JAMB Mathematics 1985

26 reviewed questions with answers and explanations.

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Question 1

Arrange 6/7, 13/15 and 0.865 in ascending order.

  1. 6/7 < 0.865 < 13/15
  2. 6/7 < 13/15 < 0.865
  3. 13/15 < 6/7 < 0.865
  4. 13/15 < 0.865 < 6/7
  5. 0.865 < 6/7 < 13/15
Answer and explanation

A: 6/7 < 0.865 < 13/15

6/7 = 0.857142...; 0.865; 13/15 = 0.866666... . Only A is ascending.

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Question 5

Evaluate 100 + 1/100 + 3/1000 + 27/10000, correct to two decimal places.

  1. 100.02
  2. 1000.02
  3. 100.22
  4. 100.01
  5. 100.51
Answer and explanation

A: 100.02

100 + 0.01 + 0.003 + 0.0027 = 100.0157, which rounds to 100.02.

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Question 6

Simplify 1/2 + 1/(2 + 1/(2 − 1/(4 + 1/5))).

  1. 3/4
  2. −1/3
  3. 169/190
  4. 13/15
  5. 1 21/169
Answer and explanation

C: 169/190

Work from the innermost fraction: 4 + 1/5 = 21/5, so 2 - 1/(21/5) = 2 - 5/21 = 37/21. Next, 2 + 1/(37/21) = 2 + 21/37 = 95/37. The whole expression is therefore 1/2 + 37/95 = 95/190 + 74/190 = 169/190, option C.

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Question 9

John gives one third of his money to Janet, who initially has ₦105. John then has one quarter of Janet’s new total. How much money did John have originally?

  1. ₦45
  2. ₦48
  3. ₦52
  4. ₦58
  5. ₦60
Answer and explanation

A: ₦45

Let original money be x. Remaining 2x/3 = (105 + x/3)/4. Thus 8x = 315 + x and x = 45. Check: John keeps 30, Janet has 120; 30 is one quarter of 120.

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Question 10

Find x if log₉(x) = 1.5.

  1. 72.0
  2. 27.0
  3. 36.0
  4. 3.5
  5. 24.5
Answer and explanation

B: 27.0

log base 9 of x = 1.5 gives x = 9^(3/2) = 27. Only B matches.

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Question 12

22½% of one Nigerian naira has the same value as 17⅒% of one unit of a foreign currency M. Express the value of one M in naira.

  1. 1 M = 15/57 naira
  2. 1 M = 2 11/57 naira
  3. 1 M = 1 18/57 naira
  4. 1 M = 38 1/4 naira
  5. 1 M = 38 43/4 naira
Answer and explanation

C: 1 M = 1 18/57 naira

22.5% of one naira has the same value as 17.1% of one M. Therefore one M = 22.5/17.1 naira = 75/57 naira = 1 18/57 naira.

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Question 13

Find the values of p for which x² − (p − 2)x + 2p + 1 = 0 has equal roots.

  1. 0, 12
  2. 1, 2
  3. 21, 0
  4. 4, 5
  5. 3, 4
Answer and explanation

A: 0, 12

Equal roots require discriminant zero: (p-2)^2 - 4(2p+1) = p^2 - 12p = p(p-12). Thus p is 0 or 12, A.

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Question 16

The cost per head of a meal is partly constant and partly inversely proportional to the number of people. It is 30 kobo for 100 people and 60 kobo for 40 people. Find the cost per head for 50 people.

  1. 15 kobo
  2. 45 kobo
  3. 20 kobo
  4. 50 kobo
  5. 40 kobo
Answer and explanation

D: 50 kobo

Cost per head C = a + b/n. Subtracting the conditions gives 30 = b(1/40 - 1/100), so b = 2000 and a = 10. At 50 people, C = 10 + 2000/50 = 50 kobo.

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Question 17

Factorise 9 − (x² − 3x − 1)².

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

A: −(x − 4)(x + 1)(x − 1)(x − 2)

9 - (x^2 - 3x - 1)^2 = (-x^2 + 3x + 4)(x^2 - 3x + 2) = -(x-4)(x+1)(x-1)(x-2).

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Question 19

Factorise abx² + 8y − 4bx − 2axy.

