Agricultural and Biosystems Engineers Licensure Examination (ABELE)
Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC · licensing
1532practice questions written to this syllabus
48subjects
Format
Paper-based, 2 days (Day 1 6 hours; Day 2 5 hours + 4 hours). Non-programmable calculator only; the Dec 2025 programme bans the Casio fx-991ES/ES Plus by name.
Mandatory to practise under RA 10915 (2016), which renamed 'Agricultural Engineer' to 'Agricultural and Biosystems Engineer'. Pass: weighted average of at least 70% with no subject below 55% (RA 10915 §18); unofficial guides that say otherwise are wrong. Scope includes aquaculture and forest engineering, laws and ethics, and PAES/PNS standards. Results: Dec 2024, 1,479 of 2,625 passed; Dec 2025, 1,475 of 2,494 passed. No verified topic-frequency data exists; Laisha's PRC questions are written to the syllabus only. Law: https://lawphil.net/statutes/repacts/ra2016/ra_10915_2016.html
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. Three pumps A, B and C deliver a, b and c m³/h. Running all three for 1 h delivers 360 m³. Running A for 2 h and B for 1 h delivers 400 m³. Running A for 1 h and C for 2 h delivers 370 m³. Solving the linear system, what is the discharge of pump A?
110 m³/h
200 m³/h
100 m³/h
150 m³/h
Answer and working
D. 150 m³/h
Equations: a + b + c = 360, 2a + b = 400, a + 2c = 370. Subtract the second from the first: −a + c = −40, so c = a − 40. Substitute in the third: a + 2a − 80 = 370, a = 150 m³/h; then b = 400 − 300 = 100 and c = 110. Check: 150 + 100 + 110 = 360. 100 and 110 are the discharges of B and C; 200 is 400/2 from ignoring B in the second equation.
Engineering Mathematics · Matrices, determinants and systems of linear equations
2. A trailer hitch pin of 20 mm diameter connects a drawbar clevis and carries a pull of 30 kN in double shear. The average shear stress in the pin is closest to:
47.7 MPa
95.5 MPa
11.9 MPa
37.5 MPa
Answer and working
A. 47.7 MPa
Double shear means two planes carry the load. A = π(20)²/4 = 314.2 mm²; τ = 30,000/(2 × 314.2) = 47.7 MPa. Using one plane gives 95.5 MPa (single-shear slip); using πd² as the area gives 30,000/(2 × 1256.6) = 11.9 MPa, and d² instead of πd²/4 gives 30,000/(2 × 400) = 37.5 MPa.
Strength of Materials · Stress, strain and Hooke's law
3. In first-angle orthographic projection, where is the top view (plan) placed relative to the front view?
Directly above the front view, aligned on the same vertical projectors
To the right of the front view, aligned on the same horizontal projectors
Anywhere on the sheet, provided it is labelled as the top view
Directly below the front view, aligned on the same vertical projectors
Answer and working
D. Directly below the front view, aligned on the same vertical projectors
In first angle the object lies between the observer and the plane; when the HP is rotated down about XY, the top view falls below the front view. Placing the top view above the front view is the third-angle arrangement, and views are never placed arbitrarily when a projection system is declared.
4. How should a hacksaw blade be fitted in its frame?
With the teeth pointing away from the handle, because it cuts on the forward stroke
With the teeth pointing towards the handle, because it cuts on the pulling stroke
Either way round, because a hacksaw blade is made to cut equally on both strokes
With the teeth facing upwards, so that chips fall clear of the cut while sawing
Answer and working
A. With the teeth pointing away from the handle, because it cuts on the forward stroke
A hand hacksaw cuts on the push (forward) stroke, so the teeth point away from the handle and pressure is relieved on the return. Teeth towards the handle is the arrangement of a pull saw, not a standard hacksaw; the blade is not symmetrical and does not cut both ways.
Workshop Technology & Manufacturing · Fitting, carpentry and smithy
5. What are the full-load line currents of a 100 kVA, 11 kV/433 V three-phase distribution transformer?
5.25 A HV; 133 A LV
133 A HV; 5.25 A LV
9.09 A HV; 231 A LV
3.03 A HV; 77 A LV
Answer and working
A. 5.25 A HV; 133 A LV
I = S/(√3 V_L): HV = 100 000/(1.732 × 11 000) = 5.25 A; LV = 100 000/(1.732 × 433) = 133 A. Dividing by V alone treats it as single-phase; dividing by 3V uses line voltage with a phase-voltage formula; the swapped pair puts the larger current on the high-voltage side.