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JAMB Mathematics CBT practice: past-question style questions
Twenty objective questions in the style of the JAMB UTME Mathematics CBT: number bases, indices, logarithms, surds, sequences, sets, matrices, calculus, statistics, probability, trigonometry and mensuration. Use the timed CBT mode, then read the worked solutions.
Questões originais escritas por professores da Studoli no estilo da prova JAMB UTME. Não são provas oficiais.
Timed mode
20 questions in 14 minutes, one question at a time. Your answers are marked when you submit or when the time runs out.
Tema: Number bases
1. Convert 101101₂ to base ten.
Solução
C. 45
- 101101₂ = 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰.
- = 32 + 0 + 8 + 4 + 0 + 1 = 45.
Tema: Indices
2. Evaluate 25^(−3/2).
Solução
B. 1/125
- A negative power means a reciprocal: 25^(−3/2) = 1 ÷ 25^(3/2).
- 25^(1/2) = 5, so 25^(3/2) = 5³ = 125.
- The answer is 1/125.
Tema: Logarithms
3. Simplify log₂8 + log₂4 − log₂2.
Solução
A. 4
- log₂8 = 3, log₂4 = 2 and log₂2 = 1.
- 3 + 2 − 1 = 4.
Tema: Indices
4. Solve 2^(x + 1) = 32.
Solução
D. x = 4
- Write 32 as a power of 2: 32 = 2⁵.
- So x + 1 = 5, which gives x = 4.
Tema: Surds
5. Simplify √12 + √27.
Solução
C. 5√3
- √12 = √(4 × 3) = 2√3.
- √27 = √(9 × 3) = 3√3.
- 2√3 + 3√3 = 5√3.
Tema: Modular arithmetic
6. Evaluate 5 × 4 (mod 7).
Solução
B. 6
- 5 × 4 = 20.
- 20 ÷ 7 = 2 remainder 6, so 20 ≡ 6 (mod 7).
Tema: Simple interest
7. Find the simple interest on ₦20,000 at 5% per annum for 3 years.
Solução
A. ₦3,000
- Simple interest = PRT ÷ 100.
- = 20,000 × 5 × 3 ÷ 100 = ₦3,000.
Tema: Quadratic equations
8. If α and β are the roots of 2x² − 7x + 3 = 0, find α + β.
Solução
D. 7/2
- For ax² + bx + c = 0, the sum of the roots is −b/a.
- Here a = 2 and b = −7, so α + β = 7/2.
Tema: Simultaneous equations
9. Solve x + y = 7 and x − y = 1.
Solução
C. x = 4, y = 3
- Add the equations: 2x = 8, so x = 4.
- Substitute: 4 + y = 7, so y = 3.
Tema: Variation
10. y varies inversely as x, and y = 6 when x = 4. Find y when x = 8.
Solução
B. 3
- Inverse variation: y = k/x.
- 6 = k/4, so k = 24.
- When x = 8, y = 24/8 = 3.
Tema: Sequences
11. Find the 5th term of the arithmetic progression 3, 7, 11, ...
Solução
A. 19
- The first term a = 3 and the common difference d = 4.
- Tₙ = a + (n − 1)d, so T₅ = 3 + 4 × 4 = 19.
Tema: Sequences
12. Find the sum of the first 4 terms of the geometric progression 2, 6, 18, ...
Solução
D. 80
- a = 2 and r = 3.
- Sₙ = a(rⁿ − 1)/(r − 1) = 2(3⁴ − 1)/(3 − 1).
- = 2 × 80 ÷ 2 = 80.
Tema: Sets
13. n(A) = 12, n(B) = 9 and n(A ∩ B) = 4. Find n(A ∪ B).
Solução
C. 17
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- = 12 + 9 − 4 = 17.
Tema: Matrices
14. Find the determinant of the matrix with rows (3, 2) and (1, 4).
Solução
B. 10
- For rows (a, b) and (c, d), the determinant is ad − bc.
- = 3 × 4 − 2 × 1 = 12 − 2 = 10.
Tema: Differentiation
15. Find the gradient of the curve y = 3x² − 4x + 1 at x = 2.
Solução
A. 8
- dy/dx = 6x − 4.
- At x = 2: 6 × 2 − 4 = 8.
Tema: Integration
16. Find ∫(4x³ + 2x) dx.
Solução
D. x⁴ + x² + c
- Raise each power by 1 and divide by the new power.
- ∫4x³ dx = x⁴ and ∫2x dx = x².
- Add the constant: x⁴ + x² + c.
Tema: Statistics
17. Find the median of 3, 8, 5, 9, 4, 7.
Solução
C. 6
- Arrange in order: 3, 4, 5, 7, 8, 9.
- There are 6 numbers, so the median is the mean of the 3rd and 4th: (5 + 7) ÷ 2 = 6.
Tema: Probability
18. A bag holds 5 red and 3 blue balls. Two balls are picked one after the other without replacement. Find the probability that both are red.
Solução
B. 5/14
- P(first red) = 5/8.
- One red ball is gone, so P(second red) = 4/7.
- 5/8 × 4/7 = 20/56 = 5/14.
Tema: Trigonometry
19. If sin θ = 3/5 and θ is acute, find tan θ.
Solução
A. 3/4
- Draw a right-angled triangle with opposite = 3 and hypotenuse = 5.
- Adjacent = √(5² − 3²) = 4.
- tan θ = opposite ÷ adjacent = 3/4.
Tema: Mensuration
20. Find the volume of a cylinder of radius 7 cm and height 10 cm. (Take π = 22/7.)
Solução
D. 1540 cm³
- V = πr²h.
- = 22/7 × 7 × 7 × 10 = 22 × 7 × 10 = 1540 cm³.