JAMB Mathematics 1994

20 reviewed questions with answers and explanations.

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Question 6

Given √2 ≈ 1.414, estimate 1/√2 to three decimal places.

  1. 0.141
  2. 0.301
  3. 0.667
  4. 0.707
Answer and explanation

D: 0.707

Rationalising gives 1/√2 = √2/2. Using the supplied approximation, 1.414 ÷ 2 = 0.707, to three decimal places.

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Question 8

For A ⊆ B in a universal set E, simplify A ∩ (A ∩ B)′.

  1. A
  2. ∅
  3. B
  4. Σ
Answer and explanation

B: ∅

Because A is a subset of B, A ∩ B = A. Therefore the expression is A ∩ A′, which has no elements and equals the empty set.

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Question 11

Factorise a²x − b²y − b²x + a²y completely.

  1. (a − b)(x + y)
  2. (y − x)(a − b)(a + b)
  3. (x − y)(a − b)(a + b)
  4. (x + y)(a − b)(a + b)
Answer and explanation

D: (x + y)(a − b)(a + b)

Grouping gives a²(x + y) − b²(x + y) = (x + y)(a² − b²). Factoring the difference of squares gives (x + y)(a − b)(a + b).

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Question 12

Find p and q if (x − 1) and (x − 3) are factors of px³ + qx² + 11x − 6.

  1. −1, −6
  2. 1, −6
  3. 1, 6
  4. 6, −1
Answer and explanation

B: 1, −6

Substituting x = 1 gives p + q = −5. Substituting x = 3 gives 27p + 9q = −27. Solving gives p = 1 and q = −6.

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Question 13

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Which equation represents the curve shown?

  1. y = (x − 3)³
  2. y = (x + 3)³
  3. y = x³ − 27
  4. y = −x³ + 27

Question 17

Solve 1/x > 2 over the real numbers.

  1. x < 1/2
  2. x < 0 or x > 1/2
  3. 0 < x < 1/2
  4. 1 < x < 2
Answer and explanation

C: 0 < x < 1/2

For negative x, 1/x is negative and cannot exceed 2. For positive x, multiplying by x gives 1 > 2x, so 0 < x < 1/2.

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Question 18

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Which inequality represents the shaded region, including its solid boundary?

  1. 2x − y − 2 ≤ 0
  2. 2x − y − 2 ≥ 0
  3. 2x − y − 2 < 0
  4. 2x − y − 2 > 0

Question 21

Find the nth term of 3, 6, 10, 15, 21, …, starting with n = 1.

  1. n(n − 1/2)
  2. n(n + 1/2)
  3. (n + 1)(n + 2)/2
  4. n(2n + 1)
Answer and explanation

C: (n + 1)(n + 2)/2

These are triangular numbers beginning with the second triangular number. Replacing the usual index by n + 1 gives (n + 1)(n + 2)/2.

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Question 28

Three angles of a quadrilateral are (3y − x − z)°, 3x° and (2z − 2y − x)°. Find the fourth.

  1. (360 − x − y − z)°
  2. (360 + x + y − z)°
  3. (180 − x + y + z)°
  4. (180 + x + y + z)°
Answer and explanation

A: (360 − x − y − z)°

The three given angles sum to x + y + z degrees. A quadrilateral has an angle sum of 360°, leaving (360 − x − y − z)°.

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Question 29

An open wooden box has uniform thickness 2 cm and internal dimensions 50 cm × 36 cm × 20 cm. Find the volume of wood.

  1. 11520 cm³
  2. 36000 cm³
  3. 38200 cm³
  4. 47520 cm³
Answer and explanation

A: 11520 cm³

The outside dimensions are 54 cm × 40 cm × 22 cm because the top is open. Subtract the inner space: 54×40×22 − 50×36×20 = 11520 cm³.

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Question 32

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Find the perpendicular height h of the triangle shown.

  1. 12/7 cm
  2. 12√6/7 cm
  3. 7/12 cm
  4. √51/2 cm

Question 33

A conical frustum has height h and its top diameter is twice its bottom diameter. Find the height of the complete cone formed by extending its sides to a vertex below.

  1. 2h
  2. 2πh
  3. πh
  4. πh/2
Answer and explanation

A: 2h

Similar cones have heights in the ratio of their diameters. If the full height is H, the removed small cone has height H/2, so h = H − H/2 and H = 2h.

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Question 35

M(4, q) is the midpoint of L(p, −2) and N(q, p). Find p and q.

  1. p = 2, q = 4
  2. p = 3, q = 1
  3. p = 5, q = 3
  4. p = 6, q = 2
Answer and explanation

D: p = 6, q = 2

The midpoint formula gives p + q = 8 and p − 2 = 2q. Substituting p = 2q + 2 into the first equation gives q = 2 and p = 6.

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Question 37

A boat on level water is viewed at a depression angle of 30° from a 10 m high cliff. Find its horizontal distance from the cliff foot.

  1. 5√3/3 m
  2. 5√3 m
  3. 10√3 m
  4. 10√3/3 m
Answer and explanation

C: 10√3 m

The depression angle equals the elevation angle from the boat. Thus tan30° = 10/d, giving d = 10/tan30° = 10√3 m.

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Question 38

Evaluate sin(−690°).

  1. √3/2
  2. −√3/2
  3. −1/2
  4. 1/2
Answer and explanation

D: 1/2

Adding two complete turns gives −690° + 720° = 30°. Sine has period360°, so sin(−690°) = sin30° = 1/2.

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Question 40

At which point does y = 2x² − 2x + 3 have gradient2?

  1. (1, 3)
  2. (2, 7)
  3. (0, 3)
  4. (3, 15)
Answer and explanation

A: (1, 3)

Differentiate to obtain dy/dx = 4x − 2. Setting this equal to2 gives x = 1. Substituting into the curve gives y = 3.

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Question 41

Integrate (1 − x)/x³ with respect to x, for x ≠ 0.

  1. (x − x²)/x⁴ + k
  2. 4/x⁴ − 3/x³ + k
  3. 1/x − 1/(2x²) + k
  4. 1/(3x³) − 1/(2x) + k
Answer and explanation

C: 1/x − 1/(2x²) + k

Write the integrand as x⁻³ − x⁻². Integrating term by term gives −1/(2x²) + 1/x + k, where k is a constant.

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Question 43

A grade pie chart has sectors A₁ = 144°, A₂ = 72°, A₃ = 64.8° and C₄ = 43.2°, with F taking the remainder. What percentage received C₄?

  1. 52.0%
  2. 43.2%
  3. 40.0%
  4. 12.0%
Answer and explanation

D: 12.0%

A full pie chart is360°. The C₄ sector represents 43.2/360 of the total, so its percentage is (43.2/360)×100 = 12.0%.

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Question 47

Class intervals 1–5, 6–10, 11–15, 16–20 and 21–25 have frequencies6,15,20,7,2. Using continuous class boundaries, estimate the median.

  1. 10.5
  2. 11.5
  3. 12.5
  4. 13
Answer and explanation

B: 11.5

There are50 observations and21 precede the median class11–15. Using lower boundary10.5 and width5 gives 10.5 + ((25 − 21)/20)×5 = 11.5.

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Question 49

A class has6 pupils aged10,27 aged11 and7 aged12. What is the probability that a randomly chosen pupil is at least11?

  1. 27/40
  2. 17/20
  3. 33/40
  4. 3/20
Answer and explanation

B: 17/20

There are6 + 27 + 7 = 40 pupils. The27 pupils aged11 and7 aged12 meet the condition, giving34/40 = 17/20.

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