WASSCE General Mathematics 2025 (Paper 1)

A man cycles a distance of $(3 𝑎) k m$ at $𝑉 k m / h$ and then walks a distance of $𝑎 k m$ at $(𝑉 - 7) k m / h$ Find the total number of hours he spent travelling.
A. $3 𝑎 𝑉 + 𝑎 𝑉 - 7$
B. $𝑉 3 𝑎 + 𝑉 - 7 𝑎$
C. $2 𝑉 - 7 4 𝑎$
D. $4 𝑎 2 𝑉 - 7$

Answer: A. $3 𝑎 𝑉 + 𝑎 𝑉 - 7$
Explanation: Total time is the sum of the time spent cycling and the time spent walking. Using the formula $T i m e = D i s t a n c e S p e e d$ :
Cycling time $= 3 𝑎 𝑉$
Walking time $= 𝑎 𝑉 - 7$
Total time $= 3 𝑎 𝑉 + 𝑎 𝑉 - 7$

2. A function is defined by $𝑔 : 𝑥 \to 2 𝑥 - 1 3 𝑥 - 2$ For what value of $𝑥$ is $𝑔$ not defined?
A. $12$
B. $- 23$
C. $- 12$
D. $23$

Answer: D. $23$
Explanation: A rational function is not defined when the denominator is equal to zero. To find this value, set $3 𝑥 - 2 = 0$ . Solving for $𝑥$ gives: $3 𝑥 = 2 ⟹ 𝑥 = 23$

3. A scientific sensor records air pollution levels in base $3$ If the highest recorded pollution level for a week was $21203$ what is the decimal equivalent?
A. $72$
B. $69$
C. $65$
D. $62$

Answer: B. $69$
Explanation: Convert $21203$ to base $10$ by multiplying each digit by its positional power of $3$ : $(2 \times 33) + (1 \times 32) + (2 \times 31) + (0 \times 30) = 54 + 9 + 6 + 0 = 69$

4. In the diagram $𝑃 𝑄 𝑅 𝑆 𝑇$ is a trapezium. $| 𝑆 𝑇 | = 3 c m$ $| 𝑅 𝑆 | = 𝑦 c m$ and $| 𝑃 𝑆 | = 4 c m$ For what values of $𝑦$ will the area of the trapezium $𝑃 𝑄 𝑅 𝑆 𝑇$ be less than or equal to $10 c m 2$ ?
A. $𝑦 \leq 1 c m$
B. $𝑦 \leq 4 c m$
C. $𝑦 \leq 2 c m$
D. $𝑦 \leq 6 c m$

Answer: A. $𝑦 \leq 1 c m$
Explanation: The area of a trapezium is $12 (s u m o f p a r a l l e l s i d e s) \times h e i g h t$ . Here, the parallel sides are $𝑦$ and $𝑦 + 3$ , and the height is $4$ . $A r e a = 12 (𝑦 + 𝑦 + 3) \times 4 = 2 (2 𝑦 + 3) = 4 𝑦 + 6$ Setting the inequality $4 𝑦 + 6 \leq 10 ⟹ 4 𝑦 \leq 4 ⟹ 𝑦 \leq 1$ .

5. The diagram shows a sector of a circle of radius $42 c m$ If the sector is folded to form a right circular cone, find the surface area of the cone. [Take $𝜋 = 227$ ]
A. $4, 158 c m 2$
B. $5, 544 c m 2$
C. $29, 106 c m 2$
D. $11, 088 c m 2$

Answer: A. $4, 158 c m 2$
Explanation: The surface area of the cone formed is equal to the area of the major sector. The angle of the major sector is $360 \circ - 90 \circ = 270 \circ$ . $A r e a = 270360 \times 227 \times 422 = 34 \times 227 \times 1764 = 4158 c m 2$

6. In the diagram, $∠ 𝑄 𝑃 𝑅 = 90 \circ$ If $𝑞 2 = 25 - 𝑟 2$ find the value of $𝑝$
A. $3$
B. $4$
C. $5$
D. $6$

Answer: C. $5$
Explanation: By Pythagoras theorem for the right-angled triangle $𝑄 𝑃 𝑅$ , $𝑝 2 = 𝑞 2 + 𝑟 2$ . Substituting the given relation $𝑞 2 = 25 - 𝑟 2$ : $𝑝 2 = (25 - 𝑟 2) + 𝑟 2 = 25$ Thus, $𝑝 = \sqrt 25 = 5$ .

