JAMB Mathematics 2000

17 reviewed questions with answers and explanations.

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Question 2

A town has population 240000 in January 1998 and grows by 2% annually. What is its population in January 2000?

  1. 480000
  2. 249696
  3. 249600
  4. 244800
Answer and explanation

B: 249696

Two annual increases occur between the dates. Compound growth gives 240000×1.02² = 249696, rather than applying the increase just once.

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Question 5

Evaluate [(2.813 × 10⁻³) × 1.063]/(5.637 × 10⁻²), first rounding each number to two significant figures, then expressing the answer to two significant figures.

  1. 0.056
  2. 0.055
  3. 0.054
  4. 0.54
Answer and explanation

B: 0.055

The rounded inputs are 2.8 × 10⁻³, 1.1 and 5.6 × 10⁻². Their quotient is 0.00308/0.056 = 0.055, already expressed to two significant figures.

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Question 13

The third term of an arithmetic progression is 4x − 2y and the ninth is 10x − 8y. Find its common difference.

  1. 19x − 17y
  2. 8x − 4y
  3. x − y
  4. 2x
Answer and explanation

C: x − y

There are six common differences between the third and ninth terms. Thus d = [(10x − 8y) − (4x − 2y)]/6 = (6x − 6y)/6 = x − y.

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Question 14

For p ≠ 1, find the inverse of p under p * q = p + q − pq, where the identity is zero.

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

C: p/(p − 1)

The inverse q satisfies p + q − pq = 0. Rearranging gives q(1 − p) = −p, so q = p/(p − 1). The denominator requires p ≠ 1.

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Question 19

If x − 1, x + 1 and x − 2 are factors of ax³ + bx² + cx − 1, find a, b and c respectively.

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

A: −1/2, 1, 1/2

The product of the factors is x³ − 2x² − x + 2. Its multiplier must be −1/2 to give constant −1, yielding coefficients a = −1/2, b = 1 and c = 1/2.

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Question 21

Solve 2 − x > x².

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

D: −2 < x < 1

Move all terms to one side to get x² + x − 2 < 0, or (x + 2)(x − 1) < 0. The product is negative strictly between its roots, so −2 < x < 1.

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Question 22

If α and β are roots of 3x² + 5x − 2 = 0, find 1/α + 1/β.

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

D: 5/2

By the root sum and product formulas, α + β = −5/3 and αβ = −2/3. Therefore 1/α + 1/β = (α + β)/(αβ) = 5/2.

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Question 27

An equilateral triangle of side √3 cm is inscribed in a circle. Find the circle’s radius.

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

C: 1 cm

For an equilateral triangle of side a, the circumradius is a/√3. This also follows by taking two-thirds of the altitude a√3/2. Here the radius is 1 cm.

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Question 28

The lines 3y = 4x − 1 and Ky = x + 3 are perpendicular. Find K.

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

A: −4/3

Their slopes are 4/3 and 1/K. Perpendicular nonvertical lines have slope product −1, so 4/(3K) = −1 and K = −4/3.

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Question 30

In a circle centred at O, diameter FH meets chord EG at right angles at N. If NH = 8 cm and EG = 24 cm, find FH.

  1. 16 cm
  2. 20 cm
  3. 26 cm
  4. 32 cm
Answer and explanation

C: 26 cm

The perpendicular diameter bisects EG, so EN = NG = 12 cm. Intersecting chords give FN × NH = EN × NG = 144. Thus FN = 18 cm and FH = 18 + 8 = 26 cm.

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Question 31

For distinct fixed points P and Q, which option most precisely describes the complete locus of X such that XP = XQ?

  1. A straight line
  2. A circle
  3. The bisector of angle PXQ
  4. The perpendicular bisector of PQ
Answer and explanation

D: The perpendicular bisector of PQ

Points equidistant from two fixed points form their perpendicular bisector. Although this is a straight line, the perpendicular-bisector description uniquely identifies which line is required.

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Question 34

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The diagram shows y = x². Find the shaded area.

  1. 64 square units
  2. 128/3 square units
  3. 64/3 square units
  4. 32 square units

Question 40

The expression ax² + bx + c equals 5 when x = 1, and its derivative is 2x + 1. Find a, b and c respectively.

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

D: 1, 1, 3

Comparing 2ax + b with 2x + 1 gives a = 1 and b = 1. Substituting x = 1 into the original expression gives 1 + 1 + c = 5, so c = 3.

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Question 43

How many distinct arrangements can be made using all the letters of MATHEMATICS?

  1. 11!/(9! × 2!)
  2. 11!/(9! × 2! × 2!)
  3. 11!/(2! × 2! × 2!)
  4. 11!/(2! × 2!)
Answer and explanation

C: 11!/(2! × 2! × 2!)

There are eleven letters. M, A and T each appear twice, while all other letters appear once. Divide 11! by 2! for each repeated pair, giving 11!/(2! × 2! × 2!).

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Question 45

An integer is chosen uniformly at random from 1 through 20. Find the probability that it is not a multiple of 4.

  1. 3/4
  2. 3/10
  3. 1/4
  4. 1/20
Answer and explanation

A: 3/4

The multiples of 4 are 4, 8, 12, 16 and 20, making five excluded outcomes. The remaining fifteen out of twenty integers give probability 15/20 = 3/4.

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Question 47

Find the population variance of x, 2x, 3x, 4x and 5x.

  1. x√2
  2. 2x²
  3. x²
  4. 3x
Answer and explanation

B: 2x²

The mean is 3x. The squared deviations are 4x², x², 0, x² and 4x². Their sum is 10x², so dividing by five gives population variance 2x².

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Question 50

Pupils’ ages in years are grouped as 10–12, 13–15, 16–18, 19–20 and 21–23, with frequencies 6, 14, 15, 10 and 5 respectively. What percentage are over 15 but under 21?

  1. 35%
  2. 45%
  3. 50%
  4. 60%
Answer and explanation

C: 50%

The relevant groups are 16–18 and 19–20, containing 15 + 10 = 25 pupils. There are 6 + 14 + 15 + 10 + 5 = 50 pupils altogether, so the percentage is 50%.

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