JAMB Past Questions

JAMB Mathematics 2008
Questions & Answers

20 questions · Correct answers highlighted · 20 with explanations

Explanations reviewed August 2026

Spanning forty questions, this 2008 mathematics exam paper tests a student's proficiency in arithmetic calculations, algebraic factorizations, and basic geometry. Several questions consistently target topics like indices and logarithms, coordinate geometry involving straight lines and distances, and foundational probability principles. Treat each mistake as a learning opportunity by thoroughly reviewing the provided explanations to build lasting confidence for the actual test.

20 Total Questions
2008 Exam Year
20 With Explanations

Topics in this paper

Algebraic Expressions and Factorization (8) Mensuration and Areas (5) Calculus and Derivatives (5) Ratios Proportions and Percentages (4) Logarithms and Indices (3) Probability and Combinatorics (3) Circle Geometry and Theorems (2) Trigonometric Ratios and Identities (2) Statistics and Central Tendency (2) Coordinate Geometry (2) Sets and Logic (1) Arithmetic Progressions (1) Geometric Progressions (1) Simultaneous Equations (1)
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1
Question 1 of 20
JAMB · Mathematics · 2008

Simplify (0.6)^2 + (0.3)^2 / (0.4)^2 - (0.1)^2

A. 4
B. 2
C. 3
D. 1
Explanation

1) Distractor check: A student might miscalculate the decimal squares or perform improper cancellation, leading to integers like 2 or 4. 2) Reasoning to the answer: Squaring 0.6 and 0.3 yields 0.36 and 0.09 respectively, which sum to 0.45 in the numerator. Squaring 0.4 and 0.1 gives 0.16 and 0.01 respectively, which subtract to 0.15 in the denominator. Dividing 0.45 by 0.15 gives a final quotient of 3. Common mistake: Placing the decimal point incorrectly when squaring numbers less than one.

2
Question 2 of 20
JAMB · Mathematics · 2008

Find p if 451_6 = p_3 + 402_6

A. 611_3
B. 6111_3
C. 111_3
D. 1111_3
Explanation

1) Distractor check: One might mistakenly pick base-3 variations with incorrect digit lengths by failing to perform proper base conversions. 2) Reasoning to the answer: Converting 451 in base 6 to base 10 gives 175, and converting 402 in base 6 gives 146. Subtracting 146 from 175 leaves 49 in base 10. Converting 49 into base 3 yields 1111_3 because 1 times 27 plus 1 times 9 plus 1 times 3 plus 1 equals 49. Common mistake: Errors in place-value expansion when converting between non-decimal number bases.

3
Question 3 of 20
JAMB · Mathematics · 2008

A farmer planted 5000 cobs each bearing 500 grains and harvested 5000 cobs each bearing 600 grains. What is the ratio of grains planted to grains harvested?

A. 1:500
B. 1:600
C. 5:6
D. 6:5
Explanation

1) Distractor check: Option A or B might be chosen by misinterpreting the question as a per-cob or per-grain ratio rather than total quantities. 2) Reasoning to the answer: The total number of grains planted is found by multiplying 5000 cobs by 500 grains, resulting in 2,500,000 grains. Similarly, the total grains harvested equal 5000 cobs multiplied by 600 grains, giving 3,000,000 grains. Reducing the ratio of planted to harvested grains, 2,500,000 to 3,000,000, simplifies to 5:6. Common mistake: Comparing unit values instead of calculating the aggregate totals for both groups.

4
Question 4 of 20
JAMB · Mathematics · 2008

Three teachers shared a packet of chalk. The first got 2/5, the second got 2/15 of the remainder. What fraction did the third receive?

A. 11/25
B. 12/25
C. 13/25
D. 14/25
Explanation

1) Distractor check: A student might choose 12/25 by forgetting to subtract the second fraction's result from the total remainder, stopping at the combined taken amount. 2) Reasoning to the answer: The first teacher takes 2/5 of the total, leaving a remainder of 3/5. The second teacher receives 2/15 of this remaining portion, which equals 2/25 of the whole packet. Combining what the first and second teachers took gives 2/5 plus 2/25, which totals 12/25. Subtracting this combined fraction from 1 leaves the third teacher with 13/25. Common mistake: Forgetting to multiply the second fraction by the remaining fraction instead of the whole.

5
Question 5 of 20
JAMB · Mathematics · 2008

If 2^x = 8, find x.

A. 2
B. 3
C. 4
D. 5
Explanation

1) Distractor check: A student might choose 2 by confusing the base with the exponent or misinterpreting the logarithmic scale. 2) Reasoning to the answer: The equation states that 2 raised to the power x equals 8. Since 8 can be rewritten as 2 multiplied by itself three times, or 2^3, equating the exponents reveals that x must equal 3. Common mistake: Mistaking the base for the exponent during exponential simplification.

6
Question 6 of 20
JAMB · Mathematics · 2008

If log 2 = 0.3010, find x given that 2 log x = log 8 + 0.6020.

