmass lifting process facilitated by sunlight or heating. A. Natural B. Convective C. Mountain lifting D. Orographic Answer : B EASY QUESTIONS 17. Indicates to what depth liquid precipitation would cover a horizontal surface in an observation period if nothing could drain, evaporate or percolate from this surface (Graf-Water, 2022). A. Precipitation B. Precipitation depth C. Rainfall depth D. Water depth Answer : B MODERATE QUESTIONS 18. The precipitation depth that corresponds to a liquid quantity of 1 liter falling into a 1 square meter of ground area. A. 0.5 mm B. 1 mm C. 2 mm D. 2.5 mm Answer : B Solution: Precipitation = (1 liter x 1,000,000 mm3 /liter) / (1 m2 x 1,000,000 mm2 /m2 ) = 1 mm MODERATE QUESTIONS 19. Probability of occurrence in any year of hydrologic event recurring once in 4 years. A. 40% B. 25% C. 20% D. 50% Answer : B Solution: Probability = (1/Return Period) x 100% = (1/4) x 100% = 25% MODERATE QUESTIONS 20. What is the probability of occurrence in any year of a hydrologic event recurring every year? A. 100% B. 10% C. 20% D. 50% Answer : A Solution: Probability = (1/Return Period) x 100 = (1/1) x 100% = 100% DIFFICULT QUESTIONS 21. Rainfall depths recorded were as follows: 8mm and 2mm for 8:00 A.M. and 2:00 P.M., respectively on 7-7-2022 and 5 mm and 9 mm for 8:00 A.M. and 2:00 P.M., respectively on 7-8- 2022. What is the total rainfall depth on 7-7- 2022? A. 0.5 mm B. 1 mm C. 2 mm D. 2.5 mm Answer : C Solution: The rainfall recorded at 8:00 AM of any day is counted as part of the total rainfall for the previous day. Rainfall on 7-7-2022 = Rainfall at 2:00 P.M. on 7-7-2022 + Rainfall at 8:00 A.M. on 7-8-2022 = 2 mm + 5 mm = 7 mm DIFFICULT QUESTIONS 22. The agrometeorological station uses electronic rain gage. The rainfall depths were recorded on June 1, 2022 as follows: 2 mm from 6:00 to 8:00 A.M., 7 mm from 8:00 to 10:00 A.M., 9 mm from 12:00 noon to 2:00 P.M. and zero rainfall for all other times of the day and the previous day. What is the rainfall intensity recorded on June 1, 2022? A. 0.75 mm/hr B. 2 mm/hr C. 4 mm/hr D. 3 mm/hr Answer : C Solution: Intensity = Depth/Time, time is from 8:00 AM of current day to 8:00 AM of the next day = (7mm from 8:00 to 10:00 A.M. + 9mm from 12:00 noon to 2:00 P.M.)/(2hr +2 hr) = 4 mm/hr DIFFICULT QUESTIONS 23. Estimate the time of concentration of a certain watershed having a time lag of 1 hour. A. 1.43 hours B. 1.34 hours C. 14.3 hours D. 13.4 hours Answer : A Solution: Using the formula for time of concentration from PAES 609:2016, TC = TL / 0.70 = 1 hr / 0.70 = 1.43 hrs II. Irrigation and Drainage Engineering (45%) EASY QUESTIONS 1. The maximum permissible water velocity for clay loam canal surface based on PAES 603:2016. A. 1.2 m/s B. 1 m/s C. 0.9 m/s D. 0.80 m/s Answer : D EASY QUESTIONS 2. The minimum permissible velocity for water with sediments in lined canals based on PAES 603:2016. A. 1.2 m/s B. 1 m/s C. 0.9 m/s D. 0.80 m/s Answer : C EASY QUESTIONS 3. Application of water in the soil to supply moisture needed for plant growth. A. Flooding B. Sprinkling C. Irrigation D. Diverting Answer : C EASY QUESTIONS 4. Loss of water from a channel during transport due to seepage and percolation. A. Channel loss B. Seepage loss C. Percolation loss D. Conveyance loss Answer : D EASY QUESTIONS 5. Depth of water flow where the energy content is at minimum hence, no other backwater forces are involved. A. Minimum depth B. Critical depth C. Energy depth D. Normal depth Answer : B EASY QUESTIONS 6. Ratio of the actual crop evapotranspiration to its potential evapotranspiration. A. Crop ratio B. ET ratio C. Crop coefficient D. Evaporation ratio Answer : C EASY QUESTIONS 7. Moisture content of the soil when gravitational water has been removed. A. Soil capacity B. Gravitational moisture C. Field capacity D. Specific capacity Answer : C EASY QUESTIONS 8. Number of days between irrigation applications. A. Irrigation interval B. Application interval C. Dry