WAEC GCE Mathematics Past Questions and Answers PDF Download – 2024 Updated Version

---

Are you about to sit for the WAEC GCE Mathematics exam and searching for relevant materials such as past questions and answers to help increase your chances of passing? This is your navigation to the link to download WAEC GCE Mathematics past questions and answers in PDF format. In this regard, these questions are important in that they prepare the candidates on the contraption. Aids in revising questions that are commonly asked and provides one with assurance before the actual examination.

In this article, we have included a link for free and available for download, WAEC GCE Mathematics past questions and answers that you can use in your revision. So, Let us begin!

Why You Need WAEC GCE Mathematics Past Questions

Success in WAEC GCE Mathematics examination does not come on a silver platter; it involves rigorous preparation and the best preparation is through the use of past papers. Here is why:

Familiarization with the Examination: Past questions orient you and leave little doubts on the structure of the WAEC GCE Mathematics examination that one knows what to expect.

Questions that Come Again and Again: Most of the questions in WAEC GCE Mathematics are such that they come back or are of the same kind. Practicing these prepares you for the eventuality that there is nothing new.

Enhance the Skills of Solving Mathematical Problems: Solving past papers enables one to hone their skills of solving mathematical questions and also aids in the management of time during the examination.

Increase Assuredness: Practicing past papers helps one to have a large and deep insight of the pattern of the examination which in turn helps one to be more ready for the examination.

WAEC GCE Mathematics past questions and answers

Question 1:

Question 1: Problem:
The third and seventh terms of an Arithmetic Progression (A.P.) are \(3 + 4\sqrt{3}\) and \(3 + 12\sqrt{3}\) respectively. Find the fifth term of the A.P.

Step-by-Step Solution:

1. Understand the Arithmetic Progression (A.P.) formula: In an A.P., the \(n\)-th term is given by: \[ T_n = a + (n – 1)d \] Where: – \(T_n\) is the \(n\)-th term, – \(a\) is the first term, – \(d\) is the common difference, – \(n\) is the position of the term.
2. Use the given information to set up equations: We know the third term and the seventh term of the A.P. – Third term: \(T_3 = 3 + 4\sqrt{3}\) \[ T_3 = a + (3 – 1)d = a + 2d \] So, \[ a + 2d = 3 + 4\sqrt{3} \quad \text{(Equation 1)} \] – Seventh term: \(T_7 = 3 + 12\sqrt{3}\) \[ T_7 = a + (7 – 1)d = a + 6d \] So, \[ a + 6d = 3 + 12\sqrt{3} \quad \text{(Equation 2)} \] 3. Solve the system of equations to find \(a\) and \(d\): Subtract Equation 1 from Equation 2: \[ (a + 6d) – (a + 2d) = (3 + 12\sqrt{3}) – (3 + 4\sqrt{3}) \] Simplifying: \[ a + 6d – a – 2d = 3 + 12\sqrt{3} – 3 – 4\sqrt{3} \] \[ 4d = 8\sqrt{3} \] \[ d = 2\sqrt{3} \] 4. Substitute \(d = 2\sqrt{3}\) into one of the equations to find \(a\): Substitute into Equation 1: \[ a + 2(2\sqrt{3}) = 3 + 4\sqrt{3} \] \[ a + 4\sqrt{3} = 3 + 4\sqrt{3} \] Now subtract \(4\sqrt{3}\) from both sides: \[ a = 3 \] So, the first term \(a = 3\) and the common difference \(d = 2\sqrt{3}\). 5. Find the fifth term of the A.P.: Now, using the formula \(T_n = a + (n – 1)d\), let’s find the fifth term \(T_5\): \[ T_5 = a + (5 – 1)d = a + 4d \] Substitute \(a = 3\) and \(d = 2\sqrt{3}\): \[ T_5 = 3 + 4(2\sqrt{3}) = 3 + 8\sqrt{3} \] Final Answer for Question 1: The fifth term of the A.P. is \(3 + 8\sqrt{3}\).

Question 2:

Question 2: Problem:
A woman sells \(\frac{7}{10}\) of her stock of cutlery at a profit of 30%. The remaining stock is sold at a loss of 8%. Find the percentage profit on the sale of the whole stock.

