WAEC Mathematics 2025 past questions and answers

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

Question 1

Bearings

The bearing of F from G is 064°. What is the bearing of G from F?

  1. A.128°
  2. B.180°
  3. C.244°
  4. D.116°
Show the answer and explanation

Answer: C (244°)

Explanation

The back bearing differs by 180°: 064° + 180° = 244°.

Open this question

Question 3

Linear and Simultaneous Equations

Ajoke will be 31 years in x years time. If Ajoke is 6 years older than Chioma, find an expression for Chioma's present age.

  1. A.25 − x
  2. B.31 + x
  3. C.25 + x
  4. D.31 − x
Show the answer and explanation

Answer: A (25 − x)

Explanation

Ajoke is now 31 − x years old, so Chioma is now 31 − x − 6 = 25 − x.

Open this question

Question 5

Mensuration of Solids

Two ball bearings have volumes 1.6 cm3 and 5.4 cm3. Find the ratio of their surface areas.

  1. A.4 : 9
  2. B.2 : 3
  3. C.3 : 8
  4. D.5 : 12
Show the answer and explanation

Answer: A (4 : 9)

Explanation

Volumes are in the ratio 16 : 54 = 8 : 27 = 2³ : 3³, so the radii are 2 : 3 and the surface areas 2² : 3² = 4 : 9.

Open this question

Question 6

Number Bases

Given that 215 = 14x, find the value of x.

  1. A.4
  2. B.5
  3. C.6
  4. D.7
Show the answer and explanation

Answer: D (7)

Explanation

215 = 10 + 1 = 11, and 14x = x + 4, so x + 4 = 11 and x = 7.

Open this question

Question 7

Algebraic Expressions

Given that x + y = 1 and x + 3y = 5, find the value of (x2 + 4xy + 3y2).

  1. A.1
  2. B.5
  3. C.10
  4. D.25
Show the answer and explanation

Answer: B (5)

Explanation

x² + 4xy + 3y² = (x + y)(x + 3y) = 1 × 5 = 5.

Open this question

Question 8

Logic

Consider the following statements:
p: The weather is warm.
q: The sun is shining.
Which of the following correctly represents the statement. "The sun is shining if and only if the weather is warm"?

  1. A.p ⇔ ~q
  2. B.~p ⇔ ~q
  3. C.~q ⇔ ~p
  4. D.q ⇔ p
Show the answer and explanation

Answer: D (q ⇔ p)

Explanation

"q if and only if p" is written q ⇔ p.

Open this question

Question 10

Probability

In a class of 42 students, 21 offer History and 28 offer Government. If each student offers at least one of the two subjects, find the probability that a student selected at random from the class offers Government only.

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

Answer: C (1/2)

Explanation

Both subjects: 21 + 28 − 42 = 7. Government only: 28 − 7 = 21, so the probability is 21/42 = ½.

Open this question

Question 11

Sets

The sets M = {x: 2 ≤ x ≤ 6} and N = {x: 4 ≤ x ≤ 8} are subsets of μ = {x: 1 ≤ x ≤ 10, where x is an integer}. Find M′ ∩ N′

  1. A.{1, 7, 8}
  2. B.{1, 8, 9}
  3. C.{1, 9, 10}
  4. D.{1, 8, 10}
Show the answer and explanation

Answer: C ({1, 9, 10})

Explanation

M′ = {1, 7, 8, 9, 10} and N′ = {1, 2, 3, 9, 10}; the common elements are 1, 9 and 10.

Open this question

Question 12

Variation

A variable P varies inversely as the square of Q. If P = 5 when Q = 6, find Q when P = 1.8.

  1. A.5
  2. B.10
  3. C.15
  4. D.20
Show the answer and explanation

Answer: B (10)

Explanation

P = k/Q²: k = 5 × 36 = 180. Then Q² = 180 ÷ 1.8 = 100, so Q = 10.

Open this question

Question 13

Mensuration of Plane Shapes

A square has diagonal length of 10 cm. Find the perimeter of the square.

  1. A.10√2 cm
  2. B.15√2 cm
  3. C.20√2 cm
  4. D.25√2 cm
Show the answer and explanation

Answer: C (20√2 cm)

Explanation

Side² + side² = 100, so side = √50 = 5√2 cm and perimeter = 4 × 5√2 = 20√2 cm.