interval D. Node interval Answer : A EASY QUESTIONS 9. Removal of excess water. A. Squeezing B. Run off C. Discharging D. Drainage Answer : D EASY QUESTIONS 10. Elevated section of open channel used for crossing natural depressions. A. Parshall flume B. Flume C. Siphon D. Elevated channel Answer : B EASY QUESTIONS 11. Surveying instrument used for determining land areas in a topographic maps. A. Aerometer B. Erometer C. Planimeter D. Lysimeter Answer : C EASY QUESTIONS 12. Elevation of water surface in a stream with reference to a certain datum. A. Stage B. Surface elevation C. Contour D. Water elevation Answer : A EASY QUESTIONS 13. Facility for determining water consumptive use of crops in an open field. A. Planimeter B. Lysimeter C. Consumeter D. Crop meter Answer : B EASY QUESTIONS 14. Time required to cover an area with one application of water. A. Irrigation interval B. Irrigation period C. Supply duration D. Application time Answer : B EASY QUESTIONS 15. At optimal emitter spacing, drip emitter spacing is ___ of the wetted diameter estimated from field tests. A. 100% B. 90% C. 80% D. 85% Answer : C EASY QUESTIONS 16. Reference crop evapotranspiration is the rate of evapotranspiration from a reference surface which is a hypothetical reference crop with an assumed crop height of 0.2 m and an albedo of _______ (AMTEC, 2020). A. 0.23 B. 0.25 C. 0.30 D. 0.32 Answer : A EASY QUESTIONS 17. Manufacturer’s coefficient of variation is the measure of the variability of discharge of a random sample of a given make, model and size of emitter, as provided by the manufacturer and before any field operations or aging has taken place determined through a discharge test of a sample of 50 emitters under a set pressure at ___ ºC. A. 200 B. 100 C. 50 D. 30 Answer : A EASY QUESTIONS 18. Which one is the flattest canal side slope? A. 1:1 B. 1:4 C. 4:1 D. 2:1 Answer : C MODERATE QUESTIONS 19. Determine the side slope angle Ө with the horizontal plane of an unlined canal with side slope ratio (run: rise) z of 2:1. A. 16.6° B. 26.6° C. 45° D. 60° Answer : B Solution: tan Ө = rise /run Ө = arctan (rise /run) = arctan (1/2) = 26.6° Notes: In specifying side slope, run is written first, ex. 4:1 means horizontal run is 4. In computing angles whether bed slope or side slope, rise is the numerator. Some fluid mechanics and hydraulics books use Ө symbol for side angle with the vertical plane, not with the horizontal. In this case subtract Ө from 90° to get angle with the horizontal plane. MODERATE QUESTIONS 20. If the most efficient concrete canal has its side angle Ө with the horizontal plane equal to 60º, what is the z value of the canal sides or the side’s horizontal run in meters per 1 meter rise? This value is commonly used in designing most efficient concrete canals. A. 0.775 B. 0.757 C. 0.577 D. 1/0.577 Answer : C Solution: tan Ө = 1 / z tan 60° = 1 / z 1.732 = 1 / z z = 0.577 DIFFICULT QUESTIONS 21.What is the top width at water surface level of the most efficient concrete open channel if the design depth is 5 meters? The design discharge is 100 m³/s and velocity is 2 m/s. A. 1.29 m B. 9.12 m C. 12.9 m D. 19.2 m Answer : C Solution: If the canal is a A = bd + zd² trapezoidal concrete and side angle A - zd² = bd is not given, then it is a Most b = (A-zd² ) / d Efficient Canal (in which Ө=60º and = [50 – 0.577(5)² ] / 5 z=0.577). Specifying it as most = 7.12 m efficient canal is often omitted in the board question. t = b + (2d / tan Ө) = 7.12 m + [2(5) / 1.732] Q = AV = 12.89 m 100 = A (2) A = 50 m² Checking 1: A = bd + zd² 50 = (7.12)(5) + 0.577(5)² tan Ө = 1 / z 50 = 35.6+14.4 50 = 50 tan 60° = 1 / z Checking 2: 1.732 = 1 / z A = d [(t + b)/2] 50 = 5 [(12.89+7.12)/2] z = 0.577 50 = 5 50 = 50 DIFFICULT QUESTIONS 22. What is the total top width of the most efficient concrete open channel if design depth is 5 meters? Design discharge is 100 m³/s and velocity is 2 m/s. Use 15% freeboard. A. 12.9 m B. 13.8 m C. 18.3 m D. 8.13 m Answer : B Solution: Use the Design Criteria: Most Efficient Canal (Ө=60º and z = 0.577) D = 1.15d Q = AV = 1.15(5) 100 = A (2) = 5.75 m A = 50 m² T = b + (2D / tan Ө) = 7.12 + [2(5.75) / 1.732] A = bd + zd² = 7.12 + 6.64 A - zd² = bd = 13.8 m b = (A-zd² ) / d = [50 – 0.577(5) ²] / 5 Checking 1: A = bd + zd² = 7.12 m 50 = (7.12)(5) + 0.577(5) ² 50 = 35.6+14.4 50 = 50 t = b + (2d / tan Ө) Checking 2: = 7.12 m + [2(5) / 1.732] A = d [(t + b)/2] 50 = 5 [(12.89+7.12)/2] = 12.89 m 50 = 5 50 = 50 DIFFICULT QUESTIONS 23. What is the base of the most efficient trapezoidal concrete open channel if discharge is 100 m³/s and velocity is 2 m/s? A. 6.14 m B. 12.8 m C. 7.21 m D. 14.6 m Answer : A Solution: Use the same approach as the previous problem but find the canal base. Q = AV A = bd + zd² 100 = A (2) b = (A-zd² ) / d A = 50 m² = [50 – 0.577(5.4)²] / 5.4 = 6.14 m A = 1.732 d² Checking 1: 50/1.732 = d² A = bd + zd² d = 5.4 m² 50 = (6.14)(5.4) + 0.577(5.4)² 50 = 33.2 + 16.8 50 = 50 DIFFICULT QUESTIONS 24. What is the bottom width for the best hydraulic cross-section (best proportion) of concrete open channel if design depth is 5 meters and side slope is 45º? A. 3 m B. 4 m C. 5 m D. 6 m Answer : B Solution: Since the concrete canal has a side angle other than 60°, then use the design criteria: Best Hydraulic Cross-Section and use the formula for concrete canals (coefficient of d is 2). b = 2 d tan (Ө/2) = 2 (5) tan (45/2) = 2 (5) tan (22.5) = 2 (5) (0.41) = 4.1 m DIFFICULT QUESTIONS 25. What is the bottom width for best hydraulic cross- section of unlined open channel for minimum seepage if design depth is 5 meters and side slope is 45º? A. 3.15 m B. 4.15 m C. 8.15 m D. 6.15 m Answer : C Solution: Since the canal is unlined or not concrete, then use the design criteria: Best Hydraulic Cross-Section and use the formula for unlined canals (coefficient of d is 4). b = 4 d tan (Ө/2) = 4 (5) tan (22.5) = 8.2 m DIFFICULT QUESTIONS 26. What is the bottom width for best hydraulic cross- section of unlined open channel with minimum seepage if design depth is 5 meters and side slope is 2:1? A. 4.72 m B. 7. 42 m C. 2.47 m D. 7.24 m Answer : A Solution: Compute for the side angle then use the design criteria: Best Hydraulic Cross-Section and then use the formula for unlined canals (coefficient of d is 4). Ө = arctan (rise /run) b = 4 d tan (Ө/2) = arctan (1/2) = 4 (5) tan (26.6/2) = 26.6º = 4 (5) tan (13.3) = 4 (5) (0.236) = 4.72 m DIFFICULT QUESTIONS 27. Estimate the width and depth of concrete-lined rectangular open channel for water velocity of 2 m/s and discharge of 10 m³/s. A. 6.1 m, 2.3 m B. 3.2 m, 1.6 m C. 2.5m, 5.0 m D. 13.6 m, 3.1 m Answer : B Solution: It is specified that the concrete canal is rectangular. Hence, it cannot qualify for Most Efficient criterion. Use the design criteria: Efficient Canal and then use the formula for concrete-lined rectangular open channels. Expressing b in terms of d: b = 2d Determining A: Solving for b and d: A = Q/V A= bd 5 = (2d)d = 10/2 d² = 5/2 = 5 m² d = 1.58 m b = 2 (1.58) = 3.16 m DIFFICULT QUESTIONS 28. What should be the base and depth of concrete- lined rectangular open channel for a cross- sectional area of 50 m²? Design for efficiency over proportion. A. 10 m, 5 m B. 12 m, 6 m C. 2.5 m, 5 m D. 3 m, 6 m Answer : A Solution: From this item until Item 32 in Irrigation and Drainage Engineering, determine what criteria to apply and what formulas to use as part of your practice in board problem analysis. A = 2 d² b = 2d Checking: 50 =2 d² b = 2(5) A = bd d=5m b = 10 m 50 = (10)(5) 50 = 50 DIFFICULT QUESTIONS 29. What should be the depth and side angle with the horizontal of concrete-lined triangular open channel for a cross-sectional area of 50 m²? A. 5 m, 16.6º B. 6 m, 26.6º C. 7 m, 45º D. 8 m, 60º Answer : C Solution: A = d² 50 = d² d = 7.1 m Ө = 45º for efficient triangular canals. DIFFICULT QUESTIONS 30.. What design depth of open channel would you recommend to carry 100 cumecs or cubic