πŸ“š WASSCE 2026 | CORE MATHEMATICS Paper 2 β€” 10 Practice Questions that may appear Section A questions = 8 marks | Section B questions = 12 marks By Sir Francis Akatsi (Accra) βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 1 (Section A β€” 8 marks) Sets + Simple Interest The universal set Β΅ = x : 1 ≀ x ≀ 20, x is a whole number. P = prime numbers in Β΅ Q = multiples of 3 in Β΅ List the elements of P and Q. (ii) Find P βˆͺ Q and P ∩ Q. (iii) Find the value of n(P βˆͺ Q)ΚΉ. (iv) Represent P and Q on a clearly labelled Venn diagram. [4 marks] Kwame invested GHS 4,500 in a savings account at a simple interest rate of 8% per annum. After a certain number of years, the total amount in the account was GHS 6,300. Calculate the simple interest earned. (ii) Find the number of years Kwame kept the money in the account. (iii) What annual rate would have yielded the same interest in 3 years? [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 2 (Section A β€” 8 marks) Number Bases + Geometric Progression Convert 10110111β‚‚ to base 10. (ii) Convert 187₁₀ to binary (base 2). Show all working. (iii) Evaluate 11011β‚‚ + 10110β‚‚, giving your answer in binary. (iv) Convert 11010β‚‚ to base 8. [4 marks] The first three terms of a geometric progression are (x + 2), (2x βˆ’ 1) and (4x βˆ’ 3). Show that (2x βˆ’ 1)Β² = (x + 2)(4x βˆ’ 3) and hence find the value of x. (ii) Write down the first three terms and state the common ratio. (iii) Find the 8th term of the progression. (iv) Find the sum of the first 10 terms. [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 3 (Section A β€” 8 marks) Ratio & Sharing + Profit, Loss and Discount Ama, Kofi and Yaw share a business profit of GHS 84,000 in the ratio 3:5:4. Calculate each person's share. (ii) Kofi uses 35% of his share to buy a fridge. How much does he spend? (iii) Ama reinvests ²⁄₃ of her share. How much does she reinvest? (iv) Yaw gives GHS 3,000 from his share to a charity and splits the rest equally with his two children. How much does each child receive? [4 marks] A shopkeeper in Kumasi bought a television set for GHS 2,400. He marked it up at GHS 3,200. During a promotion, he offered a 15% discount on the marked price. Calculate the discount amount. (ii) Find the actual selling price after the discount. (iii) Calculate the shopkeeper's actual profit or loss and express it as a percentage of the cost price, correct to 1 decimal place. (iv) A 15% VAT is charged on the discounted selling price. Find the final price paid by the customer. [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 4 (Section A β€” 8 marks) Simultaneous Equations + Age Word Problem Solve the simultaneous equations: 4x βˆ’ 3y = 18 2x + 5y = 8 [4 marks] Abena is 3 times as old as her daughter Efua. In 8 years' time, Abena will be twice as old as Efua. Write down two equations to represent this information, letting Abena's current age = a and Efua's current age = e. (ii) Solve the equations to find the current ages of both Abena and Efua. (iii) Find the sum of their ages 5 years ago. (iv) In how many years will the sum of their ages be 80? [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– https://whatsapp.com/channel/0029VabqqtT7T8bSpkb3lg3c QUESTION 5 (Section A β€” 8 marks) Mensuration β€” Sector + Pythagoras (Take Ο€ = ²²⁄₇) A sector of a circle has radius 14 cm and angle 135Β° at the centre. Calculate the arc length of the sector. (ii) Calculate the area of the sector. (iii) Find the perimeter of the sector. [4 marks] A telecommunications mast stands vertically on level ground. A support wire is attached from the top of the mast to a point P on the ground. The wire is 26 m long and the point P is 10 m from the base of the mast. Draw a clearly labelled diagram of this situation. (ii) Calculate the height of the mast. (iii) A second support wire is attached 8 m below the top of the mast to a point Q on the ground, 15 m from the base. Calculate the length of the second wire. (iv) Calculate the angle the second wire makes with the ground, correct to the nearest degree. (Use tan, sin or cos) [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– For dito dito chat on whatsapp: 