  1. (ax − 4)(bx − 2y)
  2. (ax + b)(x − 8y)
  3. (ax − 2y)(by − 4)
  4. (abx − 4)(x − 2y)
  5. (bx − 4)(ax − 2y)
Answer and explanation

A: (ax − 4)(bx − 2y)

Group as bx(ax-4) - 2y(ax-4), giving (ax-4)(bx-2y). Expansion gives abx^2 - 4bx - 2axy + 8y.

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Question 21

If the quadratic expression 3x² − 7x + R is a perfect square, find R.

  1. 49/24
  2. 49/3
  3. 49/6
  4. 49/12
  5. 49/36
Answer and explanation

D: 49/12

Complete the square: 3x^2 - 7x + R = 3(x - 7/6)^2 when R = 49/12. The discriminant check gives 49 - 12R = 0.

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Question 23

Solve the simultaneous equations 2x + y = 4 and x² + xy = −12. Give both ordered pairs (x, y).

  1. (6, -8); (-2,8)
  2. (3, -4); (-1, 4)
  3. (8, -4); (-1, 4)
  4. (-8, 6);(8, -2)
  5. (-4, 3);(4, -1)
Answer and explanation

A: (6, -8); (-2,8)

From y = 4 - 2x, x^2 + xy = -12 becomes x^2 - 4x - 12 = 0. Hence x = 6 or -2, paired with y = -8 or 8 respectively.

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Question 24

Solve the simultaneous equations 2x − 3y + 10 = 5 and 10x − 6y = 5.

  1. x = 2 1/2, y = 3 1/3
  2. x = 3 1/2, y = 2 1/3
  3. x = 2 1/4, y = 3
  4. x = 3 1/2, y = 2 1/5
  5. x = 2 1/2, y = 2 1/3
Answer and explanation

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

First rearrange 2x - 3y + 10 = 5 to get 2x - 3y = -5. Double it: 4x - 6y = -10. Subtract this from 10x - 6y = 5 to obtain 6x = 15, so x = 5/2. Substituting gives y = 10/3. Thus x = 2 1/2 and y = 3 1/3, option A.

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Question 26

In triangle XYZ, XY = 13 cm, YZ = 9 cm and XZ = 11 cm. If angle XYZ is θ, find cos θ.

  1. 4/39
  2. 43/39
  3. 209/286
  4. 1/6
  5. 43/78
Answer and explanation

E: 43/78

The angle at Y is opposite XZ = 11. Cosine rule gives (13^2 + 9^2 - 11^2)/(2*13*9) = 129/234 = 43/78.

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Question 28

Goals scored — Number of matches 0 — 3 1 — 5 2 — 7 3 — 4 4 — 1 5 — 0

Using the frequency data, find the mean number of goals per match and the mode, respectively.

  1. (1.75, 5)
  2. (1.75, 2)
  3. (1.75, 1)
  4. (2,2)
  5. (2,1)
Answer and explanation

B: (1.75, 2)

For goals 0,1,2,3,4,5 and frequencies 3,5,7,4,1,0: total frequency 20, weighted sum 35, mean 1.75, mode 2.

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Question 31

The corresponding lengths of two similar rectangular blocks are in the ratio 2:3. If the larger block has volume 351 cm³, find the volume of the smaller block.

  1. 234.00 cm³
  2. 526.50 cm³
  3. 166.00 cm³
  4. 729.75 cm³
  5. 104.00 cm³
Answer and explanation

E: 104.00 cm³

Similar solids have volume ratio equal to the cube of their length ratio. Smaller volume = 351*(2/3)^3 = 104 cubic centimetres.

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Question 32

The bearing of a bird from a hunter is N72°E. What is the bearing of the hunter from the bird?

  1. S18°W
  2. S72°W
  3. S72°E
  4. S27°E
  5. S27°W
Answer and explanation

B: S72°W

N72 degrees E is a whole-circle bearing of 72 degrees. The reverse bearing is 252 degrees, expressed as S72 degrees W.

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Question 33

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In triangle XYZ, Y, K and Z lie on a straight line, angle XKZ = 90°, XK = 15 cm, XZ = 25 cm and YK = 8 cm. Find the area of triangle XYZ.

