JAMB Mathematics 1991

A: 99.998

Dividing20 by1/5.005 gives20×5.005=100.1. Subtracting0.05×2.05=0.1025 gives99.9975, which rounds to99.998.

D: 20.7

62÷3=20.666… . The first three significant figures are2,0 and6. The next digit is6, so rounding gives20.7.

C: ₦2.80

The monthly interest is2% of₦10, or₦0.20. Four months of simple interest total₦0.80, making₦10.80 due. After paying₦8.00, Musa still owes₦2.80.

B: 16⅞%

The water volumes are3×0.20=0.60 gallon and5×0.15=0.75 gallon. Thus1.35 gallons of the8-gallon mixture is water: (1.35/8)×100=16.875%=16⅞%.

D: 1 − 2 log₁₀2

Combining logarithms gives log₁₀[(2/5)²×9÷(72/125)]=log₁₀(5/2). Since 5/2=10/4, this equals 1−2log₁₀2.

C: uvw = 12(u + v)

Inverse variation gives w = k(u + v)/(uv). Substituting the given values, 8 = k(2 + 6)/(2 × 6), so k = 12. Therefore uvw = 12(u + v). The relation applies where u, v and u + v are nonzero.

C: r − p

Group the terms: rs + tr − pt − ps = r(s + t) − p(t + s) = (r − p)(s + t). Therefore r − p is a factor.

A: −1 and 5

From the first equation, y = 8 − 3x. Substitute into the second: x² + x(8 − 3x) = 6, so x² − 4x + 3 = 0. Factor to obtain (x − 1)(x − 3) = 0. When x = 1, y = 5; when x = 3, y = −1. Thus the two values of y are −1 and 5.

A: x < 12/13

Use the common denominator 12: x/2 + x/3 + x/4 = (6x + 4x + 3x)/12 = 13x/12. Multiplying the inequality by the positive number 12/13 gives x < 12/13. The boundary value is excluded because the inequality is strict.

C: −1 and 6

Expand and collect terms: x² − 5x + 6 = 12, so x² − 5x − 6 = 0. Factor to obtain (x − 6)(x + 1) = 0. Hence x = 6 or x = −1. Substituting either value into the original equation gives 12.

B: 1 + 2x + 2√(x(1 + x))

Multiply the numerator and denominator by √(1 + x) + √x. The denominator becomes (1 + x) − x = 1. Expanding the numerator gives 1 + x + x + 2√(x(1 + x)) = 1 + 2x + 2√(x(1 + x)). For real square roots, x ≥ 0.

A: 27(1/3)^(n − 1)

This is a geometric progression with first term 27 and common ratio 9/27 = 1/3. Its nth term is ar^(n − 1), so the required term is 27(1/3)^(n − 1), for n = 1, 2, 3, …. It gives 27, 9 and 3 for the first three terms.

C: (60 − x)°

The interior angles of a quadrilateral sum to 360°. Thus 11x + 151 = 360, giving x = 19. The four angles are 65°, 136°, 41° and 118°. The smallest is 41°, represented by (60 − x)°.

C: 12√2 cm

The side length is √144 = 12 cm. A diagonal is the hypotenuse of a right triangle with both shorter sides equal to 12 cm. By Pythagoras, d² = 12² + 12² = 288, so d = 12√2 cm.

A: 8√3 cm

The diagonals of a rhombus bisect its angles and meet at right angles. In the resulting 30°–60°–90° triangle, half of the shorter diagonal is 4 cm. The longer half-diagonal is 4√3 cm, so the full longer diagonal is 8√3 cm.

C: 12.5 cm and 7.7 cm

For secants from the same external point, PM × PN = PQ × PR = PT². The common product is 5 × 12 = 60 cm². Hence PR = 60/4.8 = 12.5 cm and PT = √60 ≈ 7.7 cm.

C: sec²x

Factor out sec²x: cos²x sec²x(1 + tan²x). Since cos²x sec²x = 1 and 1 + tan²x = sec²x, the expression equals sec²x wherever the original expression is defined.

D: 10 cm²

PU = 4 cm, SW = 6 cm and RV = 2 cm. The trapezium PUWS has area ½ × (4 + 6) × 8 = 40 cm². The corner triangles UQV and VRW have areas ½ × 4 × 6 = 12 cm² and ½ × 2 × 2 = 2 cm². Hence area UVW = 64 − 40 − 12 − 2 = 10 cm².

B: 72π cm²

The two circular ends have total area 2πr² = 32π cm². The curved surface has area 2πrh = 40π cm². Adding them gives 72π cm².

D: 39.6° and 212.4°

A full circle is 360°. Education takes 11/100 × 360° = 39.6°, and food takes 59/100 × 360° = 212.4°.

A: 6.7 and 6

The frequencies total 50. The weighted total is 4×2 + 5×7 + 6×17 + 7×10 + 8×8 + 9×6 = 333, so the mean is 333/50 = 6.66, or 6.7 to one decimal place. There are 9 values below 6 and 26 values at or below 6. Both the 25th and 26th values are 6, so the median is 6.