0202440468 QUESTION 6 (Section B β€” 12 marks) Vectors + Transformation Given that a = (³₋₁), b = (⁻²₄) and c = (¹₋₃): Find 2a + 3b βˆ’ c. (ii) Find |2a + 3b βˆ’ c|, correct to 2 decimal places. (iii) Find the values of p and q such that pa + qb = (⁷₋₁₁). (iv) Find the unit vector in the direction of a + b. [6 marks] Triangle PQR has vertices P(1, 2), Q(4, 2) and R(4, 5). Find the image of triangle PQR after a translation by vector (⁻³₂). Label the image P'Q'R'. (ii) Find the image of triangle PQR after a reflection in the line y = x. Label this image P''Q''R''. (iii) Find the image of triangle PQR after an enlargement with centre (0, 0) and scale factor 2. Label this P'''Q'''R'''. (iv) State which two of the three transformations produce congruent images and give a reason. [6 marks] https://whatsapp.com/channel/0029VabqqtT7T8bSpkb3lg3c βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 7 (Section B β€” 12 marks) Angles in Polygons + Sine Rule The interior angles of a hexagon are in the ratio 2:3:3:4:4:5. Find the sum of interior angles of a hexagon. (ii) Find the value of each interior angle. (iii) State whether the hexagon is regular. Give a reason. (iv) Find the size of each exterior angle of the hexagon in (ii). [6 marks] In triangle KLM, angle K = 48Β°, angle L = 72Β° and side KL = 9.4 cm. Find angle M. (ii) Using the sine rule, find the length of side KM, correct to 1 decimal place. (iii) Find the length of side LM, correct to 1 decimal place. (iv) Calculate the area of triangle KLM, correct to 2 decimal places. [sin 48Β° = 0.7431, sin 72Β° = 0.9511, sin 60Β° = 0.8660] [6 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 8 (Section B β€” 12 marks) Grouped Frequency Table + Statistics The table shows the ages (in years) of 60 workers at a Tema factory: Age (years) | Frequency 20 – 24 | 5 25 – 29 | 9 30 – 34 | 16 35 – 39 | 18 40 – 44 | 8 45 – 49 | 4 (i) State the modal class. (ii) Calculate the mean age of the workers. [4 marks] (i) Copy and complete the cumulative frequency table for this data. (ii) On graph paper, using a scale of 2 cm to 5 years on the horizontal axis and 2 cm to 10 workers on the vertical axis, draw a cumulative frequency curve (ogive) for the data. [Remember: plot against the UPPER CLASS BOUNDARY] [4 marks] Use your ogive to estimate: the median age of the workers. (ii) the percentage of workers aged less than 33 years. (iii) the interquartile range. (iv) the number of workers who are 42 years or older. [4 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– WhatsApp for dito papers 0202440468 QUESTION 9 (Section B β€” 12 marks) Quadratic Equations + Variation (i) Solve the equation 6xΒ² + x βˆ’ 12 = 0 by factorisation. (ii) Solve the equation 3xΒ² βˆ’ 7x + 2 = 0 using the quadratic formula, giving your answers correct to 2 decimal places. (iii) The length of a rectangle is (2x + 1) cm and its width is (x βˆ’ 2) cm. If the area of the rectangle is 40 cmΒ², find the dimensions of the rectangle. [6 marks] y varies inversely as the square of x, and directly as z. When x = 2, z = 3 and y = 18. Write an expression for y in terms of x and z. (ii) Find the value of y when x = 3 and z = 4. (iii) Find the value of x when y = 8 and z = 2. (iv) What happens to y when x is doubled and z remains constant? Justify your answer mathematically. [6 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– QUESTION 10 (Section B β€” 12 marks) Indices & Logarithms + Surds (i) Simplify: 3Β²Λ£ Γ— 9Λ£ Γ· 27ˣ⁻¹, giving your answer as a power of 3. (ii) Solve for x: 4²ˣ⁺¹ = 8ˣ⁺³ (iii) Solve for x: logβ‚‚(3x βˆ’ 2) = 4 (iv) Evaluate: log₁₀ 500 + log₁₀ 2 βˆ’ log₁₀ 10, without using a calculator. [6 marks] (i) Simplify √75 + 3√12 βˆ’ 2√27, leaving your answer in surd form. (ii) Rationalise the denominator and simplify: 6/(√5 βˆ’ √2) (iii) Given that √3 = 1.732, evaluate (4 + √3)(4 βˆ’ √3) without a calculator. (iv) Expand and simplify (2√5 + √3)Β², leaving your answer in surd form. [6 marks] βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž–βž– Follow the WASSCE | BECE QUESTIONS AND MARKING UPDATES/ TRACING | CTVET channel on WhatsApp: https://whatsapp.com/channel/0029VabqqtT7T8bSpkb3lg3c

