JAMB Mathematics 1993

C: X ∩ Yᶜ ∩ Z

The shaded region lies inside both X and Z, but outside Y. It is therefore the intersection X ∩ Yᶜ ∩ Z.

B: (p−q)/[a(p+q)]

Cross multiplication gives q+aqx=p−apx. Hence a(p+q)x=p−q, so x=(p−q)/[a(p+q)], where the original relation and denominators are defined.

C: x − 5

By the factor theorem, x − a is a factor when substituting x = a gives zero. At x = 5, 15 + 35 − 50 = 0. Equivalently, 15 + 7x − 2x² = (5 − x)(2x + 3) = −(x − 5)(2x + 3), so x − 5 is a factor.

D: 1, −8

The second equation gives y = 7x − 3. Substitution into the first gives x² + 7x − 8 = 0, or (x − 1)(x + 8) = 0. Hence x = 1 or x = −8. Their corresponding y-values are 4 and −59, and both pairs satisfy the original equations.

D: x + 7

Group the terms: x³ + 7x² − x − 7 = x²(x + 7) − (x + 7) = (x² − 1)(x + 7). The polynomial quotient is x + 7 and the remainder is zero. If treated as a fraction, the original expression is undefined at x = ±1.

B: y < −3 or y > 6

Move 18 to the left and factor: y² − 3y − 18 = (y − 6)(y + 3). The product is positive when both factors have the same sign: y < −3 or y > 6. The boundary values −3 and 6 give equality, so neither is included.

B: −1, 5/3

For three consecutive geometric terms, (2k − 1)² = (k + 1)(3k + 1). Expanding and simplifying gives k² − 8k = 0, so k = 0 or 8. At k = 0 the terms are 1, −1, 1 and the ratio is −1. At k = 8 they are 9, 15, 25 and the ratio is 15/9 = 5/3.

A: 6¾ cm

The angle-bisector theorem gives QS/SR = PQ/PR. Thus 3/SR = 4/9, so SR = 27/4 cm = 6¾ cm. The mixed number6¾ is equivalent to6.75.

A: 7√3 cm

The radii to a tangent are perpendicular to it. For this internal tangent, the perpendicular separation is 5 + 2 = 7 cm. Hence ST² = 14² − 7² = 147 and ST = 7√3 cm.

C: 8 cm

The outer square has side10 cm. If the small square has sidea, the larger inner square has side4a because their areas are in ratio1:16. Their sides span10 cm, so5a = 10 and YR = 4a = 8 cm.

B: 3.50 cm

The surface area of a sphere is 4πr². Therefore r² = 154 ÷ (4 × 22/7) = 12.25. Taking the positive square root gives r = 3.50 cm. A radius cannot be negative.

C: 4.7 m²

A sector occupies θ/360 of its circle. Its area is (60/360) × π × 3² = 3π/2 m², approximately 4.7124 m². To one decimal place this is 4.7 m².

A: 45°, 225°

At a solution cos θ cannot be zero, because sin θ would then be ±1. Divide by cos θ to obtain tan θ = 1. Tangent is positive in the first and third quadrants, giving θ = 45° and 225° in the stated interval.

D: 10√3/3 cm

The third angle is 180° − 30° − 30° = 120°. The two remaining sides are equal. By the sine rule, a/sin 30° = 10/sin 120°, so a = 10 × (1/2)/(√3/2) = 10/√3 = 10√3/3 cm. This is approximately 5.77 cm.

C: 16.8 g

The modal class is 10–20 g, with frequency 27. Its preceding and following frequencies are 10 and 19. The grouped-mode estimate is L + [(f₁ − f₀)/(2f₁ − f₀ − f₂)]h = 10 + [17/(54 − 10 − 19)] × 10 = 16.8 g. This interpolates within the modal class; it is not simply the class midpoint.

C: 1/12

Only Y and Z occurring means Y and Z occur while X does not. Independence lets us multiply the three probabilities: (1 − 1/2) × (2/3) × (1/4) = 1/12. Including X would describe a different event.