JAMB Mathematics 1986

C: (1020)₃

Convert to base ten: (212)₃ = 2×9 + 1×3 + 2 = 23; (121)₃ = 1×9 + 2×3 + 1 = 16; (222)₃ = 2×9 + 2×3 + 2 = 26. Then 23 − 16 + 26 = 33. Since 33 = 1×27 + 0×9 + 2×3 + 0, the result is (1020)₃, option C. Base three uses only the digits 0, 1 and 2.

B: 5/4

First add the denominator fractions: 1/2 + 1/3 = 5/6. The given expression is (2/3) ÷ (5/6) = (2/3) × (6/5) = 4/5. Its reciprocal is 5/4, option B. The question asks for the reciprocal, not just the value of the original expression.

B: 11/216

Substitute carefully: P − Q = −3, so (P − Q)³ = −27, while S² = (−4)² = 16. The numerator is −27 + 16 = −11. The denominator is R³ = (−6)³ = −216. Dividing two negative numbers gives 11/216, option B.

B: 7⅓%

For simple interest, I = Prt/100, where r is the annual percentage rate. Therefore r = 100I/(Pt) = (100 × 55)/(150 × 5) = 22/3 = 7⅓%. The correct answer is B. Use the interest earned, ₦55, rather than the final balance.

C: (17, 11)

Let the younger age be y. Then the older age is y + 6, so y(y + 6) = 187. Rearranging gives y² + 6y − 187 = (y − 11)(y + 17) = 0. An age cannot be negative, so y = 11 and x = 17. Check: 17 − 11 = 6 and 17 × 11 = 187. The ages are (17, 11), option C.

B: a(2 + 5x)(4 − 10x + 25x²)

First factor out a: a(8 + 125x³). Since 8 = 2³ and 125x³ = (5x)³, use u³ + v³ = (u + v)(u² − uv + v²). This gives a(2 + 5x)(4 − 10x + 25x²), option B. Expanding the two factors inside a gives 8 + 125x³.

D: x > 11

Multiply both sides by 15, a positive number, so the inequality direction stays the same: 5(x + 1) − 15 > 3(x + 4). Thus 5x − 10 > 3x + 12, giving 2x > 22 and x > 11. The boundary x = 11 is excluded because the inequality is strict. Answer D.

D: (x + 2)(x + a)

Reorder the terms: x² + ax + 2x + 2a. Group them as x(x + a) + 2(x + a), then take out the common factor x + a. This gives (x + 2)(x + a), option D. Expanding this product returns x² + ax + 2x + 2a.

B: x = −1 ± √15/3

Divide by 3 and complete the square: x² + 2x = 2/3, so (x + 1)² = 5/3. Taking both square roots gives x = −1 ± √(5/3) = −1 ± √15/3, option B. Both values satisfy the original equation.

D: 12/(35x + 35)

Factor each denominator: 5x + 5 = 5(x + 1) and 7x + 7 = 7(x + 1). The common denominator is 35(x + 1), giving (7 + 5)/[35(x + 1)] = 12/(35x + 35), option D. The restriction x ≠ −1 comes from the original denominators.

D: (0, 4), (4, 0), (−1, 0)

On the y-axis, x = 0, so y = 4 and the intercept is (0, 4). On the x-axis, y = 0, giving x² − 3x − 4 = 0. Factorizing gives (x − 4)(x + 1) = 0, so x = 4 or −1. The three intercepts are (0, 4), (4, 0) and (−1, 0), option D.

C: (a + 5)(7a + 1)

Use U² − V² = (U − V)(U + V), with U = 4a + 3 and V = 3a − 2. Then U − V = a + 5 and U + V = 7a + 1. Therefore the factorization is (a + 5)(7a + 1), option C. Be careful: subtracting 3a − 2 adds 2 to the constant term.

D: 4x/(x² − 4)

Because x² − 4 = (x − 2)(x + 2), use x² − 4 as the common denominator. The numerator becomes (x + 2) + (x − 2) + 2x = 4x. Thus the expression is 4x/(x² − 4), option D. Keep x ≠ ±2 because the original expression is undefined there.

