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100+ Free OSSC Soil Conservation Extension Worker Practice Questions

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Key Facts: OSSC Soil Conservation Extension Worker Exam

200 MCQs

Total questions in the official Main Written Exam

Official OSSC Notification

180 Minutes

Total duration allowed for the Main Technical paper

Exam Guidelines

₹0 / Free

Application fee for all categories of candidates

State Gov Guidelines

0.25

Negative marking deduction for each incorrect answer

Marking Scheme

The OSSC SCEW exam features a 150-mark Prelims (120 mins) and a 200-mark Mains (180 mins) with 0.25 negative marking. Candidates with +2 Science/Vocational in Agriculture are eligible. There is no exam fee.

Sample OSSC Soil Conservation Extension Worker Practice Questions

Try these sample questions to test your OSSC Soil Conservation Extension Worker exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1If the price of urea fertilizer is increased by 25%, by how much percent must a farmer reduce its consumption so that his total expenditure remains unchanged?
A.20%
B.25%
C.15%
D.30%
Explanation: Let the original price be 100 and consumption be 100. New price becomes 125. To keep expenditure constant at 10,000, new consumption must be 10,000 / 125 = 80. Therefore, the reduction in consumption is 20%.
2A farm machine is sold at a profit of 10%. If it had been bought for 10% less and sold for Rs. 30 more, the gain would have been 25%. Find the cost price of the machine.
A.Rs. 1,200
B.Rs. 1,500
C.Rs. 1,800
D.Rs. 2,000
Explanation: Let the cost price (CP) be 100x. Selling price (SP) is 110x. New CP is 90x. New SP with 25% profit is 90x * 1.25 = 112.5x. The difference is 112.5x - 110x = 2.5x. Since 2.5x = 30, x = 12. Thus, the original CP is Rs. 1,200.
3The average paddy yield of 10 experimental plots is 15 quintals. If the highest and lowest yields are excluded, the average yield of the remaining plots is 14.5 quintals. If the highest yield is 22 quintals, what is the lowest yield?
A.6 quintals
B.8 quintals
C.10 quintals
D.12 quintals
Explanation: Total yield of all 10 plots = 150 quintals. Total yield of the 8 plots after excluding highest and lowest = 8 * 14.5 = 116 quintals. The sum of the highest and lowest yields = 150 - 116 = 34 quintals. If the highest is 22 quintals, the lowest is 34 - 22 = 12 quintals.
4Two varieties of liquid organic fertilizer containing nitrogen and phosphorus in ratios 2:1 and 3:1 are mixed in the ratio 1:2. What is the ratio of nitrogen to phosphorus in the final mixture?
A.11:5
B.13:5
C.15:7
D.17:9
Explanation: Let us take 12 units of the first and 24 units of the second fertilizer. In the first (12 units), nitrogen = 8, phosphorus = 4. In the second (24 units), nitrogen = 18, phosphorus = 6. Total nitrogen = 8 + 18 = 26, total phosphorus = 4 + 6 = 10. The ratio is 26:10, which simplifies to 13:5.
5A farmer borrowed Rs. 15,000 from a cooperative bank at a simple interest rate of 8% per annum. How much total amount will he pay to clear the debt after 3 years and 6 months?
A.Rs. 18,200
B.Rs. 19,200
C.Rs. 20,400
D.Rs. 19,800
Explanation: Simple Interest (SI) = (P * R * T) / 100. Time = 3.5 years. Interest = (15000 * 8 * 3.5) / 100 = Rs. 4,200. Total amount to be repaid = Principal + Interest = 15000 + 4200 = Rs. 19,200.
6What will be the compound interest on a sum of Rs. 20,000 for 2 years at a rate of 10% per annum, compounded annually?
A.Rs. 4,000
B.Rs. 4,100
C.Rs. 4,200
D.Rs. 4,400
Explanation: Amount A = P * (1 + R/100)^T = 20000 * (1 + 10/100)^2 = 20000 * 1.21 = Rs. 24,200. Compound Interest (CI) = A - P = 24200 - 20000 = Rs. 4,200.
7A mechanical excavator can dig a pond in 12 days, while another excavator can dig the same pond in 15 days. If both machines work together, how many days will they take to complete the excavation?
A.6.22 days
B.7.15 days
C.6.67 days
D.5.85 days
Explanation: Work rate of the first machine = 1/12 per day. Work rate of the second machine = 1/15 per day. Combined rate = 1/12 + 1/15 = 9/60 = 3/20. Total days needed = 20/3 = 6.67 days.
8A tractor travels at a constant speed of 18 km/h. How much distance in meters will the tractor cover in exactly 20 minutes?
A.5,000 meters
B.6,000 meters
C.7,500 meters
D.4,500 meters
Explanation: Speed of 18 km/h is equal to 18 * (1000 / 3600) = 5 meters per second. Time is 20 minutes = 20 * 60 = 1200 seconds. Distance covered = speed * time = 5 * 1200 = 6,000 meters.
9Which of the following is the least natural number that must be added to 1056 so that the sum is completely divisible by 23?
A.2
B.3
C.18
D.21
Explanation: Dividing 1056 by 23 yields 1056 = 23 * 45 + 21, leaving a remainder of 21. To make it completely divisible, the least number to be added is 23 - 21 = 2.
10The Highest Common Factor (HCF) of two numbers is 11 and their Least Common Multiple (LCM) is 7700. If one of the numbers is 275, find the other number.
A.279
B.308
C.318
D.440
Explanation: Using the relation Product of two numbers = HCF * LCM. The other number = (11 * 7700) / 275 = 7700 / 25 = 308.