meters/sec with a velocity of 5 mps? Use the most efficient of all trapezoidal cross-sections. A. 1.4 m B. 2.4 m C. 3.4 m D. 1.3 m Answer : C Solution: A= Q/V A = 1.732 d² = 100/5 20 = 1.732 d² = 20 m² d = 3.4 m DIFFICULT QUESTIONS 31. If the most efficient of all trapezoidal cross- sections can be used, what actual depth of open channel would you recommend to carry 100 cumecs with a velocity of 5 mps? Use 15% freeboard. A. 3.9 m B. 3.5 m C. 3.6 m D. 1.3 m Answer : A Solution: A= Q/V A = 1.732 d² D = 1.15 d = 100/5 20 = 1.732 d² = 1.15 (3.4) = 20 m² d = 3.4 m = 3.91 m DIFFICULT QUESTIONS 32. If an unlined trapezoidal canal with best hydraulic cross-section can be used, what actual depth of open channel would you recommend to carry 10 cumecs with a velocity of 1 mps? Use 2:1 side slope and 15% freeboard. A. 1.12 m B. 2.12 m C. 21.2 m D. 2.21 m Answer : B 10 = 2.944d2 Solution: d = 1.84 m A = Q/V D = 1.15 = 10/1 d = 1.15 (1.84 m) = 10 m² = 2.12m Ө = arctan (rise /run) = arctan (1/2) Checking: tan (26.565 °) = 0.5 = 26.565° b = 0.944 d b = 4 d tan (Ө/2) = 0.944 (1.84m) = 1.74 m = 0.944 d t = b + (2d / tan Ө) = 1.74 + [2(1.8 4) / 0.5] = 9.11 m A = bd + zd², A = d [(t + b)/2] 10 = 0.944d + zd² but z = 2 10 = 1.84 [(9.11+1.74)/2] 10 = 0.944d + 2d2 but z = 2 10 = 10 III. Soil and Water Conservation Engineering (30%) EASY QUESTIONS 1. Slope of the upstream face of the embankment. A. Downstream slope B. Outside slope C. Inside slope D. Upstream slope Answer : C EASY QUESTIONS 2. Inside bottom or sill of the conduit. A. Invert B. Inside base C. Inside sill D. Bottom sill Answer : A EASY QUESTIONS 3. Closed conduit designed to convey canal water in full and under pressure running condition, to convey canal water by gravity under roadways, railways, drainage channels and local depressions. A. Close siphon B. Pressurized conduit C. Siphon D. Inverted siphon Answer : D EASY QUESTIONS 4. __________ water requirement is the amount of water required in lowland rice production which includes water losses through evaporation, seepage, percolation and land soaking. A. Land preparation B. Irrigation C. Crop D. Field Answer : A EASY QUESTIONS 5. _________ water requirement is the amount of water required in lowland rice production which is a function of the initial soil moisture and the physical properties of the soil. A. Land preparation B. Irrigation C. Crop D. Land soaking Answer : D EASY QUESTIONS 6. Spacing between irrigation laterals. A. Ditch spacing B. Lateral spacing C. Horizontal spacing D. Irrigation spacing Answer : B EASY QUESTIONS 7. Deep percolation of water beyond the root zone of plants, resulting in loss of salts or nutrients. A. Vertical percolation B. Root zone percolation C. Leaching D. Salt leaching Answer : C EASY QUESTIONS 8. Canal with impermeable material (usually concrete) for channel stabilization and/or reduced seepage. A. Line canal B. Lined canal C. Unlined canal D. Impermeable canal Answer : B EASY QUESTIONS 9. Allowable pollutant-loading limit per unit of time, which the wastewater generator is permitted to discharge into any receiving body of water or land. A. Pollutant limit B. Loading limit C. Allowable limit D. Wastewater limit Answer : B EASY QUESTIONS 10. Portion of the pipe network between the mainline and the laterals. A. Diversion pipe B. Manifold C. Main-lateral pipe D. Reducer Answer : B EASY QUESTIONS 11. Spillway which is not excavated such as natural draw, saddle or drainage way. A. Surface spillway B. Flood spillway C. Natural spillway D. Earth spillway Answer : C EASY QUESTIONS 12. Constant flow depth along a longitudinal section of a channel under a uniform flow condition. A. Critical depth B. Constant depth C. Laminar depth D. Normal depth Answer : D EASY QUESTIONS 13. Maximum elevation of the water surface which can be attained by the spillway-type dam or reservoir without flow in the spillway. A. Normal storage B. Maximum Storage C. Critical elevation D. Design depth Answer : A