JAMB Mathematics 1988

D: 8.1

The exact quotient is 8.05: 0.0306 × 8.05 = 0.24633. To round to one decimal place, inspect the hundredths digit, which is 5. Round the tenths digit up, giving 8.1.

B: ₦2,400

Kehinde owns 9/(11 + 9) = 9/20 of the store. She sells (2/3) × (9/20) = 3/10 of the whole store for ₦720. Therefore the whole store is worth ₦720 ÷ (3/10) = ₦2,400.

C: 0.6532

Write 4.5 = 9/2 = 3²/2. Then log₁₀(4.5) = 2log₁₀(3) − log₁₀(2) = 2(0.4771) − 0.3010 = 0.9542 − 0.3010 = 0.6532. This uses the supplied rounded logarithms.

D: 4.50 × 10⁻⁵ m

Each leaf has two pages, so 800 pages correspond to 400 leaves. The thickness of one leaf is 18/400 = 0.045 mm. Dividing by 1,000 converts millimetres to metres: 0.045/1,000 = 0.000045 m = 4.50 × 10⁻⁵ m.

B: 1/(a² − b²)

Factor the expressions: 1/p = (a + b)²/(a − b) and 1/q = (a + b)/(a − b)². Therefore p/q = (1/q)/(1/p) = [(a + b)/(a − b)²] × [(a − b)/(a + b)²] = 1/[(a − b)(a + b)] = 1/(a² − b²). The stated exclusions keep the original expressions and their reciprocals defined.

C: 8/27

Inverse variation gives x = k/∛y. When x = 1 and y = 8, k = 1 × ∛8 = 2. For x = 3, ∛y = 2/3. Cubing both sides gives y = (2/3)³ = 8/27.

A: y/11 + 11/[y(y + 6)]

Replace every occurrence of y in the definition by y + 3. The first term becomes [(y + 3) − 3]/11 = y/11. The second denominator becomes (y + 3)² − 9 = y² + 6y = y(y + 6). Hence g(y + 3) = y/11 + 11/[y(y + 6)]. This is defined when y is neither 0 nor −6.

D: (19/24, −1/8)

Double the first equation to obtain 6x − 10y = 6. The second equation is −6x + 2y = −5. Adding gives −8y = 1, so y = −1/8. Substituting in 3x − 5y = 3 gives 3x = 19/8 and x = 19/24. The ordered pair is (19/24, −1/8).

C: 2/[(x + 1)(x + 3)]

Factor the denominators as (x + 2)(x + 3) and (x + 1)(x + 2). Over their common denominator, the numerator is (x + 1) + (x + 3) = 2(x + 2). Cancelling x + 2 gives 2/[(x + 1)(x + 3)]. The original restriction x ≠ −2 still applies, along with x ≠ −1 and x ≠ −3.

D: (2a − 7b)/(a − b)

Factor the numerator as (2a − 7b)(2a + 7b), a difference of two squares. Factor the denominator as (2a + 7b)(a − b). Cancel the common nonzero factor 2a + 7b to obtain (2a − 7b)/(a − b), option D. The original restrictions remain: a ≠ b and 2a + 7b ≠ 0.

A: x = 1.6 and x = −0.6

The quadratic formula gives x = (1 ± √5)/2. The roots are approximately 1.618 and −0.618, which round to 1.6 and −0.6. Their exact sum is 1 and their exact product is −1.

A: y = 1 − x

At a solution of x² − 2x − 1 = 0, we have x² = 2x + 1. Substituting into y = 2 + x − x² gives y = 2 + x − (2x + 1) = 1 − x. Conversely, equating 2 + x − x² and 1 − x gives the required quadratic.

B: ii only

For a regular polygon with n sides, each exterior angle is 360°/n, where n is an integer of at least 3. We get 360/35 = 72/7 and 360/115 = 72/23, neither an integer. But 360/18 = 20, so only (ii) is possible.

D: 16 m

The height above the observation point is 14√3 × tan 30° = 14√3 × (1/√3) = 14 m. Add the observation height of 2 m: the tree is 14 + 2 = 16 m tall.

B: 110π cm³

The outer radius is 12/2 = 6 cm, and the inner radius is 6 − 1 = 5 cm. Subtract the inner cylinder from the outer cylinder: volume = π(6² − 5²) × 10 = π(36 − 25) × 10 = 110π cm³.

D: 1.0

The ten values sum to 156, so the mean is x = 156/10 = 15.6. The middle two values are 15 and 16, so the median is y = 15.5. Thus x/y = 15.6/15.5 = 156/155 ≈ 1.00645, which rounds to 1.0 to one decimal place.

B: (1, 2)

Score 1 has the largest frequency, 11, so the mode is 1. There are 46 observations; the median is the average of the 23rd and 24th values. The cumulative frequencies through scores 0, 1 and 2 are 7, 18 and 24. Both middle values are therefore 2, so the median is 2. The answer is (1, 2).

A: 1/6

There are 6 × 6 = 36 equally likely ordered outcomes. A total of 10 occurs in three ways: (4,6), (5,5), (6,4). A total of 11 occurs in two ways, and 12 in one way. Thus the probability is (3 + 2 + 1)/36 = 6/36 = 1/6.