7. In the diagram $――― 𝑇 𝑋$ and $――― 𝑇 𝑌$ are tangents to the circle $𝑋 𝑌 𝑍$ at $𝑋$ and $𝑌$ respectively. $――― 𝑇 𝑋 | | ――― 𝑌 𝑍$ and $∠ 𝑋 𝑇 𝑌 = 100 \circ$ Find $∠ 𝑍 𝑋 𝑌$
A. $130 \circ$
B. $100 \circ$
C. $80 \circ$
D. $90 \circ$

Answer: B. $100 \circ$
Explanation: Since $𝑇 𝑋 = 𝑇 𝑌$ (tangents from a point), $Δ 𝑇 𝑋 𝑌$ is isosceles. $∠ 𝑇 𝑋 𝑌 = 180 - 100 2 = 40 \circ$ . Because $𝑇 𝑋 | | 𝑌 𝑍$ , the alternate angles are equal, so $∠ 𝑋 𝑌 𝑍 = ∠ 𝑇 𝑋 𝑌 = 40 \circ$ . By the alternate segment theorem, $∠ 𝑇 𝑌 𝑋 = ∠ 𝑌 𝑋 𝑍 = 40 \circ$ . In $Δ 𝑋 𝑌 𝑍$ , $∠ 𝑍 𝑋 𝑌 = 180 - (40 + 40) = 100 \circ$ .

8. In the diagram, $𝑂$ is the centre of the circle and $∠ 𝑂 𝑋 𝑍 = 19 \circ$ Find $∠ 𝑋 𝑌 𝑍$
A. $142 \circ$
B. $68 \circ$
C. $109 \circ$
D. $71 \circ$

Answer: C. $109 \circ$
Explanation: In $Δ 𝑂 𝑋 𝑍$ , $𝑂 𝑋 = 𝑂 𝑍$ (radii), so $∠ 𝑂 𝑍 𝑋 = ∠ 𝑂 𝑋 𝑍 = 19 \circ$ . The angle at the center $∠ 𝑋 𝑂 𝑍 = 180 \circ - (19 \circ + 19 \circ) = 142 \circ$ . The reflex angle $∠ 𝑋 𝑂 𝑍 = 360 \circ - 142 \circ = 218 \circ$ . The angle $∠ 𝑋 𝑌 𝑍$ is half this reflex angle: $∠ 𝑋 𝑌 𝑍 = 12 \times 218 \circ = 109 \circ$

9. The diagram shows a field $𝑀 𝑁 𝑃 𝑄 𝑅 𝑆 𝑇$ where $𝑀 𝑇 𝑆$ is a semicircle. Find the area of the field. [Take $𝜋 = 227$ ]
A. $25.7 c m 2$
B. $26.3 c m 2$
C. $27.5 c m 2$
D. $28.6 c m 2$

Answer: A. $25.7 c m 2$
Explanation: The area is the area of rectangle $𝑃 𝑁 𝑅 𝑄$ minus the area of semicircle $𝑀 𝑇 𝑆$ . Rectangle area $= 8 \times 4 = 32$ . The diameter of the semicircle is $8 - (2 + 2) = 4$ , so $𝑟 = 2$ . $A r e a = 32 - (12 \times 227 \times 22) = 32 - 6.286 \approx 25.7 c m 2$

