JAMB Mathematics 1997 past questions and answers

47 questions. Try each one, then open the answer to check yourself and read the explanation.

Question 1

Number Bases

If (1P03)4 = 11510, find P.

  1. A.0
  2. B.1
  3. C.2
  4. D.3
Show the answer and explanation

Answer: D (3)

Explanation

In base 4, 1P03 = 1×64 + P×16 + 0×4 + 3 = 67 + 16P. Setting 67 + 16P = 115 gives 16P = 48, so P = 3.

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Question 2

Fractions, Decimals and Approximation

Evaluate 64.7642 − 35.2362 correct to 3 significant figures.

  1. A.2960
  2. B.2950
  3. C.2860
  4. D.2850
Show the answer and explanation

Answer: B (2950)

Explanation

Use the difference of two squares: (64.764 − 35.236)(64.764 + 35.236) = 29.528 × 100 = 2952.8, which is 2950 to 3 significant figures.

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Question 3

Fractions, Decimals and Approximation

Find the value of (0.006)3 + (0.004)3 in standard form.

  1. A.2.8 × 10−9
  2. B.2.8 × 10−8
  3. C.2.8 × 10−7
  4. D.2.8 × 10−4
Show the answer and explanation

Answer: C (2.8 × 10−7)

Explanation

0.006³ = 2.16 × 10⁻⁷ and 0.004³ = 0.64 × 10⁻⁷. Adding gives 2.8 × 10⁻⁷.

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Question 4

Logarithms

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

  1. A.1.404
  2. B.1.790
  3. C.2.598
  4. D.2.790
Show the answer and explanation

Answer: C (2.598)

Explanation

13.5 = 27/2 = 3³/2, so log 13.5 = 3 log 3 − log 2 = 3(1.097) − 0.693 = 2.598.

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Question 5

Logarithms

Simplify log296 − 2log26

  1. A.2 − log23
  2. B.3 − log23
  3. C.log23 − 3
  4. D.log23 − 2
Show the answer and explanation

Answer: B (3 − log23)

Explanation

2log₂6 = log₂36, so the expression is log₂(96/36) = log₂(8/3) = log₂8 − log₂3 = 3 − log₂3.

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Question 6

Indices

If 8x/2 = (23/8)(43/4), find x.

  1. A.3/8
  2. B.3/4
  3. C.4/5
  4. D.5/4
Show the answer and explanation

Answer: D (5/4)

Explanation

Write everything as powers of 2: 8^(x/2) = 2^(3x/2) and 2^(3/8) × 2^(3/2) = 2^(15/8). So 3x/2 = 15/8, giving x = 5/4.

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Question 7

Surds

Simplify (2√3 + 3√5)/(3√5 − 2√3)

  1. A.(19 + 4√15)/11
  2. B.(19 + 4√15)/19
  3. C.(19 + 2√15)/11
  4. D.(19 + 2√15)/19
Show the answer and explanation

Answer: A ((19 + 4√15)/11)

Explanation

Multiply top and bottom by (3√5 + 2√3). The top becomes (2√3 + 3√5)² = 57 + 12√15 and the bottom 45 − 12 = 33. Dividing by 3 gives (19 + 4√15)/11.

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Question 8

Commercial Arithmetic

Find the simple interest rate per cent per annum at which N1000 accumulates to N1240 in 3 years.

  1. A.6%
  2. B.8%
  3. C.10%
  4. D.12%
Show the answer and explanation

Answer: B (8%)

Explanation

The interest is N240. R = 100I/(PT) = (100 × 240)/(1000 × 3) = 8%.

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Question 10

Sets

A survey of 100 students in an institution shows that 80 students speak Hausa and 20 students speak Igbo, while only 9 students speak both languages. How many students speak neither Hausa nor Igbo?

  1. A.0
  2. B.9
  3. C.11
  4. D.20
Show the answer and explanation

Answer: B (9)

Explanation

Students speaking at least one language = 80 + 20 − 9 = 91. So 100 − 91 = 9 speak neither.