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Answers

4 parts
a(iii)

7\text{7}

b(i)

1,800\text{1,800}

b(ii)

5Β years\text{5 years}

b(iii)

13.3%Β (toΒ 1Β d.p.)13.3\% \text{ (to 1 d.p.)}

Step-by-step solution

7 steps

Here's the solution to Question 1:

Part 1

Part a(i)Β· 1 step

List the elements of P and Q.

Sets

The universal set ΞΌ={x:1≀x≀20,xΒ isΒ aΒ wholeΒ number}\mu = \{x : 1 \le x \le 20, x \text{ is a whole number}\}.
So, ΞΌ={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20}\mu = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}.

  • P = {prime numbers in ΞΌ\mu}
  • Q = {multiples of 3 in ΞΌ\mu}

  1. List the elements for P and Q.

    Prime numbers in ΞΌ\mu are numbers greater than 1 that are only divisible by 1 and themselves.
    Multiples of 3 in ΞΌ\mu are numbers in ΞΌ\mu that can be divided by 3.

    P={2,3,5,7,11,13,17,19}P = \{2, 3, 5, 7, 11, 13, 17, 19\} Q={3,6,9,12,15,18}Q = \{3, 6, 9, 12, 15, 18\}

Part a(ii)Β· 1 step

Find P βˆͺ\cup Q and P ∩\cap Q.

  1. Find the union and intersection of P and Q.

    The union (P βˆͺ\cup Q) includes all unique elements from both sets.
    The intersection (P ∩\cap Q) includes elements common to both sets.

    PβˆͺQ={2,3,5,6,7,9,11,12,13,15,17,18,19}P \cup Q = \{2, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19\} P∩Q={3}P \cap Q = \{3\}

Part a(iii)

7\text{7}

Find the value of n(P βˆͺ\cup Q)ΚΉ.

  1. Find the complement of P βˆͺ\cup Q and its number of elements.

    First, count the number of elements in P βˆͺ\cup Q: n(PβˆͺQ)=13n(P \cup Q) = 13.
    The universal set has n(ΞΌ)=20n(\mu) = 20 elements.
    The complement (PβˆͺQ)ΚΉ(P \cup Q)ΚΉ consists of elements in ΞΌ\mu but not in PβˆͺQP \cup Q.

    (PβˆͺQ)ΚΉ=ΞΌβˆ’(PβˆͺQ)={1,4,8,10,14,16,20}(P \cup Q)ΚΉ = \mu - (P \cup Q) = \{1, 4, 8, 10, 14, 16, 20\} n(PβˆͺQ)ΚΉ=7n(P \cup Q)ΚΉ = 7

    The value of n(PβˆͺQ)ΚΉn(P \cup Q)ΚΉ is 7\boxed{\text{7}}.

Part a(iv)Β· 1 step

Represent P and Q on a clearly labelled Venn diagram.

  1. Draw the Venn diagram.

    • Draw a rectangle for the universal set ΞΌ\mu.
    • Draw two overlapping circles for P and Q.
    • Place the common element (3) in the intersection.
    • Place the remaining elements of P in the P-only region.
    • Place the remaining elements of Q in the Q-only region.
    • Place elements from ΞΌ\mu that are not in P or Q outside the circles but inside the rectangle.
            Β΅
+-------------------------------------+
|   1, 4, 8, 10, 14, 16, 20           |
|                                     |
|       +-------+   +-------+         |
|       |   P   |   |   Q   |         |
|       | 2, 5, |   | 6, 9, |         |
|       | 7, 11,|   | 12,15,|         |
|       | 13,17,|   | 18    |         |
|       | 19    |   |       |         |
|       +---3---+---+-------+         |
|                                     |
+-------------------------------------+

Part b(i)

1,800\text{1,800}

Calculate the simple interest earned.

Simple Interest

Kwame invested GHS 4,500 at 8% per annum. Total amount was GHS 6,300.
Principal (P) = GHS 4,500
Amount (A) = GHS 6,300
Rate (R) = 8% = 0.08

  1. Calculate the simple interest.

    Simple Interest (I) = Amount (A) - Principal (P)

    I=GHSΒ 6,300βˆ’GHSΒ 4,500I = \text{GHS } 6,300 - \text{GHS } 4,500 I=GHSΒ 1,800I = \text{GHS } \boxed{\text{1,800}}

Part b(ii)

5Β years\text{5 years}

Find the number of years Kwame kept the money in the account.

  1. Use the simple interest formula to find the time (T).

    The formula for simple interest is I=PΓ—RΓ—T100I = \frac{P \times R \times T}{100} (where R is a percentage).

    1800=4500Γ—8Γ—T1001800 = \frac{4500 \times 8 \times T}{100} 1800=45Γ—8Γ—T1800 = 45 \times 8 \times T 1800=360Γ—T1800 = 360 \times T T=1800360T = \frac{1800}{360} T=5Β yearsT = \boxed{\text{5 years}}

Part b(iii)

13.3%Β (toΒ 1Β d.p.)13.3\% \text{ (to 1 d.p.)}

What annual rate would have yielded the same interest in 3 years?

  1. Use the simple interest formula to find the new rate (R).

    We want the same interest (I = GHS 1,800) with Principal (P = GHS 4,500) and Time (T = 3 years).

    I=PΓ—RΓ—T100I = \frac{P \times R \times T}{100} 1800=4500Γ—RΓ—31001800 = \frac{4500 \times R \times 3}{100} 1800=45Γ—RΓ—31800 = 45 \times R \times 3 1800=135Γ—R1800 = 135 \times R R=1800135R = \frac{1800}{135} R=1209R = \frac{120}{9} R=403R = \frac{40}{3} R=13.3%Β (toΒ 1Β d.p.)R = \boxed{13.3\% \text{ (to 1 d.p.)}}
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