JAMB Mathematics 1990

B: −2/7

The numerator is19/4−25/4=−3/2. The denominator is(21/5)(5/4)=21/4. Therefore(−3/2)÷(21/4)=−2/7.

A: b

Factor both expressions: a²bx + abx² = abx(a + x), and a²b − b³ = b(a − b)(a + b). The only polynomial factor common to both expressions is b. The other factors involve independent variables and are not shared by both expressions.

D: 0.002172

First square the entire bracket: (3 × 10⁻³)² = 9 × 10⁻⁶. Multiplying by 241.34 gives 0.00217206. The first four significant digits are 2, 1, 7 and 2; the next digit is 0, so the answer is 0.002172. Leading zeros do not count as significant digits.

A: 1½%

For simple interest, I = Prt/100 when r is the annual percentage rate. Hence 7.50 = 100 × r × 5/100 = 5r, so r = 1.5% per year. This is option A, written as 1½%.

A: 3/16

After the first child takes ¼, the fraction left is ¾. The second child takes ¾ of that remainder, leaving ¼ of ¾ for the third child: ¼ × ¾ = 3/16. As a check, ¼ + 9/16 + 3/16 = 1.

B: ₦54,000

After the first brother receives ⅓, the remainder is ⅔. The second takes ⅔ of that remainder, leaving ⅓ × ⅔ = 2/9 of the original profit. Therefore 2/9 of the total is ₦12,000, so the total is ₦12,000 × 9/2 = ₦54,000.

A: 3 and −3

Use (a − b)² = a² + b² − 2ab = 16 − 7 = 9. Taking both square roots gives a − b = 3 or −3. Both signs are possible because exchanging a and b preserves the given equations and reverses their difference.

A: x² − x − 6

Polynomial division gives the quotient x² − x − 6 with zero remainder. Check by multiplying: (x − 1)(x² − x − 6) = x³ − x² − 6x − x² + x + 6 = x³ − 2x² − 5x + 6.

C: 9 and 25

Because √x is real, x ≥ 0. Substitute u = √x, so u ≥ 0 and x = u². Then u² − 8u + 15 = (u − 3)(u − 5) = 0, giving u = 3 or 5. Squaring gives x = 9 or 25. Both satisfy the original equation: 9 − 24 + 15 = 0 and 25 − 40 + 15 = 0.

A: 5 m

If the sides are l and w, then 2(l + w) = 24, so l + w = 12, and lw = 35. The positive numbers 5 and 7 have sum 12 and product 35. Thus the shorter side, or width, is 5 m. The perimeter check is 2(5 + 7) = 24 m.

B: y²/(x² − y²)

Use the common denominator (x + y)(x − y) = x² − y². The numerator becomes x(x − y) + y(x + y) − x² = x² − xy + xy + y² − x² = y². Therefore the result is y²/(x² − y²). The restrictions keep every original denominator nonzero.

D: log(a²¹⁰)

The terms are log a, 2 log a, 3 log a, and so on. Their sum is (1 + 2 + … + 20) log a = [20 × 21/2] log a = 210 log a = log(a²¹⁰).

D: 3(2¹⁸ − 1)

This is a geometric progression with first term 3 and common ratio 2. The sum is S₁₈ = 3(2¹⁸ − 1)/(2 − 1) = 3(2¹⁸ − 1). Multiplying the sum by 2 and subtracting the original sum also leaves 3 × 2¹⁸ − 3.

D: 10½ cm

The area of a trapezium is half the sum of its parallel sides multiplied by its height. Let PS = x cm, so QR = x/3 cm. Then 73.5 = ½(x + x/3) × 10.5 = 7x, giving x = 10.5 cm. Check: QR = 3.5 cm and ½(10.5 + 3.5) × 10.5 = 73.5 cm².

D: 29.8 cm

The arc length is (48/360) × 2π × 10.5 = 2.8π cm, approximately 8.8 cm. The perimeter includes both radii, so it is 10.5 + 10.5 + 2.8π ≈ 29.8 cm.

B: 6 cm

The marked angles give ∠XQP = ∠XZY, and the triangles share the angle at X. Thus triangles XQP and XZY are similar, with Q corresponding to Z and P corresponding to Y. Since XY = XQ + QY = 5 + 3 = 8 cm, the scale factor from the smaller triangle to the larger is XY/XP = 8/4 = 2. Therefore ZY = 2 × QP = 2 × 3 = 6 cm.

B: 5 cm

The diagonals of a rhombus bisect each other at right angles. Half-diagonals of 3 cm and 4 cm form the legs of a right triangle whose hypotenuse is a side of the rhombus. The side length is √(3² + 4²) = √25 = 5 cm.

A: 169/25

Use sin² θ = 1 − cos² θ = 1 − 144/169 = 25/169. Therefore cot² θ = cos² θ/sin² θ = 144/25, and 1 + cot² θ = 169/25. Equivalently, 1 + cot² θ = 1/sin² θ. Squared values make the result independent of the sign of sin θ.

C: 4√2 cm

The angles in a triangle sum to 180°, so angle XZY = 180° − 30° − 105° = 45°. By the sine rule, YZ/sin 30° = 8/sin 45°. Therefore YZ = 8 × (1/2)/(√2/2) = 4√2 cm.

A: 125 2/7 cm²

OS=√(6²+8²)=10 cm, so the rectangle area is 10×11=110 cm² and the semicircle radius is 5 cm. Its area is(1/2)(22/7)(25)=275/7 cm². Subtract triangle area (1/2)(6)(8)=24 cm² to obtain 110+275/7−24=125 2/7 cm².

C: 1/3

The six equally likely unordered pairs are {1,2}, {1,3}, {1,4}, {2,3}, {2,4} and {3,4}. Only {1,3} and {2,4} have even sums, so the probability is 2/6 = 1/3. An even sum requires two odd numbers or two even numbers.

A: 1/8

There are 56 − 41 + 1 = 16 integers in the inclusive range. The multiples of 9 are 45 and 54, so there are two favourable outcomes. The probability is 2/16 = 1/8.