C: v = 12/(2S² + w)

Square both sides to obtain S² = 6/v − w/2. Multiply by 2: 2S² = 12/v − w. Hence 2S² + w = 12/v and v = 12/(2S² + w), option C. The denominator must be nonzero. Substituting back gives √(S²) = S because S ≥ 0.

C: 0 and 1

Write 16ˣ = (4ˣ)² and substitute u = 4ˣ. Then u² − 5u + 4 = (u − 1)(u − 4) = 0. Therefore 4ˣ = 1 or 4, giving x = 0 or 1, option C. The exponential function 4ˣ is one-to-one, so these are all the real solutions.

D: k²

The equal ratios imply a = kb and c = kd. Substitute into the numerator: 3(kb)² − (kb)(kd) + (kd)² = k²(3b² − bd + d²). Divide by the denominator to get k², option D.

A: (−2, −3) and (1/2, 2)

At an intersection, the two expressions for y are equal: 2x + 1 = 2x² + 5x − 1. Hence 2x² + 3x − 2 = (2x − 1)(x + 2) = 0, so x = 1/2 or −2. Substituting into y = 2x + 1 gives y = 2 or −3 respectively. The points are (−2, −3) and (1/2, 2), option A.

A: b²/a²

The identity 1 + tan² θ = sec² θ gives 1 + tan² θ = 1/cos² θ. Substituting cos θ = a/b gives 1/(a²/b²) = b²/a², option A. The nonzero condition ensures that cos θ and these divisions are defined.

D: 115.47 m

Let O be the tower's base. From X, tan 30° = 50/XO, so XO = 50√3 ≈ 86.60 m. From Y, tan 60° = 50/YO, so YO = 50/√3 ≈ 28.87 m. The points are on opposite sides, so XY = XO + YO ≈ 115.47 m, option D. The old answer incorrectly added these two distances to obtain 101.03 m.

C: 816 m³

The cross-section is a trapezium. Its area is ½(6 + 11) × 12 = 102 m². A prism has the same cross-section throughout its length, so multiply by the 8 m width: volume = 102 × 8 = 816 m³, option C. The 12 m dimension belongs to the trapezium; the separate 8 m dimension gives the extent of the prism.

B: 24 cm²

For arc length s and radius r, the sector area is A = rs/2. Here A = (6 × 8)/2 = 24 cm², option B. Equivalently, θ = s/r = 4/3 radians and A = r²θ/2. The 8 cm measurement is the curved arc length, not a chord.

B: 29/36

For angle Z, the adjacent sides are XZ = 3 and YZ = 6, and the opposite side is XY = 4. By the cosine rule, 4² = 3² + 6² − 2(3)(6) cos Z. Thus cos Z = (9 + 36 − 16)/36 = 29/36, option B. Use the side opposite the requested angle as the subtracted square.

D: 36π cm²

The total surface area is the curved area plus the base: πrl + πr². With r = 2√3 and l = 4√3, the curved area is π(2√3)(4√3) = 24π cm² and the base area is π(2√3)² = 12π cm². The total is 36π cm², option D. The π factor does not cancel.

C: ₦136.00

The open box has no top. Its outside consists of a 4 × 3 = 12 m² base, two 4 × 4 = 16 m² sides and two 3 × 4 = 12 m² sides. Total area = 12 + 2(16) + 2(12) = 68 m². At ₦2.00 per m², the cost is 68 × 2 = ₦136.00, option C.

A: 3/11

There are 50 − 40 + 1 = 11 integers in the inclusive range. The primes are 41, 43 and 47: exactly three numbers. In particular, 49 = 7 × 7 is not prime. Therefore the probability is 3/11, option A.

B: 26 × 10⁴

The 15–29 years group occupies 104° of the full 360° circle. Its population is (104/360) × 900,000 = 260,000 = 26 × 10⁴. The 104° sector represents about 28.9% of the population.

A: (27, 4)

The mode is the score with the greatest frequency. Score 4 occurs 14 times, more than any other score, so m = 4. The number scoring 4 or less is S = 2 + 4 + 7 + 14 = 27. Therefore (S, m) = (27, 4).