2025 WAEC GCE Mathematics (Obj/Essay) Answers

General Mathematics Paper 2, WASSCE (SC), 2023:

1. (a) A car travels 112 km at an average speed of 70 km/h. It then continues for 60 km at an average speed of 50 km/h. Calculate the total time taken for the entire journey.
2. (b) If x = 2 and y = 3, evaluate the given algebraic expression after substituting the values.

Solution

(a) Journey time calculation

Formula: Time = Distance ÷ Speed

• First journey: 112 ÷ 70 = 1.6 hours
• Second journey: 60 ÷ 50 = 1.2 hours
• Total time: 1.6 + 1.2 = 2.8 hours

0.8 hour = 0.8 × 60 minutes = 48 minutes.
Final Answer: 2 hours 48 minutes

(b) Algebraic evaluation

Simplify the expression algebraically where possible, then substitute x = 2 and y = 3. Carry out the arithmetic carefully to arrive at the final simplified result.

Observation (Examiner’s Report)

Key Exam Tips

• Always compute each time segment separately before summing.
• Convert decimal parts of hours into minutes to give clear answers.
• Simplify algebraic expressions first to reduce substitution errors.
• Always present answers with correct units (hours, minutes, etc.).

General Mathematics Paper 2, WASSCE (SC), 2023:

Question 2

1. In a football match, tickets for children and adults were sold at D 3.00 and D 5.00 respectively. If 400 people attended the match and D 1,700.00 was collected in ticket sales:
   • (a) How many tickets were sold to adults?
   • (b) Mr. Sonko sold 250 tickets. If 175 of these were for adults, how much did he make in total sales?

Solution

(a) Tickets sold to adults

Let the number of children’s tickets = x and adults’ tickets = y.

• Total tickets: x + y = 400 … (1)
• Total sales: 3x + 5y = 1700 … (2)

From equation (1): x = 400 – y

Substitute into (2): 3(400 – y) + 5y = 1700

Simplify: 1200 – 3y + 5y = 1700 → 2y = 500 → y = 250.

Answer: 250 adult tickets were sold.

(b) Mr. Sonko’s ticket sales

• Total tickets sold = 250
• Adult tickets = 175
• Children’s tickets = 250 – 175 = 75

Sales value = (3 × 75) + (5 × 175)

= 225 + 875 = D 1,100.00

Answer: Mr. Sonko made D 1,100.00 in sales.

Observation (Examiner’s Report)

Key Exam Tips

• Translate word problems into simultaneous equations before solving.
• Always label answers with correct units (e.g., D for Dalasi).
• Check decimal places in financial calculations.
• Show working clearly to secure method marks.

General Mathematics Paper 2, WASSCE (SC), 2023:

Question 3

In the diagram, PQR is an equilateral triangle with side 18 cm.
M is the midpoint of QR. An arc of a circle with centre P touches QR at M and meets PQ at A and PR at B.
Calculate, correct to two decimal places, the area of the shaded region.
[Take π = 3.142]

Solution

1. Construct the perpendicular: Drop a perpendicular line from P to QR. Since M is the midpoint of QR, PM bisects QR, making triangle PMQ a right-angled triangle.
2. Apply Pythagoras theorem:\( PM^2 + (9)^2 = (18)^2 \)\( PM^2 + 81 = 324 \)\( PM^2 = 243 \)\( PM = 9\sqrt{3} \, cm \)
3. Area of sector PAMB:Radius = \( 9\sqrt{3} \). Angle at P = 60° (since PQR is equilateral).\( \text{Sector area} = \dfrac{60}{360} \times \pi r^2 \)= \( \dfrac{1}{6} \times 3.142 \times (9\sqrt{3})^2 \)= 127.29 cm² (approx).
4. Area of triangle PQR:Area of equilateral triangle = \( \dfrac{\sqrt{3}}{4} \times (side)^2 \)= \( \dfrac{\sqrt{3}}{4} \times (18^2) \)= 140.30 cm² (approx).
5. Area of shaded region:= Area of triangle – Area of sector= 140.30 – 127.29= 13.01 cm²

Final Answer

The area of the shaded region = 13.01 cm².

Observation (Examiner’s Report)

Exam Tips

• Always identify the geometric shapes within a diagram (triangles, sectors, etc.).
• For equilateral triangles, remember all angles = 60°.
• Apply Pythagoras theorem for perpendicular bisectors in geometry questions.
• Present answers to two decimal places when required.

General Mathematics Paper 2, WASSCE (SC), 2023: Question 4

Question

In the diagram, P, Q, R, and S are points on a circle with centre K.
KR is a bisector of ∠SRQ. Given that ∠SRQ = 41° and ∠SKR = 80°, find:

1. ∠RQP
2. ∠SPQ

(Diagram not drawn to scale)

Solution

1. Step 1: Identify triangle KRS
   Since KR is the bisector of ∠SRQ, triangle KRS is isosceles with base angles ∠KRS = ∠KSR.
   In triangle KRS:∠KRS + ∠KSR + ∠SKR = 180°2∠KSR + 80° = 180°2∠KSR = 100°∠KSR = 50°
2. Step 2: Find ∠RQP
   In cyclic quadrilateral RQPS:∠RQP + (∠KSR + ∠SRQ) = 180°∠RQP + (50° + 41°) = 180°∠RQP + 91° = 180°∠RQP = 89°
3. Step 3: Find ∠SPQ
   Since ∠SRQ = 2∠KRS = 2 × 50° = 100°
   In cyclic quadrilateral SRQP:∠SRQ + ∠SPQ = 180°100° + ∠SPQ = 180°∠SPQ = 80°

Final Answers

• ∠RQP = 89°
• ∠SPQ = 80°

Observation (Examiner’s Report)

Exam Tips

• Remember that opposite angles in a cyclic quadrilateral add up to 180°.
• In an isosceles triangle, base angles are equal.
• Always check if angle bisectors divide triangles symmetrically.
• Carefully label diagrams for clarity before solving circle geometry questions.