EASY QUESTIONS 14. Maximum elevation of the water surface which can be attained in an open channel without reaching the freeboard. A. Normal storage B. Maximum Storage C. Critical elevation D. Design depth Answer : D EASY QUESTIONS 15. In what condition is the open channel freeboard used for water conveyance? A. Maximum flow B. Emergency flow C. Inundation D. Rainy days Answer : C EASY QUESTIONS 16. The primary purpose in limiting water flow not to go below minimum velocity. A. Avoid percolation B. Avoid sedimentation C. Avoid critical depth D. Optimize flow Answer : B EASY QUESTIONS 17. Open channel flow is water flow that is conveyed in such a manner that top surface is exposed to the atmosphere such as flow in canals, ditches, drainage channels, culverts, and pipes under _____ flow conditions. A. Full B. Partially full C. Normal D. Critical Answer : B EASY QUESTIONS 18. Part of the system that impounds the runoff. A. Storage B. Reservoir C. Impounding D. Runoff collector Answer : B EASY QUESTIONS 19. Slope at the downstream face of the embankment. A. Outside slope B. Inside slope C. Side slope D. Soil gradient Answer : A EASY QUESTIONS 20. Ratio between reference evapotranspiration and water loss by evaporation from an open water surface of a pan. A. Pan coefficient B. Evaporation ratio C. Reference pan ratio D. ET ratio Answer : A EASY QUESTIONS 21. Rate of water loss by evaporation from an open water surface of a pan. A. Surface evaporation B. Sunken evaporation C. Pan evaporation D. Evaporation loss Answer : C EASY QUESTIONS 22. Vertical flow of water below the root zone which is affected by soil structure, texture, bulk density, mineralogy, organic matter content, salt type and concentration. A. Leaching B. Percolation C. Infiltration D. Seepage Answer : B EASY QUESTIONS 23. Vertical flow of water to carry salts contained in water. A. Leaching B. Percolation C. Infiltration D. Seepage Answer : A EASY QUESTIONS 24. Method to determine the rate of flow under laminar flow conditions through a unit cross sectional area of soil under unit hydraulic gradient. A. Permeability test B. Laminar test C. Flow test D. Hydraulic test Answer : A EASY QUESTIONS 25. The process by which the soil is removed from its natural place. A. Soil removal B. Runoff C. Soil erosion D. Leaching Answer : C EASY QUESTIONS 26. A kind of terrace which consists of a series of flattened areas. A. Broad-base terrace B. Bench terrace C. Conservation terrace D. Rice terrace Answer : B EASY QUESTIONS 27. The practice where legumes are plowed or incorporated into the soil. A. Legume incorporation B. Legume manuring C. Green manuring D. Manuring Answer : C EASY QUESTIONS 28. Farming practice where plowing and harrowing are done along the contour. A. Contouring B. Strip cropping C. Crop row aligning D. Contour plowing Answer : A EASY QUESTIONS 29. The simplest method of determining soil erosion over a period of time. A. Catchment method B. Erometer method C. Plumb bob method D. Pin method Answer : D EASY QUESTIONS 30. Advanced form of erosion. A. Rill erosion B. Gully erosion C. Sheet erosion D. Advanced erosion Answer : B EASY QUESTIONS 31. Dam which resists water flow of water by its weight. A. Resisting dam B. Buttress dam C. Gravity dam D. Arc dam Answer : C EASY QUESTIONS 32. Dam consisting of stones enclosed in cyclone wires which allows water passage. A. Gabion dam B. Stone dam C. Cyclone dam D. Interlink dam Answer : A EASY QUESTIONS 33. Material used to cover the soil to minimize evapotranspiration. A. Plastic B. Mulch C. Leaves D. Soil cover Answer : B EASY QUESTIONS 34. Geological formation shaped by the dissolution of a layer or layers of soluble bedrock, usually carbonate rocks such as limestone or dolomite. A. Geological layer B. Dissolved layer C. Soluble layer D. Karst topography Answer : D EASY QUESTIONS 35. Scientific name of carabao grass commonly used in vegetated open channels. A. Glerisedia sepium B. Paspalum conjugatum