Open this question

Question 14

Algebraic Expressions

Make r the subject of the relation: 1/p = b/t + c/r

  1. A.r = pct/(t − pb)
  2. B.r = pct/(t + pb)
  3. C.r = pc/(t − pb)
  4. D.r = pc/(t + pb)
Show the answer and explanation

Answer: A (r = pct/(t − pb))

Explanation

c/r = 1/p − b/t = (t − pb)/(pt), so r = pct/(t − pb).

Open this question

Question 15

Fractions, Decimals and Approximation

Write 0.0364891 in three significant figures and express the answer in standard form.

  1. A.3.65 × 102
  2. B.3.64 × 102
  3. C.3.64 × 10−2
  4. D.3.65 × 10−2
Show the answer and explanation

Answer: D (3.65 × 10−2)

Explanation

To three significant figures 0.0364891 is 0.0365 (the 4th figure, 8, rounds up), which is 3.65 × 10−2.

Open this question

Question 16

Mensuration of Solids

The radius and height of a circular base cylinder are 8 cm and 14 cm respectively. Find its curved surface area. [Take π = 22/7]

  1. A.562 cm2
  2. B.704 cm2
  3. C.352 cm2
  4. D.680 cm2
Show the answer and explanation

Answer: B (704 cm2)

Explanation

Curved surface area = 2πrh = 2 × 22/7 × 8 × 14 = 704 cm².

Open this question

Question 17

Mensuration of Solids

A cylinder and a cone have the same volume. If the height of the cone is 24 cm, find the height of the cylinder

  1. A.4 cm
  2. B.8 cm
  3. C.12 cm
  4. D.16 cm
Show the answer and explanation

Answer: B (8 cm)

Explanation

Taking the same base radius, πr²h = ⅓πr² × 24, so h = 8 cm.

Open this question

Question 18

Trigonometry

Given that sin (x − 46)° = cos 62°, find the value of x.

  1. A.46
  2. B.62
  3. C.74
  4. D.90
Show the answer and explanation

Answer: C (74)

Explanation

cos 62° = sin 28°, so x − 46 = 28 and x = 74.

Open this question

Question 19

Ratio, Percentages and Rates

The population of a town increases by 2% every year. After 2 years, the population is 83,232. Calculate, correct to the nearest thousand, the original population.

  1. A.80,000
  2. B.81,000
  3. C.82,000
  4. D.83,000
Show the answer and explanation

Answer: A (80,000)

Explanation

Two years of 2% growth multiply the population by 1.02² = 1.0404, so the original population is 83 232 ÷ 1.0404 = 80 000.

Open this question

Question 20

Fractions, Decimals and Approximation

Simplify: (2⅓ + 5/2 − 1)/(12 − 11¾)

  1. A.10⅔
  2. B.15⅓
  3. C.16¼
  4. D.17½
Show the answer and explanation

Answer: B (15⅓)

Explanation

The top is 7/3 + 5/2 − 1 = 23/6 and the bottom is ¼, so the value is 23/6 × 4 = 46/3 = 15⅓.

Open this question

Question 21

Linear and Simultaneous Equations

Two consecutive odd integers are such that the sum of 5 times the smaller and 3 times the bigger integer is 222. Find the value of the smaller integer.

  1. A.31
  2. B.29
  3. C.27
  4. D.25
Show the answer and explanation

Answer: C (27)

Explanation

5x + 3(x + 2) = 222 gives 8x = 216, so x = 27.

Open this question

Question 22

Surds

Simplify √200 − √72

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

Answer: B (4√2)

Explanation

√200 = 10√2 and √72 = 6√2, so the difference is 4√2.

Open this question

Question 23

Logarithms

Given that log416 = logy36, find the value of y.

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

Answer: C (6)

Explanation

log416 = 2, so y² = 36 and y = 6.

Open this question

Question 26

Probability

A fair dice is thrown once. Find the probability of obtaining an odd number or a prime number

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

Answer: C (2/3)

Explanation

Odd or prime gives {1, 2, 3, 5}: 4 out of 6 outcomes, so 2/3.

Open this question

Question 27

Progressions

The third and ninth terms of an Arithmetic Progression (AP) are 9 and −27 respectively. Find the fifth term.

