JAMB Mathematics 1997

22 reviewed questions with answers and explanations.

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Question 3

Express (0.006)³ + (0.004)³ in standard form.

  1. 2.8 × 10⁻⁹
  2. 2.8 × 10⁻⁸
  3. 2.8 × 10⁻⁷
  4. 2.8 × 10⁻⁶
Answer and explanation

C: 2.8 × 10⁻⁷

The cubes are 2.16×10⁻⁷ and0.64×10⁻⁷. Their sum is2.80×10⁻⁷, which is2.8×10⁻⁷ in standard form.

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Question 4

Given logₐ2 = 0.693 and logₐ3 = 1.097, find logₐ13.5.

  1. 1.404
  2. 1.790
  3. 2.598
  4. 2.790
Answer and explanation

C: 2.598

Since13.5 = 27/2 = 3³/2, logarithm rules give logₐ13.5 = 3logₐ3 − logₐ2 = 3(1.097) − 0.693 = 2.598.

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Question 5

Simplify log₂96 − 2log₂6.

  1. 2 − log₂3
  2. 3 − log₂3
  3. log₂3 − 3
  4. log₂3 − 2
Answer and explanation

B: 3 − log₂3

Combining the logarithms gives log₂(96/36) = log₂(8/3). This equals log₂8 − log₂3 = 3 − log₂3.

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Question 6

If 8^(x/2) = 2^(3/8) × 4^(3/4), find x.

  1. 3/8
  2. 3/4
  3. 4/5
  4. 5/4
Answer and explanation

D: 5/4

Writing everything in base2 gives 2^(3x/2) = 2^(3/8 + 3/2) = 2^(15/8). Thus3x/2 = 15/8 and x = 5/4.

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Question 8

At what simple interest rate per year does ₦1000 grow to ₦1240 in 3 years?

  1. 6%
  2. 8%
  3. 10%
  4. 12%
Answer and explanation

B: 8%

The interest earned is ₦240. Using I = P rt/100 gives r = 100×240/(1000×3) = 8% per year. Simple interest uses the original principal throughout.

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Question 12

Solve 2/x − 3/y = 2 and 4/x + 3/y = 10.

  1. x = 3/2, y = 1/2
  2. x = 1/2, y = 3/2
  3. x = −1/2, y = −3/2
  4. x = 1/2, y = −3/2
Answer and explanation

B: x = 1/2, y = 3/2

Adding the equations gives 6/x = 12, so x = 1/2. Substituting back gives 4 − 3/y = 2, hence y = 3/2.

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Question 13

Find the minimum of x² − 3x + 2 over the real numbers.

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

A: −1/4

Completing the square gives(x − 3/2)² − 1/4. The squared term is nonnegative, so the minimum is−1/4 at x = 3/2.

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Question 14

Make f the subject of t = √[v/(1/f + 1/g)], where the denominators are nonzero.

  1. (gv − t²)/(gt²)
  2. gt²/(gv − t²)
  3. v/√t − 1/g
  4. gv/t² − g
Answer and explanation

B: gt²/(gv − t²)

Squaring gives t² = vfg/(g + f). Rearranging yields gt² = f(gv − t²), so f = gt²/(gv − t²).

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Question 17

Let f(x) = 2x + 4 and g(x) = 6x + 7 > 0. Solve f(x)/g(x) < 1.

  1. x < −3/4
  2. x > −4/3
  3. x > −3/4
  4. x > −12
Answer and explanation

C: x > −3/4

The positive denominator permits multiplication without reversing the sign:2x + 4 < 6x + 7. Thus x > −3/4, which also satisfies 6x + 7 > 0.

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Question 18

Solve 12x² < x + 1.

  1. −1/4 < x < 1/3
  2. 1/4 < x < 1/3
  3. −1/3 < x < 1/4
  4. −1/4 < x < −1/3
Answer and explanation

A: −1/4 < x < 1/3

Rearrange to 12x² − x − 1 < 0 and factor as(4x + 1)(3x − 1) < 0. The product is negative between−1/4 and 1/3.

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Question 19

If Sₙ = n² − 1 is the sum of the first n terms, find the nth term for n ≥ 2.

  1. 4n + 1
  2. 4n − 1
  3. 2n + 1
  4. 2n − 1
Answer and explanation

D: 2n − 1

For n ≥ 2, subtract successive partial sums: aₙ = Sₙ − Sₙ₋₁ = n² − 1 − [(n − 1)² − 1] = 2n − 1. The first term is separately S₁ = 0.

