Diese Aufgaben gibt es in ihrer Originalsprache.
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.
Originale Aufgaben von Studoli-Lehrkräften im Stil der Prüfung JAMB UTME. Es sind keine offiziellen Prüfungsaufgaben.
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.
Thema: Number bases
1. Convert 101101₂ to base ten.
Lösung
C. 45
- 101101₂ = 1×2⁵ + 0×2⁴ + 1×2³ + 1×2² + 0×2¹ + 1×2⁰.
- = 32 + 0 + 8 + 4 + 0 + 1 = 45.
Thema: Indices
2. Evaluate 25^(−3/2).
Lösung
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.
Thema: Logarithms
3. Simplify log₂8 + log₂4 − log₂2.
Lösung
A. 4
- log₂8 = 3, log₂4 = 2 and log₂2 = 1.
- 3 + 2 − 1 = 4.
Thema: Indices
4. Solve 2^(x + 1) = 32.
Lösung
D. x = 4
- Write 32 as a power of 2: 32 = 2⁵.
- So x + 1 = 5, which gives x = 4.
Thema: Surds
5. Simplify √12 + √27.
Lösung
C. 5√3
- √12 = √(4 × 3) = 2√3.
- √27 = √(9 × 3) = 3√3.
- 2√3 + 3√3 = 5√3.
Thema: Modular arithmetic
6. Evaluate 5 × 4 (mod 7).
Lösung
B. 6
- 5 × 4 = 20.
- 20 ÷ 7 = 2 remainder 6, so 20 ≡ 6 (mod 7).
Thema: Simple interest
7. Find the simple interest on ₦20,000 at 5% per annum for 3 years.
Lösung
A. ₦3,000
- Simple interest = PRT ÷ 100.
- = 20,000 × 5 × 3 ÷ 100 = ₦3,000.
Thema: Quadratic equations
8. If α and β are the roots of 2x² − 7x + 3 = 0, find α + β.
Lösung
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.
Thema: Simultaneous equations
9. Solve x + y = 7 and x − y = 1.
Lösung
C. x = 4, y = 3
- Add the equations: 2x = 8, so x = 4.
- Substitute: 4 + y = 7, so y = 3.
Thema: Variation
10. y varies inversely as x, and y = 6 when x = 4. Find y when x = 8.
Lösung
B. 3
- Inverse variation: y = k/x.
- 6 = k/4, so k = 24.
- When x = 8, y = 24/8 = 3.
Thema: Sequences
11. Find the 5th term of the arithmetic progression 3, 7, 11, ...
Lösung
A. 19
- The first term a = 3 and the common difference d = 4.
- Tₙ = a + (n − 1)d, so T₅ = 3 + 4 × 4 = 19.
Thema: Sequences
12. Find the sum of the first 4 terms of the geometric progression 2, 6, 18, ...
Lösung
D. 80
- a = 2 and r = 3.
- Sₙ = a(rⁿ − 1)/(r − 1) = 2(3⁴ − 1)/(3 − 1).
- = 2 × 80 ÷ 2 = 80.
Thema: Sets
13. n(A) = 12, n(B) = 9 and n(A ∩ B) = 4. Find n(A ∪ B).
Lösung
C. 17
- n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
- = 12 + 9 − 4 = 17.
Thema: Matrices
14. Find the determinant of the matrix with rows (3, 2) and (1, 4).
Lösung
B. 10
- For rows (a, b) and (c, d), the determinant is ad − bc.
- = 3 × 4 − 2 × 1 = 12 − 2 = 10.
Thema: Differentiation
15. Find the gradient of the curve y = 3x² − 4x + 1 at x = 2.
Lösung
A. 8
- dy/dx = 6x − 4.
- At x = 2: 6 × 2 − 4 = 8.
Thema: Integration
16. Find ∫(4x³ + 2x) dx.
Lösung
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.
Thema: Statistics
17. Find the median of 3, 8, 5, 9, 4, 7.
Lösung
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.
Thema: 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.
Lösung
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.
Thema: Trigonometry
19. If sin θ = 3/5 and θ is acute, find tan θ.
Lösung
A. 3/4
- Draw a right-angled triangle with opposite = 3 and hypotenuse = 5.
- Adjacent = √(5² − 3²) = 4.
- tan θ = opposite ÷ adjacent = 3/4.
Thema: Mensuration
20. Find the volume of a cylinder of radius 7 cm and height 10 cm. (Take π = 22/7.)
Lösung
D. 1540 cm³
- V = πr²h.
- = 22/7 × 7 × 7 × 10 = 22 × 7 × 10 = 1540 cm³.