USA · NCEES (ASABE is the content partner) · licensing
1270practice questions written to this syllabus
48subjects
Format
Computer-based, 85 questions, 9.5-hour appointment (8.5 hours of exam). Once a year (next: 28 Oct 2026). Closed book with the electronic NCEES handbook and supplied searchable ASABE standards; SI and US customary units.
Sections
Common System Applications — 19–29 questions
Natural Resources & Environmental Systems — 15–23
Processing Systems — 12–18
Facilities & Structures — 12–18
Machinery Systems — 12–18
Over the years
Specifications effective October 2023; supplied-standards list effective October 2025 (including ASAE D497.7 and EP496.3 machinery management, D271.3 psychrometric data, D272.3 grain airflow resistance, EP270.5 livestock ventilation, S433.1 grain loads on bins). It tests using standards, not memorising them. Very small candidate pool: 33 first-time takers in the January 2026 pass-rate data. ASABE offers PE fee support: https://asabe.org/abelicensure. Status re-checked on ncees.org on 3 Oct 2026: still offered (not discontinued), computer-based, one day per year; next date 28 Oct 2026; first-time pass rate 91 % (33 takers, January 2026 data).
Original questions written to this exam's published syllabus and pattern. Not past papers.
1. Two grain lots are blended: 10 t at 14 per cent moisture (wet basis) and 30 t at 18 per cent moisture (wet basis). What is the moisture content of the 40 t blend, wet basis?
16.0 per cent
15.0 per cent
17.0 per cent
20.5 per cent
Answer and working
C. 17.0 per cent
Wet-basis moisture is water mass per unit wet mass, so weight by wet mass: water = 0.14 × 10 + 0.18 × 30 = 1.4 + 5.4 = 6.8 t; 6.8/40 = 17.0 per cent. The simple average (14 + 18)/2 = 16 per cent ignores the lot sizes; 15 per cent weights the wrong way round (0.14 × 30 + 0.18 × 10 = 6.0 t, /40). 20.5 per cent is the blend's dry-basis moisture, 6.8/(40 − 6.8) = 6.8/33.2.
Probability, Statistics & Data Analysis · Measures of central tendency and dispersion
2. A farm-machinery test engineer compares a mild-steel and a cast-iron specimen after tension tests. Which observation correctly identifies the mild-steel specimen?
It shows a flat granular fracture with almost no necking at all
It shows a cup-and-cone fracture after visible necking
It breaks without any yield plateau on its load–extension record
It has less elongation at fracture than the cast-iron specimen
Answer and working
B. It shows a cup-and-cone fracture after visible necking
Ductile mild steel necks and breaks with a cup-and-cone surface after large elongation. A flat, granular break with little elongation and no yield plateau is typical of brittle grey cast iron.
Strength of Materials · Stress, strain and Hooke's law
3. In third-angle projection, the view obtained by looking at the object from its right-hand side is placed where?
Below the front view, beside the top view
To the right of the front view
Above the front view, beside the top view
To the left of the front view
Answer and working
B. To the right of the front view
Third angle places each view on the side from which it is seen: the right-side view goes to the right and the top view above. Putting the right-side view on the left is the first-angle arrangement, the classic source of mirror-image errors.
4. A sensor database receives a new row every minute from 400 stations and is queried mostly by station and time range. What is the effect of adding an index on (station_id, timestamp)?
Range queries become slower, because the database must now read the index and the table
Inserts become faster, because rows are written in index order and need no further sorting
Range queries become much faster, while each insert becomes slightly slower and storage grows
Neither speed changes, because an index matters only for tables holding text columns
Answer and working
C. Range queries become much faster, while each insert becomes slightly slower and storage grows
A B-tree index lets the database locate one station's time range without scanning every row; the price is maintaining the index on each insert and the space it occupies. Indexes help numeric and time columns as much as text.
Computer Programming, Data Structures & Agricultural Informatics · Databases
5. A 48 W, 12 V DC lamp in a poultry shed is fed from a solar-charged battery. The cable out and back has a total loop resistance of 0.15 Ω. Taking the lamp current at its rated value, what percentage of the 12 V supply is lost in the cable?
5.3 %
2.5 %
5.0 %
0.26 %
Answer and working
C. 5.0 %
I = P/V = 48/12 = 4 A; drop = I × R_loop = 4 × 0.15 = 0.6 V; 0.6/12 = 5.0 %. Halving it (2.5 %) uses the one-way resistance when 0.15 Ω is already the loop value. 0.26 % comes from dividing by 230 V instead of the 12 V supply. 5.3 % divides by the lamp voltage (11.4 V) instead of the supply voltage.
Basic Electrical & Electronics Engineering · DC circuits and network theorems
Practise it, and prove it
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