JAMB Mathematics 1989

A: 4.5 × 10⁰

2,700,000 × 0.03 = 81,000. Dividing by 18,000 gives 4.5. Since 10⁰ = 1, this is 4.5 × 10⁰, option A. An exponent of 1 would instead give 45.

C: 2, 3, 5, 7

Repeated division gives 2,520 = 2³ × 3² × 5 × 7. The distinct prime factors are therefore 2, 3, 5 and 7. The number 9 is not prime because 9 = 3 × 3.

A: 10⁻⁵

To three significant figures, 0.007685 rounds to 0.00769. To four decimal places, it rounds to 0.0077. Their absolute difference is 0.007700 − 0.007690 = 0.000010 = 10⁻⁵. The leading zeros do not count as significant figures.

A: (3/q)³

Use 3log₃(q) = log₃(q³), then combine the logarithms to obtain log₃(pq³) = 3. Therefore pq³ = 3³ = 27, giving p = 27/q³ = (3/q)³. Positivity is required for the original real logarithms.

D: 0 and 1

Since 9ʸ = (3ʸ)², substitute u = 3ʸ. The equation becomes u² − 4u + 3 = (u − 1)(u − 3) = 0. Thus 3ʸ = 1 or 3, giving y = 0 or 1. The function 3ʸ is one-to-one, so there are no other real solutions.

B: R = T/[2(TS² − 1)]

Square both sides: S² = (2R + T)/(2RT). Multiply by 2RT to obtain 2RTS² = 2R + T. Collect the terms containing R: 2R(TS² − 1) = T. Therefore R = T/[2(TS² − 1)]. The original formula requires R and T to be nonzero and S ≥ 0; the rearranged denominator requires TS² ≠ 1.

B: ₦63.00

The extra 10 students cost ₦96 − ₦74 = ₦22, so the variable cost is ₦2.20 per student. The fixed cost is ₦74 − 20 × ₦2.20 = ₦30. For 15 students, the total is ₦30 + 15 × ₦2.20 = ₦63.00.

A: 2x² − x

Replace every x in the definition with x + 1: f(x + 1) = 2(x + 1)² − 5(x + 1) + 3. Expand to get 2x² + 4x + 2 − 5x − 5 + 3 = 2x² − x.

B: (2a + 3b − c)(2a + 3b + c)

Group the first, second and fourth terms: 4a² + 12ab + 9b² = (2a + 3b)². The expression is therefore (2a + 3b)² − c². Applying the difference-of-squares identity gives (2a + 3b − c)(2a + 3b + c).

A: (−12, 9/2)

Expand the square before multiplying by one half: ½(9y² − 24xy + 16x²) = 8x² − 12xy + (9/2)y². Comparing coefficients gives K = −12 and L = 9/2.

D: (2, −1)

Substitute u = 1/x and v = 1/y. The equations become 2u − 3v = 4 and 4u + v = 1. Add the first equation to three times the second to obtain 14u = 7, so u = 1/2. Then v = 1 − 4(1/2) = −1. Thus x = 2 and y = −1. Substitution gives 1 + 3 = 4 and 2 − 1 = 1, checking both equations.

A: 25/16

Write the perfect square as (2x + c)² = 4x² + 4cx + c². Matching the coefficient of x gives 4c = 5, so c = 5/4 and Q = c² = 25/16. Hence the expression is (2x + 5/4)².

A: r > abc/(bc + ac + ab)

Factor out r: r(1/a + 1/b + 1/c) > 1. Since a, b and c are positive, their reciprocal sum is positive, so division keeps the inequality direction unchanged. Therefore r > 1/(1/a + 1/b + 1/c) = abc/(bc + ac + ab).

B: b, e, g, j, m

A zero gradient means a horizontal tangent. The labelled peaks b, g and m and troughs e and j have horizontal tangents. Crossing the x-axis alone does not make the gradient zero.

A: (1, 10)

There are three equal steps from −8 to 19. The common difference is [19 − (−8)]/3 = 27/3 = 9. The terms are −8, 1, 10, 19, so m = 1 and n = 10.

B: 3 : 7

PQ = 4 + 3 = 7 cm. Similar triangles PHK and PQR give PK/PR = PH/PQ = 4/7. The remaining part is KR/PR = 1 − 4/7 = 3/7, so KR : PR = 3 : 7.

C: 3√3 cm

Angle X is 180° − 30° − 30° = 120°. By the sine rule, YZ/sin 120° = XZ/sin 30°. Thus YZ = 3(√3/2)/(1/2) = 3√3 cm. The side opposite 120° must be the longest, consistent with this result.

D: 120 cm

Let the perpendicular sides be5k and12k. By Pythagoras, the hypotenuse is13k. Since13k=52, k=4. The sides are20,48 and52 cm, giving perimeter120 cm.

D: 10(cot 35° − cot 55°) km

The angle of depression equals the angle of elevation from the landmark. At angle θ, the horizontal distance to the landmark is 10/tan θ = 10 cot θ. As the aircraft approaches, this distance decreases. The distance flown is therefore 10(cot 35° − cot 55°) km, approximately 7.28 km. This uses the constant-height, level-ground geometry described in the question.

D: 54 years

The total age of all 21 people is 21 × 14 = 294 years. The total for the other 20 people is 20 × 12 = 240 years. The grandfather is therefore 294 − 240 = 54 years old.

C: 1/3

The three possible pairs are(1,3), (1,6) and(3,6). Their sums are4,7 and9. Only one sum is even, so the probability is1/3.