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Question 11

Polynomials

If the function f(x) = x3 + 2x2 + qx − 6 is divisible by x + 1, find q

  1. A.−5
  2. B.−2
  3. C.2
  4. D.5
Show the answer and explanation

Answer: A (−5)

Explanation

By the factor theorem f(−1) = 0: −1 + 2 − q − 6 = 0, so −5 − q = 0 and q = −5.

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Question 12

Linear and Simultaneous Equations

Solve the simultaneous equations 2/x − 3/y = 2, 4/x + 3/y = 10

  1. A.x = 3/2, y = 1/2
  2. B.x = 1/2, y = 3/2
  3. C.x = −1/2, y = −3/2
  4. D.x = 1/2, y = −3/2
Show the answer and explanation

Answer: B (x = 1/2, y = 3/2)

Explanation

Adding the equations gives 6/x = 12, so x = 1/2. Then 4/x = 8, so 3/y = 2 and y = 3/2.

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Question 13

Quadratic Equations

Find the minimum value of x2 − 3x + 2 for all real values of x.

  1. A.−1/4
  2. B.−1/2
  3. C.1/4
  4. D.1/2
Show the answer and explanation

Answer: A (−1/4)

Explanation

x² − 3x + 2 = (x − 3/2)² − 1/4, which is smallest when x = 3/2, giving −1/4.

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Question 14

Algebraic Expressions

Make f the subject of the formula t = √(v/(1/f + 1/g))

  1. A.(gv − t2)/gt2
  2. B.gt2/(gv − t2)
  3. C.v/(1/t2 − 1/g)
  4. D.gv/(t2 − g)
Show the answer and explanation

Answer: B (gt2/(gv − t2))

Explanation

Square: t² = vfg/(f + g). Then t²f + t²g = vfg, so f(vg − t²) = gt² and f = gt²/(gv − t²).

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Question 15

Quadratic Equations

What value of g will make the expression 4x2 − 18xy + g a perfect square?

  1. A.9
  2. B.9y2/4
  3. C.81y2
  4. D.81y2/4
Show the answer and explanation

Answer: D (81y2/4)

Explanation

(2x − 9y/2)² = 4x² − 18xy + 81y²/4, so g = 81y²/4.

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Question 16

Polynomials

Find the value of K if (5 + 2r)/((r + 1)(r − 2)) expressed in partial fraction is K/(r − 2) + L/(r + 1) where K and L are constants.

  1. A.3
  2. B.2
  3. C.1
  4. D.−1
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Answer: A (3)

Explanation

5 + 2r = K(r + 1) + L(r − 2). Putting r = 2 gives 9 = 3K, so K = 3.

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Question 17

Inequalities

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

  1. A.x < −3/4
  2. B.x > −4/3
  3. C.x > −3/4
  4. D.x > −12
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Answer: C (x > −3/4)

Explanation

Since g(x) > 0 we can multiply through: 2x + 4 < 6x + 7, so −3 < 4x and x > −3/4.

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Question 18

Inequalities

Find the range of value of x which satisfies the inequality 12x2 < x + 1.

  1. A.−1/4 < x < 1/3
  2. B.1/4 < x < 1/3
  3. C.−1/3 < x < 1/4
  4. D.−1/4 < x < −1/3
Show the answer and explanation

Answer: A (−1/4 < x < 1/3)

Explanation

12x² − x − 1 < 0 factorises as (4x + 1)(3x − 1) < 0, which holds between the roots: −1/4 < x < 1/3.

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Question 19

Progressions

Sn is the sum of the first n terms of a series given by Sn = n2 − 1. Find the nth term.

  1. A.4n + 1
  2. B.4n − 1
  3. C.2n + 1
  4. D.2n − 1
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Answer: D (2n − 1)

Explanation

Tn = Sn − Sn−1 = (n² − 1) − ((n − 1)² − 1) = 2n − 1.

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Question 20

Progressions

The nth term of a sequence is given by 31−n. Find the sum of the first three terms of the sequence.

  1. A.13/9
  2. B.1
  3. C.1/3
  4. D.1/9
Show the answer and explanation

Answer: A (13/9)

Explanation

The terms are 3⁰ = 1, 3⁻¹ = 1/3 and 3⁻² = 1/9. Their sum is 13/9.