10. Mr. Gyan bought a car for $16, 800.00$ and later sold it at $85 %$ of the original cost. He spent $3, 930.75$ from the money he received from the sale of the car and invested the remaining at $18 %$ per annum simple interest. How much interest did he earn in $9$ years?
A. $16, 765.79$
B. $20, 848.19$
C. $32, 018.32$
D. $67, 657.00$

Answer: A. $16, 765.79$
Explanation: First, find the selling price of the car: $0.85 \times 16, 800 = 14, 280$ After spending $3, 930.75$ , the remaining amount invested is $14, 280 - 3, 930.75 = 10, 349.25$ Using the simple interest formula $𝐼 = 𝑃 𝑅 𝑇$ : $𝐼 = 10, 349.25 \times 0.18 \times 9 = 16, 765.785$ Rounded to two decimal places, the interest is $16, 765.79$ .

11. A town $𝑃$ is due south of town $𝑄$ and town $𝑅$ is on a bearing of $125 \circ$ from $𝑄$ If town $𝑅$ is $20 k m$ due east of $𝑃$ find $| 𝑃 𝑄 |$ .
A. $15 k m$
B. $14 k m$
C. $12 k m$
D. $20 k m$

Answer: B. $14 k m$
Explanation: The positions form a right-angled triangle $𝑃 𝑄 𝑅$ with the right angle at $𝑃$ (since $𝑃$ is south of $𝑄$ and $𝑅$ is east of $𝑃$ ). The bearing of $125 \circ$ from $𝑄$ implies the interior angle $∠ 𝑃 𝑄 𝑅 = 180 \circ - 125 \circ = 55 \circ$ . Using trigonometry: $t a n (55 \circ) = 𝑃 𝑅 𝑃 𝑄 ⟹ 𝑃 𝑄 = 20 t a n (55 \circ) \approx 14 k m$

12. If $𝑉 = 𝜋 𝑟 2 ℎ$ and $𝑆 = 2 𝜋 𝑟 ℎ$ express $𝑉$ in terms of $𝑟$ and $𝑆$ .
A. $𝑉 = 𝑟 2 𝑆$
B. $𝑉 = 2 𝑟 𝑆$
C. $𝑉 = 12 𝑟 𝑆$
D. $𝑉 = 2 𝑟 𝑆$

Answer: C. $𝑉 = 12 𝑟 𝑆$
Explanation: The volume of a cylinder is $𝑉 = 𝜋 𝑟 2 ℎ$ , which can be rewritten as $𝑉 = 12 𝑟 (2 𝜋 𝑟 ℎ)$ . Substituting the expression for surface area $𝑆 = 2 𝜋 𝑟 ℎ$ into this formula gives: $𝑉 = 12 𝑟 𝑆$

13. The mean and range of $1.20, 1.00, 0.90, 1.40, 0.80, 0.80, 1.20$ and $1.10$ are $𝑚$ and $𝑟$ respectively. Find the value of $(𝑚 + 𝑟)$
A. $1.65$
B. $2.45$
C. $2.55$
D. $1.85$

Answer: A. $1.65$
Explanation: The mean $𝑚$ is the sum of the values divided by $8$ : $𝑚 = 8.40 8 = 1.05$ The range $𝑟$ is the difference between the highest and lowest values: $𝑟 = 1.40 - 0.80 = 0.60$ Therefore, $𝑚 + 𝑟 = 1.05 + 0.60 = 1.65$

14. Solve: $l o g 3 (2 𝑥 - 1) l o g 3 243 = 25$ .
A. $𝑥 = 5$
B. $𝑥 = 6$
C. $𝑥 = 3$
D. $𝑥 = 4$

Answer: A. $𝑥 = 5$
Explanation: Note that $l o g 3 243 = l o g 3 35 = 5$ . The equation becomes $l o g 3 (2 𝑥 - 1) 5 = 25$ , which simplifies to $l o g 3 (2 𝑥 - 1) = 2$ . Converting to exponential form: $2 𝑥 - 1 = 32 ⟹ 2 𝑥 - 1 = 9 ⟹ 2 𝑥 = 10 ⟹ 𝑥 = 5$