  1. A.18
  2. B.3
  3. C.−3
  4. D.−6
Show the answer and explanation

Answer: C (−3)

Explanation

6d = −27 − 9 = −36, so d = −6 and a = 21. T5 = 21 + 4(−6) = −3.

Open this question

Question 28

Commercial Arithmetic

A profit of 8% was made when an article was sold for $500.00. At what price should it be sold to make a profit of 16%?

  1. A.$537.04
  2. B.$540.00
  3. C.$573.04
  4. D.$580.00
Show the answer and explanation

Answer: A ($537.04)

Explanation

Cost price = 500 ÷ 1.08 = $462.96. For 16% profit: 462.96 × 1.16 ≈ $537.04.

Open this question

Question 29

Commercial Arithmetic

John sold an article for ₦105,000.00 at a loss of 4%. Find the cost price of the article.

  1. A.₦100,800.00
  2. B.₦105,400.00
  3. C.₦109,200.00
  4. D.₦109,375.00
Show the answer and explanation

Answer: D (₦109,375.00)

Explanation

At a 4% loss the selling price is 96% of the cost: cost = 105 000 ÷ 0.96 = ₦109,375.

Open this question

Question 30

Trigonometry

In ΔPQR, |QR| = 2 cm, ∠PRQ = 60° and ∠PQR = 90°. Find |PR|

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

Answer: B (4 cm)

Explanation

PR is the hypotenuse: cos 60° = QR/PR, so PR = 2 ÷ ½ = 4 cm.

Open this question

Question 31

Mensuration of Solids

The volume of a cone of height 18 cm is 8,316 cm3. Find the base radius of the cone. [Take π = 22/7]

  1. A.21 cm
  2. B.14 cm
  3. C.28 cm
  4. D.42 cm
Show the answer and explanation

Answer: A (21 cm)

Explanation

⅓ × 22/7 × r² × 18 = 8316 gives r² = 441, so r = 21 cm.

Open this question

Question 32

Coordinate Geometry

Two towns X and Y are located at points (−2, −5) and (3, 7) respectively. Calculate the distance between the two towns.

  1. A.114 units
  2. B.25 units
  3. C.13 units
  4. D.12 units
Show the answer and explanation

Answer: C (13 units)

Explanation

Distance = √(5² + 12²) = √169 = 13 units.

Open this question

Question 34

Angles and Polygons

The interior angles of a triangle are in the ratio 2 : 5 : 8. Find the difference between the smallest and largest angles

  1. A.12°
  2. B.24°
  3. C.48°
  4. D.72°
Show the answer and explanation

Answer: D (72°)

Explanation

The angles are 2/15, 5/15 and 8/15 of 180°: 24°, 60° and 96°. The difference is 96° − 24° = 72°.

Open this question

Question 35

Indices

Given that (x−n)4 = 1/x6, find the value of n.

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

Answer: C (3/2)

Explanation

x−4n = x−6, so 4n = 6 and n = 3/2.

Open this question

Question 36

Trigonometry

The angle of elevation of the top P of a ladder PQ from its base Q, 12 m away from a wall is 36°. Find, correct to the nearest metre, the length of the ladder.

  1. A.9 m
  2. B.12 m
  3. C.15 m
  4. D.17 m
Show the answer and explanation

Answer: C (15 m)

Explanation

cos 36° = 12/length, so the length = 12 ÷ 0.809 ≈ 14.8 m ≈ 15 m.

Open this question

Question 37

Triangles and Quadrilaterals

In the diagram, ∠PMQ = 34° and ∠NQM = 28°. Find ∠QTN.

Diagram for this question
  1. A.152°
  2. B.146°
  3. C.124°
  4. D.118°
Show the answer and explanation

Answer: D (118°)

Explanation

∠PNT = ∠PMT = 34° (angles in the same segment). In triangle QTN, ∠QTN = 180° − 28° − 34° = 118°.

Open this question

Question 38

Triangles and Quadrilaterals

In the diagram, ∠PMQ = 34° and ∠NQM = 28°. Find ∠MPN.

Diagram for this question
  1. A.62°
  2. B.56°
  3. C.34°
  4. D.28°
Show the answer and explanation

Answer: A (62°)

Explanation

∠MPN is an exterior angle of triangle QPM, so it equals 28° + 34° = 62°.