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Question 21

Binary Operations

Two binary operations ∗ and ⊗ are defined as m∗n = mn − n − 1 and m ⊗ n = mn + n − 2 for all real numbers m, n. Find the value of 3 ⊗ (4 ∗ 5)

  1. A.60
  2. B.57
  3. C.54
  4. D.42
Show the answer and explanation

Answer: C (54)

Explanation

4 ∗ 5 = 20 − 5 − 1 = 14. Then 3 ⊗ 14 = 42 + 14 − 2 = 54.

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Question 22

Binary Operations

If x∗y = x + y − xy, find x, when (x∗2) + (x∗3) = 68

  1. A.24
  2. B.22
  3. C.−12
  4. D.−21
Show the answer and explanation

Answer: D (−21)

Explanation

x∗2 = x + 2 − 2x = 2 − x and x∗3 = 3 − 2x. So 5 − 3x = 68, giving x = −21.

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Question 23

Matrices and Determinants

Determine x + y if (2 −3 / −1 4)(x / y) = (−1 / 8)
(The 2 × 2 matrix has rows 2, −3 and −1, 4; the column vectors are (x, y) and (−1, 8).)

  1. A.3
  2. B.4
  3. C.7
  4. D.12
Show the answer and explanation

Answer: C (7)

Explanation

The equations are 2x − 3y = −1 and −x + 4y = 8. Solving gives y = 3 and x = 4, so x + y = 7.

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Question 24

Matrices and Determinants

Find the non-zero positive value of x which satisfies the equation
| x 1 0 |
| 1 x 1 | = 0
| 0 1 x |

  1. A.2
  2. B.√3
  3. C.√2
  4. D.1
Show the answer and explanation

Answer: C (√2)

Explanation

Expanding along the first row: x(x² − 1) − 1(x − 0) = x³ − 2x = 0. So x(x² − 2) = 0 and the positive non-zero root is √2.

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Question 25

Circle Theorems

Each of the base angles of an isosceles triangle is 58° and all the vertices of the triangles lie on a circle. Determine the angles which the base of the triangle subtends at the centre of the circle.

  1. A.128°
  2. B.116°
  3. C.64°
  4. D.58°
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Answer: A (128°)

Explanation

The apex angle is 180° − 2(58°) = 64°. The angle at the centre is twice the angle at the circumference, so 128°.

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Question 26

Angles and Polygons

From the figure above, FK//GR and FH = GH, ∠RFK = 34° and ∠FGH = 47°. Calculate the angle marked x

Diagram for this question
  1. A.42°
  2. B.52°
  3. C.64°
  4. D.72°
Show the answer and explanation

Answer: B (52°)

Explanation

In the figure F, G are P, Q. Since PK ∥ QR, ∠PRQ = ∠KPR = 34°. PH = QH makes ∠QPH = 47°. In triangle PQR, ∠QPR = 180° − 47° − 34° = 99°, so x = 99° − 47° = 52°.

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Question 27

Circle Theorems

The figure above shows circles of radii 3cm and 2cm with centre at X and Y respectively. The circles have a transverse common tangent of length 25cm. Calculate XY.

Diagram for this question
  1. A.√630cm
  2. B.√626cm
  3. C.√615cm
  4. D.√600cm
Show the answer and explanation

Answer: B (√626cm)

Explanation

Drawn as in the figure, the tangent, the difference of the radii and XY form a right-angled triangle: XY² = 25² + (3 − 2)² = 626, so XY = √626 cm.

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Question 28

Mensuration of Plane Shapes

A chord of a circle of diameter 42cm subtends an angle of 60° at the centre of the circle. Find the length of the minor arc.

  1. A.22cm
  2. B.44cm
  3. C.110cm
  4. D.220cm
Show the answer and explanation

Answer: A (22cm)

Explanation

Arc length = (60/360) × 2πr = (1/6) × 2 × (22/7) × 21 = 22 cm.

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Question 29

Mensuration of Plane Shapes

An arc of a circle subtends an angle 70° at the centre. If the radius of the circle is 6cm, calculate the area of the sector subtended by the given angle.

  1. A.22cm2
  2. B.44cm2
  3. C.66cm2
  4. D.88cm2
Show the answer and explanation

Answer: A (22cm2)

Explanation

Area = (70/360) × (22/7) × 6² = 22 cm².