C. Cyperus rotundos D. Leucaena leucocepala Answer : B MODERATE QUESTIONS 36. Philippine geographical constant for determining terrace vertical interval. A. 0.5 B. 0.8 C. 1 D. 1.5 Answer : A MODERATE QUESTIONS 37. Constant for soil erodibility and cover conditions for erodible soil and poor cover. A. 0.5 B. 0.8 C. 1 D. 1.5 Answer : C MODERATE QUESTIONS 38. Estimate the peak runoff rate from a 20-hectare drainage basin if rainfall depth for 6 hours reached 60 mm. Use typical runoff coefficient of 0.35. A. 2 cms B. 0.02 cms C. 0.2 cms D. 2.2 cms Answer : C Solution: I = 60 mm/6 hrs = 10 mm/hr q = 0.0028 CIA = 0.0028 (0.35)(10 mm/hr)(20 has) = 0.2 cms MODERATE QUESTIONS 39. Estimate the runoff volume from a 20-hectare drainage basin if rainfall depth for 6 hours reached 60 mm and duration of runoff is about 3 hours. Use typical runoff coefficient of 0.35. A. 1,080 cms B. 1,800 cms C. 1,008 cms D. 8,100 cms Answer : A Solution: I = 60 mm/6 hrs = 10 mm/hr q = 0.0028 CIA = 0.0028 (0.35)(10 mm/hr)(20 has) = 0.2 cms Q = 0.5qT = 0.5 (0.2 cms) (3 hrs) (3600 s/hr) = 1,080 cms DIFFICULT QUESTIONS 40. A trapezoidal concrete gravity dam has vertical upstream wall. Water depth in the upstream side is 50 meters above the dam base. Freeboard of 15% based on design depth. What is the force of water exerted against dam per meter of dam width at deepest section? A. 1,100 tonnes B. 1,150 tonnes C. 1,200 tonnes D. 1,250 tonnes Answer : D Solution: P = 0.5 x Water density x H² = 0.5 x1 tonne/m³ x (50 m)² = 1,250 tonnes DIFFICULT QUESTIONS 41. A trapezoidal concrete gravity dam has vertical upstream wall. Water depth in the upstream side is 50 meters above the dam base. The base of dam is 78 meters along water flow. What is the uplift force of seepage water below the dam per meter of dam width at deepest section? A. 1,900 tonnes B. 1,950 tonnes C. 2,000 tonnes D. 2,050 tonnes Answer : B Solution: U = 0.5 x Water density x H x B = 0.5 x 1 tonne/m3 x 50m x 78m = 1,950 tonnes DIFFICULT QUESTIONS 42. A trapezoidal concrete gravity dam has inclined upstream wall. Resisting moment relative to the dam toe is 263,953.2 tonne-meters while overturning moment is 96,233.3 tonne-meters per meter of dam width at deepest section. What is the safety factor against overturning? Is it safe? A. 0.36, unsafe B. 2.74, safe C. 3. 96, safe D. 5.48, excessively safe Answer : B Solution: FSo = RM/OM = 263953.2/96233.3 = 2.74 IV. Mathematics, Agricultural Statistics and Operations Research (10%) EASY QUESTIONS 1. Based on PhilMech’s 2011 and 2012 agricultural statistics of Asian countries, the Philippine farm mechanization showed an upswing trend which recorded an average mechanization level of 1.23 hp/ha; rice and corn farms registered the highest level of available farm power at 2.31 hp/ha (Philippine Official Gazette, 2013). For the Philippines to level up with Japan, the leader in Asia, agricultural and biosystems engineers should know Japan’s statistics. What was the average level of farm mechanization in Japan as of 2012? A. 7 hp/ha B. 8 hp/ha C. 9 hp/ha D. 10 hp/ha Answer : A EASY QUESTIONS 2. Based on the agricultural statistics survey of PhilMech in 2011 and 2012, ____ percent of the total farm power is available for use in production operations while the remaining percentage is used for postharvest operations. A. 60 B. 70 C. 80 D. 90 Answer : B EASY QUESTIONS 3. An applied science that is concerned with quantitative decision problems that generally involve the allocation and control of limited resources (IEOR-Columbia, 2022). A. Operations research B. Management C. Control D. Decision making Answer : A EASY QUESTIONS 4. You are the project engineer of an ABE company having a project involving 3 constraints, 5 stages and 18 locations. To minimize project duration, which technique will you use? A. Dynamic programming B. Linear programming C. PERT/CPM D. GANTT Charting Answer : C EASY QUESTIONS 5. The equivalence of ln e(y+z)/2. A. 0.5(y+z) B. (ey+ez )/2 C. log 2 x e (y+z) D. ln [(y+z)/2] Answer : A Solution: From the rules on natural logarithm, ln e x = x ln e(y+z)/2 = (x+y)/2 = 0.5(x+y) EASY QUESTIONS 6. The factorial of 10 less factorial of 9. A. 3,628,800 B. 362,880 C. 3,265,902 D. 3,265,920 Answer : D Solution: 10! - 9! = 3,628,800 – 362,880 = 3,265,920 EASY QUESTIONS 7. In an experiment, the total degrees of freedom increase as ___________ increases. A. Level of significance B. Replication C. Error D. Mean square Answer : B EASY QUESTIONS 8. In a rice yield experiment, pre-experiment data shows varying soil fertility in the field. What experimental design should be used such that the effects of blocking can be determined? A. RCBD B. CRD C. Latin square D. DMRT Answer : A EASY QUESTIONS 9. A ranch is composed of 2,500 heads of Holstein- Friesian and 7,500 heads of Brown Swiss cattle. Thirty representative heads are to be selected at random for use in an experiment. What is the probability that a Brown Swiss will be randomly picked up at first selection? A. 25% B. 50% C. 75% D. 100% Answer : C Solution: Probability = Number of heads of Brown Swiss / Total Number of Heads x 100% = 7,500 / (2,500+7,500) x 100% = 75% MODERATE QUESTIONS 10. A single-factor experiment using completely randomized experimental design results to the following statistical values: treatment sum of squares of 2,510.5, error mean square error of 13.4, error degrees of freedom of 12 and total degrees of freedom of 14. How many treatments are involved? A. 3 B. 4 C. 5 D. 6 Answer : A Solution: Tr DF = Total DF – Error DF = 14-12 = 2 No. of treatments = Tr DF + 1 = 2+1 = 3 MODERATE QUESTIONS 11. In random sampling for the location of different treatments in the field, the labels for Treatments 1, 2, 3, 4 and 5 were placed inside a box. Treatment 3 came out in the first pick. What is the probability that you can pick Treatment 1 in the second pick? A. 20% B. 25% C. 33.3% D. 50% Answer : B Solution: Probability = 1/No. of Treatments Left Unpicked = 1/(5-1) = 0.25 = 25% MODERATE QUESTIONS 12. The return period of a flood stage of 5 meters or more is 2 years. What is the probability that this event will be reached next year? A. 2% B. 20% C. 50% D. 75% Answer : C Solution: Probability = 1/ return Period =½ = 0.5 = 50% DIFFICULT QUESTIONS 13. In Analysis of Variance, what is the F value of an experiment having treatment sum of squares of 2,510.5, error mean square error of 13.4, treatment degrees of freedom of 2 and error degrees of freedom of 12? A. 15.61 B. 15.16 C. 93.86 D. 93.68 Answer : D Solution: Tr MS = TrSS/Tr DF = 2510.5/2 = 1255.25 F = Tr MS/MSE = 1255.25/13.4 = 93.68 DIFFICULT QUESTIONS 14. You need to present a graph of x values vs. y values. The x values are 0, 1, 2, 3, 4 and 5. The y values are ln 0, ln 1, ln 2, ln 3, ln 4 and ln 5, respectively. Since ln 0 results to a mathematical error, what value will you use for ln 0? Use 2 decimal places in your computations. A. -6.91 B. -4.61 C. Infinity D. 6.91 Answer : B Solution: Since infinity cannot be used for graphing and computation is limited to 2 decimal places, use ln 0.01 (closest to ln 0) instead of ln 0, then ln 0.01 = -4.61 DIFFICULT QUESTIONS 15. In a regression equation Y = 2.5 + 3X, where X is the flood stage in meters at a certain location and Y is the estimated amount in million Php of damaged crops downstream. If flood stage is 20 meters, what is the estimated amount of damaged crops in million Php? A. 62.5 B. 65.2 C. 26.5 D. 25.6 Answer : A Solution: Y = 2.5 + 3X = 2.5 + 3(20) = Php62.5 million V. Agricultural and Fishery Sciences (5%) EASY QUESTIONS 1. Per Department of Agriculture Administrative Order No. 4, series of 2022, the minimum inorganic fertilizer rates should be ___ bags/hectare (3 bags urea plus three bags 14-14- 14 or three bags urea plus two bags 16-20-0 and one bag 0-0-60). A. 5 B. 6 C. 7 D. 8 Answer : B EASY QUESTIONS 2. Per BFAR standard, the intensive stocking density of tilapia is __ heads