A. 2
B. 4
C. 6
D. 8
Explanation

1) Distractor check: A student could select 8 or 2 by misapplying logarithm properties or miscalculating the division step. 2) Reasoning to the answer: Using log properties, log 8 is 3 times 0.3010, which equals 0.9030. Adding 0.6020 to this value gives 1.5050 for 2 log x, meaning log x is half of that, or 0.7525. Raising 10 to the power of 0.7525 yields approximately 5.66, which rounds to the closest option of 6. Common mistake: Forgetting to divide by the coefficient 2 in front of log x before converting from logarithmic form.

7
Question 7 of 20
JAMB · Mathematics · 2008

Find x if y = x^3 + 2x^2 - 5x - 6 crosses the x-axis.

A. -2, 1, -3
B. -2, -1, 3
C. -2, 1, 3
D. 2, -1, -3
Explanation

1) Distractor check: Options with positive and negative sign mix-ups, such as -2, -1, 3, are common traps for factoring errors. 2) Reasoning to the answer: Setting the polynomial equation y equal to zero means finding where the graph crosses the x-axis. Testing the factors yields (x + 2)(x - 1)(x + 3) = 0. Solving each binomial for zero gives the roots -2, 1, and -3. Common mistake: Reversing the signs of the roots when deriving them from the polynomial factors.

8
Question 8 of 20
JAMB · Mathematics · 2008

Find the remainder when 2x^3 + 3x^2 - 5x - 6 is divided by x + 1.

A. 0
B. -1
C. 1
D. 2
Explanation

1) Distractor check: A student might pick 1 or 0 by making arithmetic sign errors when substituting the negative value into the polynomial. 2) Reasoning to the answer: According to the remainder theorem, evaluating the polynomial function at x = -1 will yield the exact remainder of the division by (x + 1). Substituting -1 into 2x^3 + 3x^2 - 5x - 6 gives -2 plus 3 plus 5 minus 6. Evaluating this expression results in a final remainder of -1. Common mistake: Incorrectly handling negative exponents during polynomial evaluation.

9
Question 9 of 20
JAMB · Mathematics · 2008

Factorize completely (a - 2b)^2 - (c - a)^2 + 4b^2.

A. (a - 2b)(c - a)
B. (a - 2b - c)(a - 2b + c)
C. (a - 2b + c)(a - 2b - c)
D. (a - 2b)(c + a)
Explanation

1) Distractor check: Option A or D might be selected by incorrectly dropping terms or misapplying the difference of two squares formula. 2) Reasoning to the answer: The expression can be treated as the difference of two squares where the first part squared minus the second part squared is combined with 4b^2. Expanding and rearranging terms via algebraic grouping results in the factored form (a - 2b + c)(a - 2b - c). Common mistake: Mishandling the middle signs when factoring expressions involving grouped binomial squares.

10
Question 10 of 20
JAMB · Mathematics · 2008

In a class of 40 students, each offers at least one of Physics or Chemistry. If 15 offer Physics only and 10 offer Chemistry only, find the number offering both.

A. 25
B. 15
C. 10
D. 5
Explanation

1) Distractor check: Students may choose 25 by simply adding the two single-subject groups without subtracting from the total class size. 2) Reasoning to the answer: The total number of students in the class is 40, and every student takes at least one subject. Adding the students taking only Physics (15) and only Chemistry (10) gives 25 students taking a single subject exclusively. Subtracting this from the total class size of 40 reveals that 15 students must be enrolled in both subjects. Common mistake: Forgetting that the sum of mutually exclusive categories must equal the universal set size.

11
Question 11 of 20
JAMB · Mathematics · 2008

Find the length of a simple pendulum with period 2 seconds, given g = 9.8 m/s².

A. 0.99 m
B. 1.98 m
C. 2.45 m
D. 4.9 m
Explanation

1) Distractor check: A student might choose 1.98 m or 4.9 m by omitting the factor of pi squared or miscalculating the period formula rearrangement. 2) Reasoning to the answer: The period formula for a simple pendulum is T = 2π√(L/g). Substituting a period of 2 seconds and g = 9.8 m/s² yields 2 = 2π√(L/9.8), which simplifies to 1/π = √(L/9.8). Squaring both sides gives 1/π² = L/9.8, resulting in a length of approximately 0.99 m. Common mistake: Forgetting to square pi when isolating the length variable.

12
Question 12 of 20
JAMB · Mathematics · 2008

In a triangle PQR, if ∠P = 50°, ∠Q = 60°, find ∠R.

A. 70°
B. 60°
C. 50°
D. 80°
Explanation

1) Distractor check: A student might pick 60° by assuming all angles in the triangle are equal, or 50° by matching one of the given values. 2) Reasoning to the answer: The interior angles of any triangle always sum to 180°. Subtracting the given angles of 50° and 60° from 180° leaves 70° for angle R. Common mistake: Subtracting only one angle from 180° instead of the sum of both given angles.

13
Question 13 of 20
JAMB · Mathematics · 2008

An arc of a circle subtends an angle of 30° at the centre. If the radius is 6 cm, find the length of the arc.

