JAMB Mathematics 1987

B: 240 cm

The length must be a multiple of both 40 and 48. Since 40 = 2³ × 5 and 48 = 2⁴ × 3, their least common multiple is 2⁴ × 3 × 5 = 240 cm. This gives six 40 cm strips or five 48 cm strips. A length of 480 cm also works, but is not the least.

C: 1.1

First round each input: 0.02174 becomes 0.022, 1.2047 becomes 1.2, and 0.023789 becomes 0.024. Now calculate (0.022 × 1.2) / 0.024 = 0.0264 / 0.024 = 1.1. The instruction requires rounding the inputs before doing the multiplication and division.

D: 24:25

Let the original side length be s. The rectangle has length 1.2s and width 0.8s, so its area is 0.96s². Compared with the original area s², the ratio is 0.96:1 = 96:100 = 24:25. The area decreases by 4%.

A: ₦2,000.00

The profit is ₦15,000 − ₦5,000 = ₦10,000. There are 2 + 3 = 5 equal shares, so each share is ₦2,000. The brothers receive ₦4,000 and ₦6,000, a difference of ₦2,000.

A: ₦37.00

Total annual wages are 52 × ₦30 = ₦1,560. The first 44 weeks account for 20 × ₦20 + 24 × ₦36 = ₦400 + ₦864 = ₦1,264. The remaining eight weeks therefore account for ₦296, giving an average of ₦296 ÷ 8 = ₦37 per week.

C: 12%

An 8% profit means ₦1.35 is 108% of the cost. The cost is ₦1.35 ÷ 1.08 = ₦1.25. Selling for ₦1.10 gives a loss of ₦0.15. The percentage loss is (₦0.15 ÷ ₦1.25) × 100 = 12%.

B: 3.0 × 10⁻³

35/6 = 5.8333… rounds to 5.83 at three significant figures and 5.833 at three decimal places. Their difference is 5.833 − 5.830 = 0.003 = 3.0 × 10⁻³. Standard form requires a coefficient from 1 up to, but not including, 10. Option A has the same numerical value but is not written in the requested standard form.

A: 1/5

The numerator is log₂(6/3) = log₂(2) = 1. The denominator is log₂(8) − 2log₂(1/2) = 3 − 2(−1) = 5. Therefore the expression is 1/5, option A.

D: 3√2/28

Combine the coefficients and square roots: (6/14)√[(14 × 21)/(24 × 98)] = (3/7)√(1/8). Since √(1/8) = √2/4, the result is 3√2/28. The numbers before the radical signs are multiplying coefficients; all the radicals here are square roots.

D: 50 kobo

For one extra person, the increase is Q(n + 1) − Q(n) = [15 + 0.5(n + 1)] − [15 + 0.5n] = 0.5 naira. Since one naira equals 100 kobo, the extra cost is 50 kobo.

B: PR² = 98

Combining the variations gives P proportional to 1/R², so PR² is constant. Substituting P = 2 and R = 7 gives PR² = 2 × 49 = 98.

C: y = 1/∛(Z − x²)

Subtract x² to obtain Z − x² = 1/y³. Taking reciprocals gives y³ = 1/(Z − x²), so y = 1/∛(Z − x²). This requires Z ≠ x². A cube root does not distribute over subtraction: ∛(Z − x²) is not generally ∛Z − ∛(x²).

C: 1, −2

A monic perfect square with middle term 2(m + 1)x is [x + (m + 1)]². Its constant term must satisfy m + 3 = (m + 1)². This gives m² + m − 2 = (m + 2)(m − 1) = 0, so m = −2 or 1. These give (x − 1)² and (x + 2)² respectively.

C: −12, 2

From x + y = 5, x = 5 − y. Substitution gives (5 − y)² − 2y² = 1, so 25 − 10y − y² = 1. Rearranging gives y² + 10y − 24 = (y + 12)(y − 2) = 0. Thus y = −12 or 2; the matching x values are 17 and 3.

B: x < −3

Expand the bracket: x − 1 > 4x + 8. Rearranging gives −3x > 9. Dividing by −3 reverses the inequality, so x < −3. The boundary x = −3 is excluded because the original inequality is strict.

C: (x − y)/(2x − y)

Factor the numerator as (x − y)(x + y). The denominator is (2x − y)(x + y), since expanding gives 2x² + xy − y². Cancel x + y to obtain (x − y)/(2x − y). The original restrictions remain: x + y ≠ 0 and 2x − y ≠ 0.

B: 4:5

Triangles PQR and PTS are similar because QR is parallel to TS. Their corresponding lengths have ratio 2:3, so their areas have ratio 4:9. The trapezium is the part of the larger triangle outside the smaller triangle: it has 9 − 4 = 5 area parts. Thus triangle PQR : trapezium QRST = 4:5.

B: 20 cm

The hypotenuse is XZ = 9 + 16 = 25 cm. Similarity of triangles XYZ and YTZ gives YZ/XZ = TZ/YZ, so YZ² = XZ × TZ = 25 × 16 = 400. Therefore YZ = 20 cm. Equivalently, the altitude satisfies YT² = 9 × 16 = 144, and Pythagoras gives YZ² = 144 + 16² = 400.

D: 5 cm

The outer area is (2h)(3h) = 6h². The opening has width 2h − 4 and height 3h − 2, so the shaded area is 6h² − (2h − 4)(3h − 2) = 16h − 8. Set 16h − 8 = 72: then 16h = 80 and h = 5 cm. Check: the outer area is 150 cm² and the opening is 6 × 13 = 78 cm², leaving 72 cm².

C: 202°30′, 157°30′

Daylight lasts 13 hours 30 minutes, or 13.5 hours. Darkness lasts 24 − 13.5 = 10.5 hours. Earth rotates through 360°/24 = 15° per hour. The daylight angle is 13.5 × 15 = 202.5° = 202°30′, and the darkness angle is 10.5 × 15 = 157.5° = 157°30′. The two angles add to 360°.

A: 3/5

A square, a rectangle and a rhombus each have two pairs of parallel opposite sides. The ordinary kite and trapezium do not. There are three qualifying figures among five equally likely choices, so the probability is 3/5.