Open this question

Question 39

Measures of Dispersion

The height of 4 orange seedlings: are 2 cm, 5 cm, 7 cm and 10 cm. Calculate the variance

  1. A.7.2
  2. B.6.0
  3. C.8.5
  4. D.9.1
Show the answer and explanation

Answer: C (8.5)

Explanation

The mean is 24/4 = 6. Squared deviations 16, 1, 1, 16 add to 34, so the variance is 34/4 = 8.5.

Open this question

Question 40

Circle Theorems

In the diagram, P, Q, R and S are points on the circle with centre O. ∠QRS = (5x)° and ∠QOS = (6x + 24)°. Find ∠QPS.

Diagram for this question
  1. A.75°
  2. B.87°
  3. C.105°
  4. D.150°
Show the answer and explanation

Answer: A (75°)

Explanation

∠QPS = ½∠QOS = 3x + 12, and opposite angles of the cyclic quadrilateral give 5x + 3x + 12 = 180, so x = 21. ∠QPS = 3(21) + 12 = 75°.

Open this question

Question 41

Circle Theorems

In the diagram, Q, R, S and T are points on the circle with centre O. ∠STR = 37° and ∠QRT = 59°. Find ∠SPR.

Diagram for this question
  1. A.121°
  2. B.96°
  3. C.84°
  4. D.59°
Show the answer and explanation

Answer: B (96°)

Explanation

∠SQR = ∠STR = 37° (same segment). ∠SPR is an exterior angle of triangle PQR, so it equals 37° + 59° = 96°.

Open this question

Question 42

Coordinate Geometry

What is the gradient of the line 7x − 5y + 3 = 0?

  1. A.7/5
  2. B.5/7
  3. C.−5/7
  4. D.−7/5
Show the answer and explanation

Answer: A (7/5)

Explanation

Rearranged, y = (7/5)x + 3/5, so the gradient is 7/5.

Open this question

Question 43

Angles and Polygons

In the diagram SR ∥ UW, ∠WVT = x, ∠VUT = y, ∠RST = 45° and ∠VTU = 20°. Find the angle marked x.

Diagram for this question
  1. A.20°
  2. B.65°
  3. C.45°
  4. D.135°
Show the answer and explanation

Answer: C (45°)

Explanation

S, V and T lie on one line and SR ∥ VW, so ∠WVT and ∠RST are corresponding angles: x = 45°.

Open this question

Question 44

Angles and Polygons

In the diagram SR ∥ UW, ∠WVT = x, ∠VUT = y, ∠RST = 45° and ∠VTU = 20°. Find the angle marked y.

Diagram for this question
  1. A.20°
  2. B.25°
  3. C.45°
  4. D.135°
Show the answer and explanation

Answer: B (25°)

Explanation

∠UVT = 180° − 45° = 135° (angles on the straight line UVW). In triangle UVT, y = 180° − 135° − 20° = 25°.

Open this question

Question 45

Ratio, Percentages and Rates

Kwakye, Sabina and Owusu shared an amount of $12,000.00. Kwakye had 20% of the amount and the remaining amount was shared between Sabina and Owusu in the ratio 5 : 3 respectively. How much did Sabina receive?

  1. A.$2,400.00
  2. B.$3,600.00
  3. C.$6,000.00
  4. D.$9,600.00
Show the answer and explanation

Answer: C ($6,000.00)

Explanation

Kwakye gets $2,400, leaving $9,600. Sabina gets 5/8 × 9600 = $6,000.

Open this question

Question 46

Logic

The Venn diagram illustrates the information about the sets, P = {Pharmacists}, C = {Smart People} and F = {My Friends}. Which of the following is a valid conclusion from the diagram?

Diagram for this question
  1. A.All pharmacists are my friends.
  2. B.Some of my friends are smart.
  3. C.All smart people are my friends.
  4. D.None of my friends is smart.
Show the answer and explanation

Answer: B (Some of my friends are smart.)

Explanation

F overlaps C, so some of my friends are smart. P is not wholly inside F and C is not inside F, so A and C are not valid.

Open this question

Question 47

Measures of Central Tendency

The table shows the ages of students in a class.
Age (Years): 13, 14, 15, 16, 17
Number of Students: 3, 14, 13, 8, 3
Find, correct to the nearest whole number, the average age of the students.

