JAMB Mathematics 2001

18 reviewed questions with answers and explanations.

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Question 3

Evaluate 21.05347 − 1.6324 × 0.43 to three decimal places.

  1. 20.351
  2. 20.352
  3. 20.980
  4. 20.981
Answer and explanation

B: 20.352

Apply multiplication before subtraction: 1.6324 × 0.43 = 0.701932. The difference is 20.351538, which rounds to 20.352 at three decimal places.

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Question 4

Evaluate (0.14² × 0.275)/(7 × 0.02), correct to three decimal places.

  1. 0.033
  2. 0.039
  3. 0.308
  4. 0.358
Answer and explanation

B: 0.039

The numerator is 0.0196 × 0.275 = 0.00539 and the denominator is 0.14. Their quotient is 0.0385, which rounds up to 0.039 at three decimal places.

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Question 5

Given p = 1 + √2 and q = 1 − √2, evaluate (p² − q²)/(2pq).

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

C: −2√2

We have p + q = 2, p − q = 2√2 and pq = −1. Hence p² − q² = 4√2, and division by 2pq = −2 gives −2√2.

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Question 10

Solve m² + n² = 29 and m + n = 7. Give the ordered pairs (m,n).

  1. (5,2) and (5,3)
  2. (5,3) and (3,5)
  3. (2,3) and (3,5)
  4. (2,5) and (5,2)
Answer and explanation

D: (2,5) and (5,2)

Squaring the sum gives 49 = 29 + 2mn, so mn = 10. Thus m and n are roots of t² − 7t + 10 = 0, namely 2 and 5, in either order.

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Question 11

Divide a^(3x) − 26a^(2x) + 156a^x − 216 by a^(2x) − 24a^x + 108, where the divisor is nonzero and a > 0.

  1. a^x − 18
  2. a^x − 6
  3. a^x − 2
  4. a^x + 2
Answer and explanation

C: a^x − 2

Let t = a^x. The numerator equals (t − 2)(t² − 24t + 108), as expansion confirms. Cancelling the nonzero divisor gives t − 2 = a^x − 2.

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Question 12

Find all integer pairs (x,y) satisfying 3y + 5x ≤ 15, with 0 < y < 3 and x > 0.

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

C: (1,1), (1,2), (2,1)

The possible y values are 1 and 2. For y = 1, x can be 1 or 2; for y = 2, only x = 1 works. These give the three listed pairs.

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Question 18

For K = [[2,1],[3,4]], find K² + K + I, where I is the 2 × 2 identity matrix.

  1. [[9,8],[22,23]]
  2. [[10,7],[21,24]]
  3. [[7,2],[12,21]]
  4. [[6,3],[13,20]]
Answer and explanation

B: [[10,7],[21,24]]

Matrix multiplication gives K² = [[7,6],[18,19]]. Adding K and the identity matrix gives [[10,7],[21,24]]. The identity contributes one only to each diagonal entry.

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Question 25

A chord ST is equal in length to the radius r of its circle. Find the length of the minor arc ST.

  1. πr/2
  2. πr/3
  3. πr/6
  4. πr/12
Answer and explanation

B: πr/3

The two radii and the chord form an equilateral triangle, so the minor central angle is 60° = π/3 radians. Arc length rθ is therefore πr/3.

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Question 27

Find the complete locus of a point whose perpendicular distance from y = 4 is a constant k ≥ 0.

  1. y = 4 + k
  2. y = k − 4
  3. y = k ± 4
  4. y = 4 ± k
Answer and explanation

D: y = 4 ± k

Distance from a horizontal line is the absolute difference of y coordinates. Thus |y − 4| = k, giving the two parallel lines y = 4 + k and y = 4 − k.

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Question 29

P(−6,1) and Q(6,6) are the endpoints of a circle’s diameter. Calculate its radius.

  1. 3.5 units
  2. 6.5 units
  3. 7.0 units
  4. 13.0 units
Answer and explanation

B: 6.5 units

The diameter length is √[(6 + 6)² + (6 − 1)²] = √169 = 13 units. The radius is half the diameter, giving 6.5 units.

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Question 33

Differentiate (2x + 5)²(x − 4) with respect to x.

  1. (2x + 5)(6x − 11)
  2. (2x + 5)(2x − 13)
  3. 4(2x + 5)(x − 4)
  4. 4(2x + 5)(4x − 3)
Answer and explanation

A: (2x + 5)(6x − 11)

The product rule gives 4(2x + 5)(x − 4) + (2x + 5)². Factoring out 2x + 5 leaves 4x − 16 + 2x + 5 = 6x − 11.

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Question 36

Find dV/dr for a sphere when its radius r = 1.

  1. 4π
  2. 8π
  3. 12π
  4. 24π
Answer and explanation

A: 4π

A sphere has volume V = 4πr³/3. Differentiating with respect to its radius gives dV/dr = 4πr². At r = 1, the rate is 4π.

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Question 37

Find the dimensions of the rectangle with greatest area for a fixed positive perimeter p.

  1. A square of side p/4
  2. A square of side p/2
  3. A square of side p
  4. A square of side 2p
Answer and explanation

A: A square of side p/4

If one side is x, the other is p/2 − x. The area x(p/2 − x) is a downward-opening quadratic with its maximum at x = p/4. Both sides are therefore p/4.

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Question 38

Evaluate ∫2(2x − 3)^(2/3) dx.

  1. 2x − 3 + k
  2. 2(2x − 3) + k
  3. (6/5)(2x − 3)^(5/3) + k
  4. (3/5)(2x − 3)^(5/3) + k
Answer and explanation

D: (3/5)(2x − 3)^(5/3) + k

Let u = 2x − 3, so du = 2 dx. The integral becomes ∫u^(2/3) du = (3/5)u^(5/3) + k. Substituting back gives the stated result.

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Question 40

Cars counted by colour were: yellow 3, white 4, red 8, green 2, blue 6 and black 2. What fraction of the cars were yellow?

  1. 4/15
  2. 1/5
  3. 3/25
  4. 2/25
Answer and explanation

C: 3/25

The counts total 3 + 4 + 8 + 2 + 6 + 2 = 25 cars. Three were yellow, so the fraction is 3/25. Each colour count contributes once to the total.

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Question 43

Scores 4, 7, 8, 11, 13 and 8 have respective frequencies 3, 5, 2, 7, 2 and 1. Find the mean score.

  1. 11.0
  2. 9.5
  3. 8.7
  4. 7.0
Answer and explanation

C: 8.7

Multiply each score by its frequency. The total score is 12 + 35 + 16 + 77 + 26 + 8 = 174 across 20 observations, giving mean 174/20 = 8.7.

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Question 46

If Q₁, Q₂ and Q₃ denote the lower quartile, median and upper quartile of a distribution, which expression gives its interquartile range?

  1. Q₃ − Q₁
  2. Q₃ − Q₂
  3. Q₂ − Q₁
  4. (Q₃ − Q₁)/2
Answer and explanation

A: Q₃ − Q₁

The interquartile range measures the span of the middle half of the data. Subtract the lower quartile from the upper quartile: IQR = Q₃ − Q₁.

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Question 49

A bowl contains 1 blue, 2 black, 4 yellow, 5 white and 3 brown beads. A bead is chosen uniformly at random. Find the probability it is blue or white.

  1. 1/15
  2. 1/3
  3. 2/5
  4. 7/15
Answer and explanation

C: 2/5

There are 15 beads in total. The blue and white groups contain 1 + 5 = 6 beads, so their combined probability is 6/15 = 2/5.

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