per square meter. A. 3 B. 4-8 C. >8 D. 3-5 Answer : C EASY QUESTIONS 3. Intensive fish stocking density needs _________ feeds. A. Natural feeds B. Commercial C. Organic feeds D. Plankton Answer : B EASY QUESTIONS 4. Per BFAR standard, the extensive stocking density of tilapia is __ heads per square meter. A. 3 B. 4-8 C. >8 D. 3-5 Answer : A EASY QUESTIONS 5. Tilapia fingerlings need feeds equivalent to ____% of body weight. A. 2-5 B. 5-10 C. 10-15 D. 15-30 Answer : C EASY QUESTIONS 6. Tilapia feeding frequency. A. Once in 2 days B. Once daily C. 2-3 times daily D. 3-4 times daily Answer : C EASY QUESTIONS 7. A marketable size tilapia weighs ___ grams/head. A. >76 B. >81 C. >86 D. >91 Answer : D EASY QUESTIONS 8. Tilapia fry needs feeds equivalent to ____% of body weight. A. 2-5 B. 5-10 C. 10-15 D. 15-30 Answer : D MODERATE QUESTIONS 9. How many mango seedlings are needed to be purchased for 1 hectare area if the planting density is 10 meters by 10 meters. Give 5% allowance in the number of seedlings purchased. A. 100 B. 105 C. 150 D. 525 Answer : B Solution: No. of Seedlings = Safety Factor [Area in ha x 10,000 m2 /ha / Planting Density in m x m] = 1.05 [1 ha x 10,000 m2 /ha / (10 m x 10 m)] = 105 seedlings/hectare MODERATE QUESTIONS 10. If the recommended rate of nitrogen is 90 kg/hectare, how many kilograms of urea fertilizer are to be applied per hectare using 46-0-0 urea fertilizer alone? A. 195 B. 196 C. 198 D. 200 Answer : B Computation procedure: To determine the fertilizer rate for a particular nutrient, multiply the rate of the desired nutrient by 100 and divide by the percentage of the nutrient in the fertilizer (Manitoba Soil Fertility Guide, 2022). Solution: Fertilizer rate = (Rate of Desired Nutrient x 100) / Percentage of the Nutrient in the Fertilizer = (90 kg/ha x 100) / 46 = 195.65 kg/ha MODERATE QUESTIONS 11. Based on the experimental field’s soil analysis, the recommended fertilization rate for P2O5 is 45 kg/ha. What should be the fertilization application rate of P2O5 in kilograms/hectare if using 0-22-0 fertilizer alone? A. 200 B. 202 C. 205 D. 207 Answer : C Solution: Fertilizer rate = (Rate of Desired Nutrient x 100) / Percentage of the Nutrient in the Fertilizer = (45 kg/ha x 100) / 22 = 204.5 kg/ha MODERATE QUESTIONS 12. Based on the farm’s soil analysis, the recommended fertilization rate for K2O is 16.8 kg/ha. What should be the weight per unit area in kilograms/hectare of K2O to be applied if using 0- 0-60 fertilizer alone? A. 28 B. 82 C. 2.8 D. 8.2 Answer : C Solution: Fertilizer rate = (Rate of Desired Nutrient x 100) / Percentage of the Nutrient in the Fertilizer = (16.8 kg/ha x 100) / 60 = 28 kg/ha DIFFICULT QUESTIONS 13. If the recommended rate of nitrogen is 90 kg/hectare, how many bags of urea fertilizer are needed per hectare using 46-0-0 urea containing 50 kilograms per bag? A. 195 B. 196 C. 198 D. 200 Answer : B Solution: Computation procedure: To obtain the number of bags of fertilizer, divide the fertilizer requirement in kg/hectare by the number of kilograms contained per bag, then round up (not round off) to the nearest integer Fertilizer rate = (Rate of Desired Nutrient x 100) / % of the Nutrient in the Fertilizer = (90 kg/ha x 100) / 46 = 195.65 kg/ha No. of Bags = Fertilizer Requirement in kg/hectare / No. of Kilograms Contained per Bag = 195.56 kg/ha / 50 kg/bag = 3.91 bags/ha, round up to 4 bags/ha DIFFICULT QUESTIONS 14. If tilapia stocking relies only on natural food source, how many tilapia fingerlings should be stocked in each of 50 experimental cages, each cage has a diameter of 1 meter and depth of 1 meter? Mortality safety factor is 7% for small cages. A. 2 B. 3 C. 126 D. 162 Answer : B Solution: Tilapia extensive stocking density (relies only on natural food source): 3 heads/m² Area of cage = 3.14 (1 m)²/4 = 0.785 m² Stocks = Density x Area x [1+ safety factor in ratio] = 3 x 0.785 x (1.07) = 3 heads/cage  END OF AREA III By Engr. Arthur It. Tambong, FPSAE