  1. A.13
  2. B.14
  3. C.15
  4. D.16
Show the answer and explanation

Answer: C (15)

Explanation

Σfx = 39 + 196 + 195 + 128 + 51 = 609 and Σf = 41, so the mean is 609/41 ≈ 14.9 ≈ 15.

Open this question

Question 48

Measures of Central Tendency

The table shows the ages of students in a class.
Age (Years): 13, 14, 15, 16, 17
Number of Students: 3, 14, 13, 8, 3
What is the median age?

  1. A.14
  2. B.15
  3. C.16
  4. D.17
Show the answer and explanation

Answer: B (15)

Explanation

With 41 students the median is the 21st. Cumulative frequencies are 3, 17, 30, …, so the 21st student is 15.

Open this question

Question 49

Quadratic Equations

Solve: d2 − 4d − 96 = 0.

  1. A.d = 8, 12
  2. B.d = −8, −12
  3. C.d = −8, 12
  4. D.d = 8, −12
Show the answer and explanation

Answer: C (d = −8, 12)

Explanation

(d − 12)(d + 8) = 0, so d = 12 or d = −8.

Open this question

Theory question 101

Sets

Given that μ = {x: 1 < x < 20, where x is an integer}, P = {x: x is a multiple of 3} and Q = {x: x is a prime number} where P and Q are subsets of μ, find: (a) P ∩ Q′; (b) P′ ∪ Q; (c) (P ∪ Q)′.

Show the model answer
Model answer

μ = {2, 3, …, 19}; P = {3, 6, 9, 12, 15, 18}; Q = {2, 3, 5, 7, 11, 13, 17, 19}
(a) P ∩ Q′ = {6, 9, 12, 15, 18} (3 is prime, so it is not in Q′)
(b) P′ = {2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19}; P′ ∪ Q = {2, 3, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19}
(c) P ∪ Q = {2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}; (P ∪ Q)′ = {4, 8, 10, 14, 16}

Explanation

List μ, P and Q first; remember 3 is both a multiple of 3 and prime.

Open this question

Theory question 102

Quadratic Equations

The product of the ages of Adu and Tanko is 9 less than Akorfa's age. If Tanko is 4 years older than Adu and Akorfa's age is six times Tanko's age, find Akorfa's age.

Show the model answer
Model answer

Let Adu = x, Tanko = x + 4, Akorfa = 6(x + 4)
x(x + 4) + 9 = 6(x + 4) ⇒ x² − 2x − 15 = 0 ⇒ (x − 5)(x + 3) = 0 ⇒ x = 5
Tanko = 9, so Akorfa = 54 years

Explanation

Ages are positive, so reject x = −3.

Open this question

Theory question 103

Differentiation

A company installs solar panels in its premises to reduce its electricity cost. The monthly savings on electricity in $, is modelled by S = 200 + 50x − 2x2, where x is the number of months after installation (a) At what time will the savings on electricity stop increasing? (b) Find the maximum savings.

Show the model answer
Model answer

(a) dS/dx = 50 − 4x = 0 ⇒ x = 12.5 months (12½ months after installation)
(b) S = 200 + 50(12.5) − 2(12.5)² = 200 + 625 − 312.5 = $512.50

Explanation

Savings stop increasing where the gradient dS/dx is zero.

Open this question

Theory question 104

Trigonometry

The diagram shows a tower TR and an observer at O. |OR| = 84 m and the angle of elevation of the top of the tower T from O is 57°.
(a) Calculate, correct to three significant figures, the height of the tower.
(b) The observer at O, moved away from the tower in the same straight line until the angle of elevation of T is 49°. Find, correct to two decimal places, how far the observer moved backwards.

Diagram for this question
Show the model answer
Model answer

(a) TR = 84 tan 57° ≈ 129.35 m ≈ 129 m (3 s.f.)
(b) New distance from R = 129.35 ÷ tan 49° ≈ 112.44 m
Distance moved = 112.44 − 84 ≈ 28.44 m

Explanation

Keep the unrounded height for part (b).

Open this question

Theory question 105

Measures of Central Tendency

The data represent the scores obtained by 9 applicants in an interview arranged in ascending order: (3x + 2), 22, (4x − 2), 23, 25, (5x − 4), 29, 29 and (x2 − 7). (a) Given that the range is 9, find the: (i) value of x; (ii) mean mark of the applicants. (b) If four of the applicants who obtained the highest score were selected, determine the pass mark.