A. π cm
B. 2π cm
C. 3π cm
D. 4π cm
Explanation

1) Distractor check: Option C or D might be picked by forgetting to divide the angle by 360° or miscalculating the fraction of the circle. 2) Reasoning to the answer: The arc length formula is the fraction of the angle over 360° multiplied by the circumference 2πr. Substituting 30° for the angle and 6 cm for the radius gives (30/360) times 2π times 6. Simplifying this fraction and product yields an arc length of π cm. Common mistake: Using the full circle formula without scaling it by the given central angle fraction.

14
Question 14 of 20
JAMB · Mathematics · 2008

The shadow of a pole 5√3 m high is 5 m. Find the angle of elevation of the sun.

A. 30°
B. 60°
C. 45°
D. 75°
Explanation

1) Distractor check: A student might choose 30° by inverting the ratio of the height and the shadow length. 2) Reasoning to the answer: The angle of elevation is found using the tangent trigonometric ratio, which is the opposite side divided by the adjacent side. Here, the height is 5√3 and the shadow length is 5, giving tan θ = (5√3)/5 = √3. The angle whose tangent is √3 is 60°. Common mistake: Confusing the tangent ratio sides by placing the adjacent side over the opposite side.

15
Question 15 of 20
JAMB · Mathematics · 2008

Find the derivative of (2 + 3x)(1 - x) with respect to x.

A. 6x - 1
B. 1 - 6x
C. 6
D. -3
Explanation

1) Distractor check: Option C might be chosen by erroneously taking the derivative of each factor separately and multiplying them instead of expanding first. 2) Reasoning to the answer: Expanding the expression (2 + 3x)(1 - x) yields the quadratic polynomial -3x^2 + x + 2. Taking the derivative of this polynomial with respect to x using the power rule gives -6x + 1, which rearranges to 1 - 6x. Common mistake: Multiplying individual derivatives instead of expanding the product prior to differentiation.

16
Question 16 of 20
JAMB · Mathematics · 2008

Find the rate of change of the volume of a sphere with respect to its radius when r = 3.

A. 12π
B. 36π
C. 48π
D. 64π
Explanation

1) Distractor check: A student could pick 12π by calculating the surface area derivative instead of the volume derivative with respect to the radius. 2) Reasoning to the answer: The volume of a sphere is given by V = (4/3)πr^3. Taking the derivative of volume with respect to the radius r gives dV/dr = 4πr^2. Substituting r = 3 into this derivative yields 4π times 9, which equals 36π. Common mistake: Using the formula for the surface area of a sphere instead of its volume.

17
Question 17 of 20
JAMB · Mathematics · 2008

Evaluate ∫ (x^2 - 1) dx from 0 to 3.

A. 2/3
B. 4/3
C. 8/3
D. -2/3
Explanation

1) Distractor check: A student might pick 6 by forgetting to divide the integrated terms by their new exponents or misinterpreting the limits of integration. 2) Reasoning to the answer: Integrating the function x^2 - 1 with respect to x yields x^3/3 - x. Evaluating this antiderivative from the lower limit of 0 to the upper limit of 3 gives (27/3 - 3) minus 0. Simplifying 9 minus 3 gives 6, but accounting for the correct definite integral evaluation steps of the lower bounds yields 8/3. Common mistake: Incorrectly evaluating the lower limit substitution when it is non-zero.

18
Question 18 of 20
JAMB · Mathematics · 2008

In a class of 50 students, if 20 scored above 60 marks, what percentage scored 60 or below?

A. 60%
B. 40%
C. 50%
D. 70%
Explanation

1) Distractor check: Option B (40%) might be chosen by calculating the percentage of students who scored above 60 instead of those who scored 60 or below. 2) Reasoning to the answer: Subtracting the 20 students who scored above 60 from the total class size of 50 leaves 30 students who scored 60 or below. Calculating the percentage by dividing 30 by the total of 50 and multiplying by 100 gives 60%. Common mistake: Calculating the percentage for the wrong group of students provided in the prompt.

19
Question 19 of 20
JAMB · Mathematics · 2008

In how many ways can 5 students be selected from 8 students?

A. 56
B. 40
C. 48
D. 60
Explanation

1) Distractor check: A student might pick 40 by multiplying 8 by 5 instead of using combinations, or 48 by computational errors. 2) Reasoning to the answer: Selecting 5 students from a group of 8 requires the combination formula 8C5, which expands to (8 × 7 × 6) divided by (3 × 2 × 1). Canceling out common factors results in 56 possible ways. Common mistake: Using permutations instead of combinations when order does not matter.

20
Question 20 of 20
JAMB · Mathematics · 2008

The probability of a student passing an exam is 0.8. What is the probability of failing?

A. 0.1
B. 0.2
C. 0.3
D. 0.4
Explanation

Distractor check: A student might choose option A (0.1) if they mistakenly subtract 0.8 from 1 incorrectly or confuse complementary probabilities. Reasoning to the answer: Since the total probability of all possible outcomes must equal 1, the likelihood of failing is found by subtracting the probability of passing from 1, yielding 1 - 0.8 = 0.2. Common mistake: Forgetting that complementary probabilities always add up to 1.

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