About the OSSC Soil Conservation Extension Worker Exam

The OSSC Soil Conservation Extension Worker (SCEW) exam is conducted by the Odisha Staff Selection Commission to recruit technical extension workers for the Soil Conservation department. This practice test contains 100 multiple-choice questions matching the official syllabus, including Mathematics, General Science, and Agriculture/Soil Science subjects. Each question includes a detailed explanation to ensure robust preparation.

Questions

200 scored questions

Time Limit

180 minutes

Passing Score

Selection based on merit

Exam Fee

No fee / waived for all categories (Odisha Staff Selection Commission (OSSC))

OSSC Soil Conservation Extension Worker Exam Content Outline

50%

Mathematics

Arithmetic, algebra, geometry, mensuration, trigonometry, and statistics matching matriculation level standard.

25%

General Science

Core concepts of physics, chemistry, and biology with applications in agriculture and plant sciences.

25%

Agriculture & Soil Science

Odisha soils, agronomic practices, crop production, soil and water conservation techniques, and farm machinery.

How to Pass the OSSC Soil Conservation Extension Worker Exam

What You Need to Know

  • Passing score: Selection based on merit
  • Exam length: 200 questions
  • Time limit: 180 minutes
  • Exam fee: No fee / waived for all categories

Keys to Passing

  • Complete 500+ practice questions
  • Score 80%+ consistently before scheduling
  • Focus on highest-weighted sections
  • Use our AI tutor for tough concepts

OSSC Soil Conservation Extension Worker Study Tips from Top Performers

1Dedicate significant preparation time to Mathematics, as it carries 50% weight in the Main Technical paper.
2Build a strong understanding of Odisha's soil profiles, local crops, and specific agro-climatic zones.
3Review basic concepts of biology and chemistry, particularly soil chemistry, plant nutrition, and plant physiology.
4Practice solving arithmetic, mensuration, and algebra problems under time constraints to improve speed.
5Attempt full-length mock exams and analyze negative marks to minimize careless errors on exam day.

Frequently Asked Questions

What is the educational qualification for OSSC SCEW?

Candidates must have passed +2 Science or +2 Vocational course in Agriculture-related subjects (such as crop production, horticulture, or farm machinery repair) from a recognized Board or Council.

Is there an application fee for OSSC SCEW?

No, the Odisha Government has waived the application fee for all OSSC recruitments, making the application process free for all candidates.

What is the exam pattern for the OSSC SCEW recruitment?

The recruitment consists of a Preliminary Exam (150 marks, 120 minutes), a Main Written Exam (200 marks, 180 minutes), and Certificate Verification. Both exams are objective multiple-choice tests.

Is there negative marking in the OSSC SCEW exam?

Yes, there is negative marking of 0.25 marks for each wrong answer in both the Preliminary and Main examinations.

What is the age limit for applying to this post?

Applicants must be between 21 and 38 years of age. Relaxations of 5 years apply for SC, ST, SEBC, and women candidates, and 10 years for PwD candidates.