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Question 30

Mensuration of Solids

Find the volume of the prism above.

Diagram for this question
  1. A.990cm3
  2. B.880cm3
  3. C.550cm3
  4. D.495cm3
Show the answer and explanation

Answer: D (495cm3)

Explanation

The cross-section is a trapezium with parallel sides 8 cm and 10 cm and height 5 cm: area = ½(8 + 10) × 5 = 45 cm². Volume = 45 × 11 = 495 cm³.

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Question 31

Mensuration of Solids

A cone with the sector angle of 45° is cut out of a circle of radius r cm. Find the base radius of the cone.

  1. A.r/16cm
  2. B.r/8cm
  3. C.r/4cm
  4. D.r/2cm
Show the answer and explanation

Answer: B (r/8cm)

Explanation

The arc of the sector becomes the base circumference: (45/360) × 2πr = 2πR, so R = r/8.

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Question 32

Loci

A point P moves so that it is equidistant from point L and M. If LM is 16cm, find the distance of P from LM when P is 10cm from L.

  1. A.12cm
  2. B.10cm
  3. C.8cm
  4. D.6cm
Show the answer and explanation

Answer: D (6cm)

Explanation

P lies on the perpendicular bisector of LM, 8 cm from L along LM. By Pythagoras the distance is √(10² − 8²) = 6 cm.

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Question 33

Coordinate Geometry

The angle between the positive horizontal axis and a given line is 135°. Find the equation of the line if it passes through the point (2, 3)

  1. A.x − y = 1
  2. B.x + y = 1
  3. C.x + y = 5
  4. D.x − y = 5
Show the answer and explanation

Answer: C (x + y = 5)

Explanation

Gradient = tan 135° = −1. So y − 3 = −(x − 2), which gives x + y = 5.

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Question 34

Coordinate Geometry

Find the distance between the point Q(4, 3) and the point common to the lines 2x − y = 4 and x + y = 2.

  1. A.3√10
  2. B.3√5
  3. C.√26
  4. D.√13
Show the answer and explanation

Answer: D (√13)

Explanation

Adding the equations gives 3x = 6, so x = 2 and y = 0. Distance from (4, 3) to (2, 0) = √(2² + 3²) = √13.

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Question 35

Trigonometry

The angle of elevation of a building from a measuring instrument placed on the ground is 30°. If the building is 40m high, how far is the instrument from the foot of the building?

  1. A.20/√3 m
  2. B.40/√3 m
  3. C.20√3m
  4. D.40√3m
Show the answer and explanation

Answer: D (40√3m)

Explanation

tan 30° = height/distance, so distance = 40 ÷ (1/√3) = 40√3 m.

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Question 36

Trigonometry

In a triangle XYZ, if ∠XYZ is 60°, XY = 3cm and YZ = 4cm. Calculate the length of the side XZ.

  1. A.√23cm
  2. B.√13cm
  3. C.2√5cm
  4. D.2√3cm
Show the answer and explanation

Answer: B (√13cm)

Explanation

By the cosine rule, XZ² = 3² + 4² − 2(3)(4)cos 60° = 25 − 12 = 13, so XZ = √13 cm.

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Question 37

Trigonometry

In the figure above, XYZ is a triangle with XY = 5cm, XZ = 2cm and XZ is produced to E marking the angle YZE = 150°. If the angle XYZ = θ calculate the value of sin θ.

Diagram for this question
  1. A.3/5
  2. B.1/2
  3. C.2/5
  4. D.1/5
Show the answer and explanation

Answer: D (1/5)

Explanation

∠XZY = 180° − 150° = 30°. By the sine rule, sin θ/XZ = sin 30°/XY, so sin θ = 2 × ½ ÷ 5 = 1/5.

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Question 38

Differentiation

Differentiate (6x3 − 5x2 + 1)/3x2 with respect to x

  1. A.2 + 2/(3x3)
  2. B.2 + 1/(6x)
  3. C.2 − 2/(3x3)
  4. D.2 − 1/(6x)
Show the answer and explanation

Answer: C (2 − 2/(3x3))

Explanation

Divide first: 2x − 5/3 + (1/3)x⁻². Differentiating gives 2 − (2/3)x⁻³ = 2 − 2/(3x³).

