JAMB Mathematics 1983

D: (5,5)

Put the values in ascending order: 3, 5, 5, 8, 9. The median is the middle value, 5. The mode is the most frequent value, also 5. Therefore (M, D) = (5, 5), option D.

A: x = 5, y = 2

Equating the first two expressions gives y = x − 3. Equating the last two gives 2x + y = 12. Substitute: 2x + x − 3 = 12, so x = 5 and y = 2. Check: all three expressions equal 17.

C: 3 and –2

Dividing by x − 1 gives x² − x − 6 = (x − 3)(x + 2). Thus the polynomial is (x − 1)(x − 3)(x + 2). Apart from x = 1, the roots are 3 and −2: option C.

B: 4√6cm

In right triangle PQR, QR is the hypotenuse. PR = 8 sin 60° = 4√3 cm. In right triangle PRS, PS is the hypotenuse, so cos 45° = PR/PS. Therefore PS = 4√3 ÷ (√2/2) = 4√6 cm.

A: ℓ = −6, k = −9

A factor x + 2 gives f(−2) = 0, so −8ℓ + 8k + 24 = 0, or ℓ − k = 3. A factor x − 1 gives f(1) = 0, so ℓ + 2k = −24. Substitute ℓ = k + 3 to get 3k = −27. Thus k = −9 and ℓ = −6, option A. Both −2 and 1 make the cubic equal to zero.

A: y = 10x²/31 + 52/(31√x)

Write y = ax² + b/√x. The first condition gives a + b = 2. The second gives 16a + b/2 = 6, or 32a + b = 12. Subtracting gives 31a = 10, so a = 10/31 and b = 52/31. Therefore y = 10x²/31 + 52/(31√x), option A. Substitution gives 2 at x = 1 and 6 at x = 4.

D: x/[(x + 3)(x + 7)]

Factor x² − 9 = (x − 3)(x + 3), x² − 3x = x(x − 3), and x² − 49 = (x − 7)(x + 7). Cancel the common factors x − 3 and x − 7 to obtain x/[(x + 3)(x + 7)], option D. The original expression excludes x = −7, −3, 3 and 7; cancellation does not remove those restrictions.

E: u = 8/V³

Write w = a/V and u = bw³. Substitution gives u = ba³/V³ = k/V³. Since u = 1 when V = 2, k = 1 × 2³ = 8. Therefore u = 8/V³, option E.

C: 6, -1

From y + 5x − 2 = 0, y = 2 − 5x. Substitute into x² + y − 8 = 0: x² − 5x − 6 = 0. Factor: (x − 6)(x + 1) = 0, so x = 6 or x = −1. These are the two x-values in option C; the corresponding y-values are −28 and 7.

D: 30°, 50°, 70°, 80°, 130°

The total is 6 + 10 + 14 + 16 + 26 = 72. Each unit represents 360° ÷ 72 = 5°. Multiply each value by 5 to obtain 30°, 50°, 70°, 80° and 130°. They sum to 360°, so option D is correct.

C: 6/13

MATRICULATION contains 13 letters. Its vowels are A, I, U, A, I and O: six letters. Each letter is equally likely to be drawn, so the probability is 6/13, option C. Repeated vowels each count because each letter is a separate item.

C: 4.47

First round each number to three significant figures, as requested: 59.81789 becomes 59.8 and 0.0746829 becomes 0.0747. Their product is 4.46706. Rounded to three significant figures, this is 4.47, option C.

D: m = 4, n = -2

Rewrite the left side as 2^(m − 2n) × 3^(n − m). The right side is 2^8 × 3^(−6). Equate the exponents: m − 2n = 8 and n − m = −6. Solving gives n = −2 and m = 4, option D. Substituting these values gives (16/81) × (16/9) = 256/729.

C: (3a − 2b)(3a + 2b)(9a² + 4b²)

Write the expression as (9a²)² − (4b²)². The difference-of-squares rule gives (9a² − 4b²)(9a² + 4b²). Factor the first bracket again: (3a − 2b)(3a + 2b)(9a² + 4b²), option C. The conjugate factors need opposite signs; repeating the minus factor does not give the original expression.

A: 25 cm

PR = 2 × 3.5 = 7 cm. Both base angles of triangle PRT are 60°, so its third angle is also 60° and PT = TR = 7 cm. The semicircular arc has length πr = (22/7) × 3.5 = 11 cm. The outer perimeter is 7 + 7 + 11 = 25 cm. PR is inside the combined figure and is not part of its outer perimeter.

C: 25.26

Add the five values: 24.57 + 25.63 + 24.32 + 26.01 + 25.77 = 126.30. Divide by 5 to obtain 25.26, option C.