UGC-NET (June 2025) · India · National Testing Agency
Original practice questions written to the published syllabus · not past papersDesigning and running production: location, layout, planning, inventory, quality and productivity, applied to processing plants and agribusiness operations.
In the exams: agribusiness management PhD entrance and fellowship examinations, production and operations (heavy)
UGC-NET Management (subject 17)
Production and operations management at MBA level: process-product matrix, capacity measures, layout, aggregate planning, scheduling rules, work study, inventory classification, MRP and ERP, JIT and lean, TQM thinkers, SPC and acceptance sampling, ISO standards, BPR, value engineering and maintenance. Mostly one-step conceptual and matching questions with occasional short numericals.
How our questions for UGC-NET spread across the units of Production & Operations Management. Difficulty: 33 easy, 68 medium, 23 hard.
| TQM, statistical process control and acceptance sampling | 16 | |
| Master production schedule | 9 | |
| Business process re-engineering, value engineering and lean | 9 | |
| Work study | 8 | |
| Standards: ISO, HACCP, national standards bodies and grade marks | 8 | |
| Maintenance management | 8 | |
| Productivity | 7 | |
| Facility location | 7 | |
| Plant layout | 7 | |
| Aggregate planning: level, chase and mixed | 7 | |
| Materials management, MRP, MRP II and ERP | 7 | |
| Inventory management | 7 | |
| JIT and Kanban | 7 | |
| Process types: job, batch and continuous | 6 | |
| Forecasting | 6 | |
| Process-product matrix | 5 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A sugar mill's crushing line has four sequential stages with hourly capacities of 120 t (cane preparation), 100 t (milling), 90 t (clarification) and 110 t (evaporation). What is the line's capacity, and where should the next investment go?
C. 90 t per hour; expand clarification, the bottleneck
In a sequential process the system capacity equals the slowest stage, here clarification at 90 t/h. Averaging stages (105) or taking the first stage (120) overstates capacity; adding capacity elsewhere adds nothing until the bottleneck is relieved, after which milling at 100 t/h becomes the new bottleneck.
Further reading: Stevenson, Operations Management
As a processing operation moves from job-shop towards continuous-flow production, which change is expected?
B. Unit variable cost falls while fixed cost and capital intensity rise
Continuous processes spread high fixed cost over large volumes, so unit variable cost falls, but flexibility falls and work-in-process is lower. Craft skills matter most in job shops; continuous plants need monitoring skills.
Further reading: Stevenson, Operations Management
A dal mill compares two sites. Site A: fixed cost 3,000,000 currency units a year, variable cost 400 per tonne. Site B: fixed cost 5,000,000 currency units a year, variable cost 300 per tonne. At what annual volume do the two sites cost the same?
A. 20,000 tonnes
Set 3,000,000 + 400V = 5,000,000 + 300V, so V = (difference in fixed cost)/(difference in variable cost) = 2,000,000/100 = 20,000 tonnes. Adding the fixed costs (80,000) or the variable costs (2,857) instead of subtracting them gives the wrong crossover; 5,000 divides the fixed-cost gap by Site A's full variable cost of 400 instead of the 100 gap.
Further reading: Heizer, Render and Munson, Operations Management
A custom flour mill has five orders on one machine, with processing time and due date in days: P 6 and 10, Q 2 and 6, R 8 and 18, S 3 and 15, T 9 and 23. Using the shortest-processing-time rule, what is the mean flow time?
C. 13.0 days
SPT order Q, S, P, R, T gives completion times 2, 5, 11, 19, 28; mean = 65/5 = 13.0. 15.4 is FCFS and 13.6 is EDD; 28/5 = 5.6 is mean processing time, not flow time.
Further reading: Stevenson, Operations Management
A pilot work-sampling study finds a forklift idle about 25% of the time. How many random observations are needed to estimate idle time within plus or minus 5 percentage points at 95% confidence (z = 1.96)?
B. 289
n = z^2 p(1-p)/e^2 = 1.96^2 x 0.25 x 0.75/0.05^2 = 288.1, rounded up to 289. Reading 5% as relative to p (e = 0.0125) gives about 4,610; forgetting to square z gives 147. 385 uses the conservative p = 0.5 instead of the pilot estimate of 0.25.
Further reading: ILO, Introduction to Work Study
A dairy cooperative is choosing an ERP system. Which feature is most specific to its needs rather than to general manufacturing?
D. Procurement from many farmers with quality-based payment and lot traceability
General ERP covers accounts, payroll and sales. Agribusinesses also need grower accounts with payment on fat and SNF or sugar content, batch traceability and shelf-life control, which standard packages may lack without adaptation.
Further reading: Heizer, Render and Munson, Operations Management
Heijunka in the Toyota Production System refers to:
C. Levelling production volume and mix over time
Heijunka is level, mixed-model scheduling. Stopping the line is jidoka; mistake-proofing is poka-yoke; card signals are kanban.
Further reading: Ohno, Toyota Production System
Which list gives Womack and Jones's five lean principles in order?
D. Specify value, identify the value stream, flow, pull, perfection
Lean Thinking (1996) set out value, value stream, flow, pull and perfection. PDCA is the Shewhart-Deming cycle, DMAIC is Six Sigma, and 5S is workplace organisation.
Further reading: Womack and Jones (1996)
In a flour mill, lorries queue for two hours at the weighbridge before unloading. In lean terms this is primarily:
D. Waiting waste
Idle time of people, vehicles or material is waiting waste. Over-processing adds work the customer does not value; defects are rework or scrap.
Further reading: Ohno, Toyota Production System
For the same filler (subgroups of 5, average range 8 g, D3 = 0, D4 = 2.114), what are the R chart limits?
D. UCL 16.912 g, LCL 0
R chart: UCL = D4 x R-bar = 2.114 x 8 = 16.912; LCL = D3 x R-bar = 0. The A2 factor belongs to the X-bar chart (8 +/- 4.616 gives 12.616 and 3.384); a symmetric lower limit (8 - 8.912 = -0.912) is wrong because D3 = 0 for n = 5; 24 uses 3 x R-bar.
Further reading: Montgomery, Introduction to Statistical Quality Control
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