Show the model answer
Model answer

(a)(i) Range: (x² − 7) − (3x + 2) = 9 ⇒ x² − 3x − 18 = 0 ⇒ (x − 6)(x + 3) = 0; x = −3 would make the first score negative, so x = 6
Scores: 20, 22, 22, 23, 25, 26, 29, 29, 29
(ii) Mean = 225 ÷ 9 = 25
(b) The four highest scores are 29, 29, 29 and 26, so the pass mark is 26

Explanation

Substitute x back in and check the list is in ascending order.

Open this question

Theory question 106

Commercial Arithmetic

In a certain year, the consumption pattern of electricity charges in a town was as follows: the cost of the first 30 units was $1.00 per unit; the cost of the next 30 units was $7.00 per unit, the cost of each additional unit was $5.00. (a) If Amaka used 420 units of electricity in January that year, calculate the amount paid. (b) If Amaka paid $2,740.00 in the month of February, calculate the number of units of electricity consumed. (c) Find, correct to two decimal places, the percentage change in units of electricity consumed by Amaka in January and February.

Show the model answer
Model answer

(a) 30 × $1 + 30 × $7 + 360 × $5 = 30 + 210 + 1800 = $2,040
(b) Extra = 2740 − 2040 = $700 ⇒ 700 ÷ 5 = 140 more units; units = 420 + 140 = 560
(c) (560 − 420)/420 × 100% ≈ 33.33% increase

Explanation

Charge each block of units at its own rate.

Open this question

Theory question 107

Linear and Simultaneous Equations

Yaro drove from a town Gaja to Banga. After 2 hours in the journey, he observed that he had covered 80 km and realized that if he continued driving at same average speed, he would end up being late for 15 minutes. If he decided to increase the average speed by 10 km/h, he would arrive at Banga 36 minutes earlier. Find the distance between Gaja and Banga.

Show the model answer
Model answer

Speed so far = 80 ÷ 2 = 40 km/h. Let the distance be d km and the planned time T h
At 40 km/h: d/40 = T + ¼
At 50 km/h for the rest: 2 + (d − 80)/50 = T − 36/60 = T − 0.6
Subtract: d/40 − 2 − (d − 80)/50 = 0.85 ⇒ d/200 − 0.4 = 0.85 ⇒ d = 250 km
(Check: 250/40 = 6.25 h = 6 h + 15 min; 2 + 170/50 = 5.4 h = 6 h − 36 min)

Explanation

Use the planned (on-time) journey time as the link between the two cases.

Open this question

Theory question 108

Constructions

(a) Using a ruler and a pair of compasses only, construct: (i) a quadrilateral PQRS such that |PQ| = 8.5 cm, |QR| = 7.5 cm, ∠QPS = 60°, ∠PQR = 105° and S is a point on the locus, L1, which is equidistant from PQ and QR. (ii) locus L2, of points equidistant from P and Q. (iii) locate the point K, which is the point of intersection of L1 and L2. (b) Measure |KS|

Show the model answer
Model answer

(a)(i) Draw PQ = 8.5 cm; construct ∠PQR = 105° (90° + 15°) and mark QR = 7.5 cm; construct ∠QPS = 60° at P
L1 is the bisector of ∠PQR; S is where it meets the 60° arm from P; join RS
(ii) L2 is the perpendicular bisector of PQ
(iii) K is where L2 meets L1
(b) |KS| ≈ 1.0 cm (accept 0.9 – 1.1 cm)

Explanation

Points equidistant from two lines lie on the bisector of the angle between them.

Open this question

Theory question 109

Commercial Arithmetic

(a) Mrs Otoo spends 1/9 of her monthly salary on rent, 1/2 on food, 1/4 on clothes and still had $195.00 left. How much does she earn in a month? (b) A sector of a circle of radius 6 cm subtends an angle of 105° at the centre. Calculate the: (i) perimeter; (ii) area, of the sector [Take π = 22/7]

Show the model answer
Model answer

(a) Spent 1/9 + 1/2 + 1/4 = 31/36, so 5/36 of salary = $195 ⇒ salary = 195 × 36 ÷ 5 = $1,404
(b)(i) Arc = 105/360 × 2 × 22/7 × 6 = 11 cm; perimeter = 6 + 6 + 11 = 23 cm
(ii) Area = 105/360 × 22/7 × 6² = 33 cm²

Explanation

The perimeter of a sector includes the two radii.