15. The first $4$ terms of an Arithmetic Progression (A.P) are $8, 𝑥, 𝑦$ and $17$ Find the value of $(𝑥 + 𝑦)$ .
A. $25$
B. $20$
C. $40$
D. $31$

Answer: A. $25$
Explanation: In an Arithmetic Progression, the sum of terms equidistant from the beginning and the end is constant. Therefore, for the sequence $8, 𝑥, 𝑦, 17$ , the sum $𝑥 + 𝑦$ is equal to the sum of the first and fourth terms: $𝑥 + 𝑦 = 8 + 17 = 25$

16. Given that $l o g 10 5 = 𝑥$ and $l o g 10 11 = 𝑦$ express $l o g 10 275$ in terms of $𝑥$ and $𝑦$ .
A. $2 (𝑥 + 𝑦)$
B. $2 𝑥 𝑦$
C. $𝑥 + 2 𝑦$
D. $2 𝑥 + 𝑦$

Answer: D. $2 𝑥 + 𝑦$
Explanation: Express $275$ as a product of its factors: $275 = 25 \times 11 = 52 \times 11$ . Using the laws of logarithms: $l o g 10 275 = l o g 10 (52 \times 11) = 2 l o g 10 5 + l o g 10 11 = 2 𝑥 + 𝑦$

17. Find the coefficient of $𝑥 2$ in the expansion of $(4 𝑥 + 5) (3 - 2 𝑥)$ .
A. $15$
B. $8$
C. $- 8$
D. $2$

Answer: C. $- 8$
Explanation: Expand the expression using the distributive property: $(4 𝑥 + 5) (3 - 2 𝑥) = 12 𝑥 - 8 𝑥 2 + 15 - 10 𝑥$ Simplify the terms: $- 8 𝑥 2 + 2 𝑥 + 15$ . The coefficient of $𝑥 2$ is the numerical factor multiplying the $𝑥 2$ term, which is $- 8$ .

18. A train is moving at $60$ miles per hour. If it passes a sign post in $15$ seconds, find, in yards, the length of the train. [Take $1 m i l e = 1760 y a r d s$ ]
A. $360 y a r d s$
B. $440 y a r d s$
C. $420 y a r d s$
D. $400 y a r d s$

Answer: B. $440 y a r d s$
Explanation: First, convert the speed to yards per second. Since $1 m i l e = 1760 y a r d s$ , $60 m p h = 60 \times 1760 3600 y a r d s / s e c = 883 y a r d s / s e c$ . The length of the train is the distance covered in $15$ seconds: $L e n g t h = 883 \times 15 = 440 y a r d s$

19. A school has a total population of $1, 800$ students. Out of this number $49$ are in SHS 1, $1336$ are in SHS 2 and the rest are in SHS 3. Calculate the number of students in SHS 3.
A. $350$
B. $650$
C. $800$
D. $500$

Answer: A. $350$
Explanation: Calculate the total fraction for SHS 1 and SHS 2: $49 + 1336 = 16 + 13 36 = 2936$ The fraction of students in SHS 3 is $1 - 2936 = 736$ The number of students is $736 \times 1800 = 350$

20. A car consumes $36$ gallons of fuel over a distance of $2, 268 k m$ Find, in km per litre, the fuel consumption. [Take $1 g a l l o n = 4.5 l i t r e s$ ]
A. $50$
B. $14$
C. $15$
D. $63$

Answer: B. $14$
Explanation: First, convert the fuel from gallons to litres: $36 g a l l o n s \times 4.5 l i t r e s / g a l l o n = 162 l i t r e s$ . Fuel consumption in km per litre is the distance divided by the total litres: $2268 \div 162 = 14 k m / l$