PracticeVideosBlogFlashcardsEspañol
All Practice Exams

100+ Free JCE Mathematics (Syllabus 13) Practice Questions

Botswana Examinations Council Junior Certificate Examination Mathematics (Syllabus 13) practice questions are available now; exam metadata is being verified.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
Not published by BEC Pass Rate
100+ Questions
100% Free

Loading practice questions...

Same family resources

Explore More Botswana Junior Certificate Examination

Continue into nearby exams from the same family. Each card keeps practice questions, study guides, flashcards, videos, and articles in one place.

Botswana Junior Certificate Examination — Agriculture (16)

Practice Questions

Botswana Junior Certificate Examination — Art (33)

Practice Questions

Botswana Junior Certificate Examination — Commerce and Accounting (26)

Practice Questions

Botswana Junior Certificate Examination — Commerce and Office Procedures (25)

Practice Questions

Botswana Junior Certificate Examination — Design and Technology (17)

Practice Questions

Botswana Junior Certificate Examination — English (12)

Practice Questions
2026 Statistics

Key Facts: JCE Mathematics (Syllabus 13) Exam

Syllabus 13

Official BEC Code

Botswana Examinations Council

100

Practice MCQs

BEC JCE Assessment Adaptation

1h + 2h

Exam Duration (P13/1 + P13/2)

BEC Examination Timetable

40% / 60%

Paper Weighting (P13/1 / P13/2)

BEC JCE Mathematics Syllabus

Grade C

Credit Passing Standard

BEC Grading Standard

BEC JCE Mathematics (Syllabus 13) is Botswana's national Form 3 secondary mathematics qualification. It features Paper 13/1 (40 multiple-choice items, 1 hour, 40%) and Paper 13/2 (structured problem-solving, 2 hours, 60%). Grade C or higher is required for senior secondary (BGCSE) progression. The 100 questions on this page are an English-language MCQ study adaptation, not an official BEC paper simulation.

Sample JCE Mathematics (Syllabus 13) Practice Questions

Try these sample questions to test your JCE Mathematics (Syllabus 13) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Evaluate the expression: -12 + 4 × (-5) - (-8)
A.-24
B.-36
C.-8
D.0
Explanation: Following the order of operations (BODMAS/BIDMAS), perform multiplication first: 4 × (-5) = -20. Then substitute back into the expression: -12 + (-20) - (-8) = -12 - 20 + 8 = -32 + 8 = -24.
2Calculate 2 1/3 ÷ 1 3/4 and give the answer in its simplest fraction form.
A.1 1/3
B.1 1/4
C.4 1/12
D.7/12
Explanation: Convert the mixed numbers to improper fractions: 2 1/3 = 7/3 and 1 3/4 = 7/4. Dividing fractions requires multiplying by the reciprocal: 7/3 ÷ 7/4 = 7/3 × 4/7 = 28/21 = 4/3 = 1 1/3.
3Find the Highest Common Factor (HCF) of 48, 72, and 108.
A.12
B.6
C.24
D.432
Explanation: Express each number as a product of prime factors: 48 = 2^4 × 3, 72 = 2^3 × 3^2, and 108 = 2^2 × 3^3. The HCF takes the lowest power of common prime factors: 2^2 × 3^1 = 4 × 3 = 12.
4Express 0.000456 in standard form (scientific notation).
A.4.56 × 10^-4
B.4.56 × 10^-3
C.45.6 × 10^-5
D.4.56 × 10^4
Explanation: Standard form requires a number A such that 1 ≤ A < 10 multiplied by a power of 10. Moving the decimal point 4 places to the right gives 4.56, so the exponent is -4: 4.56 × 10^-4.
5Simplify the expression (3^4 × 3^-2) ÷ 3^5 and express the answer with a positive index.
A.1/27
B.1/9
C.3
D.27
Explanation: Using index laws: numerator is 3^4 × 3^-2 = 3^(4 - 2) = 3^2. Dividing gives 3^2 ÷ 3^5 = 3^(2 - 5) = 3^-3. Expressed as a positive index, 3^-3 = 1 / (3^3) = 1/27.
6The population of a village in Botswana increased from 4,000 to 4,860 over a three-year period. Calculate the percentage increase in population.
A.21.5%
B.17.7%
C.20.0%
D.86.0%
Explanation: First calculate the actual increase: 4,860 - 4,000 = 860. The percentage increase is (Increase ÷ Original Value) × 100 = (860 ÷ 4,000) × 100 = 0.215 × 100 = 21.5%.
7A retail shop in Gaborone sells a smartphone for P1,840 after applying a 20% discount. What was the original price of the smartphone before the discount?
A.P2,300
B.P2,208
C.P2,160
D.P1,472
Explanation: The discounted price P1,840 represents (100% - 20%) = 80% of the original price. Let original price be x: 0.80x = 1,840. Therefore, x = 1,840 ÷ 0.80 = P2,300.
8Calculate the simple interest earned on an investment of P6,500 deposited at an annual interest rate of 6% for 4 years.
A.P1,560
B.P1,300
C.P8,060
D.P1,703
Explanation: Simple interest formula: I = (P × R × T) / 100. Substituting P = 6,500, R = 6, and T = 4 gives I = (6,500 × 6 × 4) / 100 = 156,000 / 100 = P1,560.
9Mpho invests P10,000 in a commercial bank account compounding annually at a rate of 5% per annum. What is the total value of the investment at the end of 2 years?
A.P11,025
B.P11,000
C.P1,025
D.P10,500
Explanation: Compound interest total amount formula: A = P(1 + r/100)^n. For P = 10,000, r = 5, n = 2: A = 10,000 × (1.05)^2 = 10,000 × 1.1025 = P11,025.
10A trader buys a box of oranges for P150 and sells all the oranges for P195. Calculate the percentage profit made by the trader.
A.30%
B.23%
C.45%
D.25%
Explanation: Profit = Selling Price - Cost Price = P195 - P150 = P45. Percentage profit = (Profit ÷ Cost Price) × 100 = (45 ÷ 150) × 100 = 0.30 × 100 = 30%.

About the JCE Mathematics (Syllabus 13) Practice Questions

Verified exam format metadata for Botswana Examinations Council Junior Certificate Examination Mathematics (Syllabus 13) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.