Open this question

Theory question 110

Variation

(a) The cost, C, of feeding some students in a class is partly constant and partly varies as the number of students, n, in the class. For 8 students, the cost is $70.00 and for 10 students the cost is $90.00. Find: (i) an expression for C in terms of n; (ii) the cost of feeding 12 students.
(b) In the diagram. A, B, C and D are points on a circle. AB ∥ DC, AC and BD intersect at X and ∠BDC = 40°. Find: (i) ∠ACD; (ii) ∠AXD

Diagram for this question
Show the model answer
Model answer

(a)(i) C = a + bn: a + 8b = 70 and a + 10b = 90 ⇒ b = 10, a = −10, so C = 10n − 10
(ii) C = 10(12) − 10 = $110
(b)(i) ∠ABD = ∠BDC = 40° (alternate angles, AB ∥ DC); ∠ACD = ∠ABD = 40° (angles in the same segment)
(ii) ∠AXD is an exterior angle of triangle XDC: ∠AXD = 40° + 40° = 80°

Explanation

'Partly constant and partly varies' means C = a + bn.

Open this question

Theory question 111

Matrices and Determinants

(a) Given that P = (2 4 / −9 1) and Q = (1 −1 / 3 −2), find PQ + 2Q
(b) A bag contains 8 red balls and some white balls, all of the same size. If the probability of drawing at random a white ball from the bag is half the probability of drawing a red ball, find the number of white balls in the bag.

Show the model answer
Model answer

(a) PQ = (2+12 −2−8 / −9+3 9−2) = (14 −10 / −6 7)
2Q = (2 −2 / 6 −4)
PQ + 2Q = (16 −12 / 0 3)
(b) Let w = white balls: w/(8 + w) = ½ × 8/(8 + w) ⇒ w = 4 white balls

Explanation

Multiply rows of P by columns of Q.

Open this question

Theory question 112

Progressions

(a) The eighth term of an Arithmetic Progression (A.P) is 46 and the sum of the first eight terms is 200. Find the: (i) first term; (ii) sum of the first 12 terms.
(b) The points X(70°S, 60°E) and Y(7°S, 60°E) lie on the surface of the earth. (i) Illustrate the information in a diagram. (ii) Find the distance between X and Y along the meridian. [Take π = 22/7 and R = 6,400 km]

Show the model answer
Model answer

(a) a + 7d = 46; S8 = 4(2a + 7d) = 200 ⇒ 2a + 7d = 50 ⇒ a = 4, d = 6
(i) First term = 4
(ii) S12 = 6(2 × 4 + 11 × 6) = 6 × 74 = 444
(b)(i) Sketch the earth with the equator and the meridian 60°E; mark Y at 7°S and X at 70°S on it
(ii) Angle between them = 70° − 7° = 63°; distance = 63/360 × 2 × 22/7 × 6400 = 7,040 km

Explanation

Both points are south, so subtract the latitudes.

Open this question

Theory question 113

Measures of Dispersion

The following are the marks scored by 20 students in a test:
15, 11, 17, 25, 13, 15, 16, 22, 24, 27,
20, 22, 15, 16, 15, 19, 22, 24, 22, 11
(a) Prepare a frequency table for the distribution using class intervals 10 − 12, 13 − 15, 16 − 18 … .
(b) Calculate the variance of the distribution.
(c) If the pass mark for the test was 16, find the probability, that a student selected at random from the class failed.

Show the model answer
Model answer

(a) 10–12: 2; 13–15: 5; 16–18: 3; 19–21: 2; 22–24: 6; 25–27: 2 (total 20)
(b) Class marks 11, 14, 17, 20, 23, 26; Σfx = 373, mean = 18.65
Σf(x − x̄)² = 117.05 + 108.11 + 8.17 + 3.65 + 113.54 + 108.05 ≈ 458.55; variance = 458.55 ÷ 20 ≈ 22.93
(c) Students scoring below 16: 11, 11, 13, 15, 15, 15, 15 = 7, so P(failed) = 7/20

Explanation

Use the class marks (mid-points) as x.

Open this question