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Question 39

Differentiation

d/dx cos (3x2 − 2x) is equal to

  1. A.−sin (6x − 2)
  2. B.−sin (3x2 − 2x)
  3. C.(6x − 2)sin(3x2 − 2x)
  4. D.−(6x − 2)sin(3x2 − 2x)
Show the answer and explanation

Answer: D (−(6x − 2)sin(3x2 − 2x))

Explanation

By the chain rule, the derivative of cos u is −sin u × du/dx, with du/dx = 6x − 2.

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Question 40

Differentiation

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

  1. A.0
  2. B.1
  3. C.2
  4. D.3
Show the answer and explanation

Answer: C (2)

Explanation

dy/dx = 1/√x + 1/x². At x = 1 this is 1 + 1 = 2.

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Question 41

Integration

Integrate 1/x + cos x with respect to x

  1. A.−1/x + sin x + k
  2. B.ln x + sin x + k
  3. C.ln x − sin x + k
  4. D.−1/x sin x + k
Show the answer and explanation

Answer: B (ln x + sin x + k)

Explanation

∫(1/x)dx = ln x and ∫cos x dx = sin x, so the result is ln x + sin x + k.

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Question 43

Data Presentation

The pie chart above shows the income of a civil servant in a month. If his monthly income is N6 000, find his monthly basic salary.

Diagram for this question
  1. A.N2 000
  2. B.N2 600
  3. C.N3 100
  4. D.N3 450
Show the answer and explanation

Answer: A (N2 000)

Explanation

Basic salary angle = 360° − (69° + 60° + 61° + 50°) = 120°. So basic = (120/360) × 6000 = N2 000.

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Question 44

Data Presentation

In an examination, the result of a certain school is as shown in the histogram above. How many candidates did the school present?

Diagram for this question
  1. A.12
  2. B.16
  3. C.18
  4. D.19
Show the answer and explanation

Answer: D (19)

Explanation

Add the frequencies (heights) of the bars: 3 + 5 + 8 + 1 + 2 = 19.

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Question 45

Measures of Central Tendency

Age: 20, 25, 30, 35, 40, 45
No. of people: 3, 5, 1, 1, 2, 3
Find the median age of the frequency distribution in the table above.

  1. A.20
  2. B.25
  3. C.30
  4. D.35
Show the answer and explanation

Answer: B (25)

Explanation

There are 15 people, so the median is the 8th. The cumulative frequencies are 3, 8, …, so the 8th person is aged 25.

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Question 46

Measures of Central Tendency

The following are the scores of ten students in a test of 20 marks: 15, 16, 17, 13, 16, 8, 5, 16, 19, 17. What is the modal score?

  1. A.13
  2. B.15
  3. C.16
  4. D.19
Show the answer and explanation

Answer: C (16)

Explanation

16 appears three times, more than any other score.

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Question 47

Measures of Dispersion

Find the standard deviation of the following data −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5.

  1. A.2
  2. B.3
  3. C.√10
  4. D.√π
Show the answer and explanation

Answer: C (√10)

Explanation

The mean is 0. The sum of squares is 2(1 + 4 + 9 + 16 + 25) = 110, so the variance is 110/11 = 10 and the standard deviation is √10.

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Question 48

Measures of Dispersion

Find the difference between the range and the variance of the following set of numbers 4, 9, 6, 3, 2, 8, 10, 5, 6, 7 where Σd2 = 60.

  1. A.2
  2. B.3
  3. C.4
  4. D.6
Show the answer and explanation

Answer: A (2)

Explanation

Range = 10 − 2 = 8. Variance = Σd²/n = 60/10 = 6. The difference is 8 − 6 = 2.

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Question 49

Probability

In a basket of fruits, there are 6 grapes, 11 bananas and 13 oranges. If one fruit is chosen at random, what is the probability that the fruit is either a grape or a banana?

  1. A.17/30
  2. B.11/30
  3. C.6/30
  4. D.5/30
Show the answer and explanation

Answer: A (17/30)

Explanation

There are 30 fruits and 6 + 11 = 17 are grapes or bananas, so the probability is 17/30.

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