NCEES PE Agricultural and Biological Engineering · USA · NCEES (ASABE is the content partner)
Original practice questions written to the published syllabus · not past papersNCEES PE Agricultural and Biological Engineering is a licensing and registration exam run by NCEES (ASABE is the content partner), for candidates in USA. 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.
Source: ncees.org.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Machinery-systems problems that use sprayer and spreader rate equations, field capacity, and economic analysis (annualised cost, break-even area) of precision equipment.
Using positioning, sensing and variable-rate control to manage fields by the metre instead of by the hectare.
Our 20 NCEES PE Agricultural and Biological Engineering questions in this subject: 1 easy, 11 medium, 8 hard, across 7 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to NCEES PE Agricultural and Biological Engineering and are not shown on any other exam's page.
An RTK receiver is specified at 10 mm + 1 ppm horizontal. The expected horizontal accuracy at a 25 km baseline is about
D. 35 mm
1 ppm of 25 km = 25 × 10⁶ mm × 10⁻⁶ = 25 mm; add the 10 mm constant: 35 mm. 11 mm treats 1 ppm as 1 mm; 25 mm drops the constant; 260 mm treats ppm as parts per hundred thousand.
Why is GNSS vertical accuracy usually poorer than horizontal accuracy?
D. All visible satellites are above the horizon, so vertical geometry is weaker
No satellites are below the receiver, so vertical geometry (VDOP) is worse than horizontal. Height is computed from the ranges like the other coordinates; tyre pressure changes are small and are not the reason for the GNSS characteristic.
Auto-steer reduces a 12 m drill's overlap from 0.6 m to 0.05 m. At 10 km/h and field efficiency 0.70 (unchanged), effective field capacity rises by about
D. 4.8 %
Effective widths are 11.4 m and 11.95 m. Capacities are 10 × 11.4 × 0.7 / 10 = 7.98 ha/h and 10 × 11.95 × 0.7 / 10 = 8.37 ha/h; ratio 1.048, so +4.8 %. 0.5 % is roughly the new 0.05 m overlap as a fraction of the width, 5.0 % is the old 0.6 m overlap as a fraction of the 12 m width, and 12 % confuses width with a percentage.
A combine measures 12 kg/s wet grain flow at 6 km/h with a 7.5 m cutting width. The wet yield is
B. 9.6 t/ha
Speed = 6/3.6 = 1.667 m/s; area rate = 1.667 × 7.5 = 12.5 m²/s; yield = 12 / 12.5 = 0.96 kg/m² = 9.6 t/ha. 0.96 t/ha forgets ×10 for kg/m² to t/ha; 2.67 uses speed in km/h without conversion (12/(6×7.5)×10 = 2.67 t/ha). 96 t/ha multiplies by 100 instead of 10.
10 t of maize is harvested at 22 % moisture (wet basis). Its mass at the 15 % standard moisture is
D. 9.18 t
Dry matter is conserved: 10 × (100 − 22)/(100 − 15) = 10 × 78/85 = 9.18 t. 9.30 t subtracts the 7-point difference directly; 10.90 t inverts the ratio; 8.50 t is just 10 × 0.85.
A sprayer without speed compensation is calibrated at 200 L/ha at 8 km/h. The operator speeds up to 10 km/h at the same pressure. The applied rate becomes
A. 160 L/ha
At constant flow, rate is inversely proportional to speed: 200 × 8/10 = 160 L/ha. 250 L/ha inverts the ratio; 200 L/ha assumes speed compensation that the sprayer lacks. 179 L/ha applies the square-root nozzle-pressure law to speed (200 × √0.8).
Why does direct injection of chemical concentrate give a delayed rate response compared with changing carrier flow at the nozzle?
C. The concentrate must travel through the boom plumbing from the injection point to the nozzles
With injection, a change in concentration only reaches the nozzles after the carrier transports it along the boom, causing a transport delay. Injection avoids tank mixing; the pump runs continuously; nozzles are not changed.
A consultant writes a nitrogen prescription on a 5 m × 5 m grid for a spinner spreader with a 24 m throw and a 10 m rate-transition distance. What is the main problem?
B. The grid is far finer than the machine can apply, so the detail is lost
The spreader treats a 24 m-wide band and needs about 10 m to change rate, so 5 m cells cannot be honoured; the prescription should match machine resolution. Finer grids do not make the spreader hunt less; spinner spreaders routinely apply N by map.
Why does individual-nozzle control reduce overlap more than a boom divided into six 6 m sections?
B. The unwanted wedge at an angled edge is limited to about one nozzle spacing
When a boom meets covered ground at an angle, each section stays on until its whole width is covered; narrower sections leave smaller wedges, down to one nozzle spacing. Sections switch on the move; all systems need a coverage map; nozzle control is on/off, not half-rate.
A variable-rate controller kit costs 9000 currency units with a 5-year life, no salvage and 8 % interest. Expected savings are 18 units/ha on 150 ha per year. What is the annual net benefit?
D. +446 units/year
CRF(8 %, 5 yr) = 0.2505; annual cost = 9000 × 0.2505 = 2254; savings = 18 × 150 = 2700; net = 2700 − 2254 = +446. +900 ignores interest (2700 − 1800); −446 reverses the sign; +2700 ignores the cost.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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