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Question 20

The nth term is 3^(1−n), starting at n = 1. Find the sum of the first three terms.

  1. 13/9
  2. 1
  3. 1/3
  4. 1/9
Answer and explanation

A: 13/9

The first three terms are 1,1/3 and 1/9. Expressing them over denominator 9 gives 9/9 + 3/9 + 1/9 = 13/9.

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Question 29

A sector has radius 6 cm and central angle 70°. Find its area using π = 22/7.

  1. 22 cm²
  2. 44 cm²
  3. 66 cm²
  4. 88 cm²
Answer and explanation

A: 22 cm²

The area is the fraction 70/360 of the full circle: (70/360)×(22/7)×6² = 22 cm². The central angle determines the fraction of the circle included.

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Question 30

A right prism is 11 cm long. Its trapezoidal cross-section has parallel sides 8 cm and 10 cm separated perpendicularly by 5 cm. Find its volume.

  1. 990 cm³
  2. 880 cm³
  3. 550 cm³
  4. 495 cm³
Answer and explanation

D: 495 cm³

The trapezoid area is½(8 + 10)×5 = 45 cm². Multiplying by the prism length gives 45×11 = 495 cm³.

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Question 33

A line makes an angle 135° with the positive x-axis and passes through(2,3). Find its equation.

  1. x − y = 1
  2. x + y = 1
  3. x + y = 5
  4. x − y = 5
Answer and explanation

C: x + y = 5

The gradient is tan 135° = −1. Using the point(2,3), y − 3 = −(x − 2), which rearranges to x + y = 5.

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Question 36

In triangle XYZ, angle XYZ = 60°, XY = 3 cm and YZ = 4 cm. Find XZ.

  1. √23 cm
  2. √13 cm
  3. 2√5 cm
  4. 2√3 cm
Answer and explanation

B: √13 cm

The cosine rule gives XZ² = 3² + 4² − 2×3×4×cos 60° = 25 − 12 = 13. Therefore XZ = √13 cm. The positive square root is required because a side length is positive.

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Question 37

In triangle XYZ, XY = 5 cm and XZ = 2 cm. Extending XZ through Z to E gives exterior angle YZE = 150°. If angle XYZ = θ, find sinθ.

  1. 3/5
  2. 1/2
  3. 2/5
  4. 1/5
Answer and explanation

D: 1/5

The interior angle XZY is 180° − 150° = 30°. By the sine rule, sinθ/2 = sin 30°/5, so sinθ = (2/5)×(1/2) = 1/5.

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Question 38

Differentiate (6x³ − 5x² + 1)/(3x²), for x ≠ 0.

  1. 2 + 2/(3x³)
  2. 2 + 1/(6x)
  3. 2 − 2/(3x³)
  4. 2 − 1/(6x)
Answer and explanation

C: 2 − 2/(3x³)

Divide each numerator term by 3x² to get 2x − 5/3 + (1/3)x⁻². Differentiating gives 2 − (2/3)x⁻³.

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Question 40

Find the gradient of y = 2√x − 1/x at x = 1.

  1. 0
  2. 1
  3. 2
  4. 3
Answer and explanation

C: 2

Differentiating gives dy/dx = 1/√x + 1/x². At x = 1, the gradient is 1 + 1 = 2. The derivative of−x⁻¹ contributes the positive term x⁻².

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Question 41

Integrate 1/x + cos x with respect to x, for x > 0.

  1. −1/x² + sin x + k
  2. ln x + sin x + k
  3. ln x − sin x + k
  4. −1/x² − sin x + k
Answer and explanation

B: ln x + sin x + k

For x > 0, the integral of 1/x is ln x and the integral of cos x is sin x. Adding the constant gives ln x + sin x + k.

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Question 43

A₦6000 monthly income is divided in a pie chart: Housing 69°, Transport 60°, Meal 61°, Others 50° and Basic the remainder. Find the Basic amount.

  1. ₦2000
  2. ₦2600
  3. ₦3100
  4. ₦3450
Answer and explanation

A: ₦2000

The listed sectors total 240°, leaving 120° for Basic. This is 120/360 = 1/3 of the income, so the amount is ₦6000/3 = ₦2000.

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Question 49

A basket contains 6 grapes,11 bananas and 13 oranges. One fruit is selected uniformly at random. Find the probability of a grape or banana.

  1. 17/30
  2. 11/30
  3. 6/30
  4. 5/30
Answer and explanation

A: 17/30

There are 30 fruits altogether and 17 are grapes or bananas. The favourable outcomes are disjoint, so the probability is 17/30.

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