  1. 180sq.cm
  2. 210sq.cm
  3. 160sq.cm
  4. 320sq.cm
  5. 390sq.cm

Question 35

An integer is chosen uniformly at random from 1 to 10 inclusive. What is the probability that it is prime or a multiple of 3?

  1. 7/10
  2. 3/5
  3. 4/5
  4. 1/2
  5. 3/10
Answer and explanation

B: 3/5

Primes in 1..10 are 2,3,5,7. Multiples of three are 3,6,9. Their union is 2,3,5,6,7,9: six outcomes out of ten, so 3/5. Count 3 only once; 1 is not prime.

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Question 36

Find the area of a regular hexagon inscribed in a circle of radius 8 cm.

  1. 16√3 cm²
  2. 96√3 cm²
  3. 192.3 cm²
  4. 16 cm²
  5. 32 cm²
Answer and explanation

B: 96√3 cm²

The six triangles formed by joining the centre to the vertices are equilateral, each with side 8. Each has area sqrt(3)*8^2/4 = 16sqrt(3). Total = 96sqrt(3) square centimetres, B.

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Question 37

M, N, Q and P lie in that order on a circle. Lines MN and PQ meet at X, with M–N–X and P–Q–X in that order. Lines NQ and MP meet at Y, with N–Q–Y and M–P–Y in that order. Angle MNQ = 86° and angle NQP = 122°. Find x = angle NXQ and y = angle QYP.

  1. (28°, 36°)
  2. (36°, 28°)
  3. (43°, 61°)
  4. (61°, 43°)
  5. (36°, 43°)
Answer and explanation

A: (28°, 36°)

In triangle NXQ, angle XNQ = 180° − 86° = 94° and angle NQX = 180° − 122° = 58°. Thus x = 180° − 94° − 58° = 28°. Opposite angles of cyclic quadrilateral MNQP sum to 180°, so angle MPQ = 180° − 86° = 94°. In triangle PQY, angle QPY = 180° − 94° = 86° and angle PQY = 180° − 122° = 58°. Therefore y = 180° − 86° − 58° = 36°.

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Question 38

If cos θ = √3/2 and θ is acute (0° < θ < 90°), calculate cot(90° − θ) / sin² θ.

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

A: 4√3/3

For an acute angle, cos(theta) = sqrt(3)/2 implies theta = 30 degrees. cot(90-theta)/sin^2(theta) = cot(60)/sin^2(30) = (1/sqrt(3))/(1/4) = 4sqrt(3)/3.

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Question 41

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GHIJKLMN is a cube of side length a. Find the length of HN shown in the diagram.

  1. 3√a
  2. 3a
  3. 3a2
  4. a√2
  5. a√3

Question 42

PQRS is a trapezium of area 14 cm², with PQ parallel to RS. If PQ = 4 cm and RS = 3 cm, find the area of triangle SQR in cm².

  1. 7.0
  2. 6.0
  3. 5.2
  4. 5.0
  5. 4.1
Answer and explanation

B: 6.0

Trapezium area = (4+3)h/2 = 14, hence height h = 4. Triangle SQR uses base SR = 3 and the same perpendicular height, giving area 3*4/2 = 6 square centimetres, B.

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Question 46

Two points X and Y are both at latitude 60°S, with longitudes 147°E and 153°W respectively. Find, to the nearest kilometre, the shorter distance between them along this parallel of latitude. Take the equatorial circumference 2πR = 4 × 10⁴ km.

  1. 28,850 km
  2. 16,667 km
  3. 8,333 km
  4. 6,667 km
  5. 3,333 km
Answer and explanation

E: 3,333 km

The longitude difference measured the long way is 147° + 153° = 300°. The shorter separation across the date line is 360° − 300° = 60°. The circumference of the 60°S parallel is 40000 cos 60° = 20000 km. The distance is (60/360) × 20000 = 3333.333… km, which rounds to 3,333 km.

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Question 47

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In the diagram, the circle has radius 3 units. Two radii enclose an angle of 120°. Find the area of the shaded minor segment.

  1. 3π
  2. 9√3/4
  3. 3(π − 3√3/4)
  4. 3(√3 − π)/4
  5. π + 9√3/4

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