Step-by-Step Solution:**

1. Let’s assume the total value of the stock is \(100\) units (for simplicity):
– The woman sells \(\frac{7}{10}\) of the stock, which means she sells: \[ \frac{7}{10} \times 100 = 70 \text{ units} \] – The remaining stock is: \[ 100 – 70 = 30 \text{ units} \] 2. Calculate the profit from the \(70\) units sold: The profit is 30%, so the selling price of the \(70\) units will be: \[ 70 + 30\% \text{ of } 70 = 70 + \left( \frac{30}{100} \times 70 \right) \] \[ = 70 + 21 = 91 \text{ units} \] 3. Calculate the loss from the \(30\) units sold: The loss is 8%, so the selling price of the \(30\) units will be: \[ 30 – 8\% \text{ of } 30 = 30 – \left( \frac{8}{100} \times 30 \right) \] \[ = 30 – 2.4 = 27.6 \text{ units} \] 4. Calculate the total amount made from selling the whole stock: Total selling price = Selling price of \(70\) units + Selling price of \(30\) units: \[ 91 + 27.6 = 118.6 \text{ units} \] 5. Calculate the percentage profit: The total cost of the stock was 100 units. The total selling price is 118.6 units. So the profit made is: \[ \text{Profit} = 118.6 – 100 = 18.6 \text{ units} \] Now, calculate the percentage profit: \[ \text{Percentage profit} = \left( \frac{18.6}{100} \right) \times 100 = 18.6\% \] Final Answer for Question 2: The percentage profit on the sale of the whole stock is \(18.6\%\).

Question 3: Finding the Radius of a Cylindrical Tank Given: – Height \(h = 16\) m – Total cost to fill the tank = ₦30,800,000 – Cost per liter = ₦20 Step 1: Calculate the volume of the tank in liters. \[ \text{Volume in liters} = \frac{\text{Total Cost}}{\text{Cost per liter}} \] \[ \text{Volume in liters} = \frac{30,800,000}{20} = 1,540,000 \text{ liters} \] Since 1 cubic meter = 1000 liters: \[ \text{Volume in cubic meters} = \frac{1,540,000}{1000} = 1540 \, \text{m}^3 \] Step 2: Use the formula for the volume of a cylinder. \[ V = \pi r^2 h \] Substitute \(V = 1540\) and \(h = 16\): \[ 1540 = \pi r^2 (16) \] \[ r^2 = \frac{1540}{16\pi} \] \[ r^2 = \frac{1540}{50.2655} \quad (\text{using } \pi \approx 3.1416) \] \[ r^2 \approx 30.63 \] \[ r \approx \sqrt{30.63} \approx 5.5 \, \text{m} \] Answer: The radius of the cylinder is approximately 5.5 meters.

Question 4a: Find the Variance Given scores: \[ (7n – 2), \, 7n, \, (7n + 2) \] Step 1: Find the mean. The formula for the mean (\(\mu\)) is: \[ \mu = \frac{\text{Sum of all values}}{\text{Number of values}} \] Sum of scores: \[ (7n – 2) + 7n + (7n + 2) = 7n – 2 + 7n + 7n + 2 = 21n \] Mean: \[ \mu = \frac{21n}{3} = 7n \] Step 2: Calculate the variance. The formula for variance (\(\sigma^2\)) is: \[ \sigma^2 = \frac{1}{N} \sum (x_i – \mu)^2 \] Here, \(N = 3\), and we subtract the mean from each value: – For \(x_1 = 7n – 2\): \[ (x_1 – \mu)^2 = ((7n – 2) – 7n)^2 = (-2)^2 = 4 \] – For \(x_2 = 7n\): \[ (x_2 – \mu)^2 = (7n – 7n)^2 = 0^2 = 0 \] – For \(x_3 = 7n + 2\): \[ (x_3 – \mu)^2 = ((7n + 2) – 7n)^2 = (2)^2 = 4 \] Sum of squared deviations: \[ 4 + 0 + 4 = 8 \] Variance: \[ \sigma^2 = \frac{8}{3} \] Answer: The variance is \(\frac{8}{3}\).

Question 4b: Solve for \(x\) in \(\log_{10}(3x – 7) – \log_{10}(5) = 1\) Step 1: Apply the logarithm subtraction rule. \[ \log_{10} \left(\frac{3x – 7}{5}\right) = 1 \] Step 2: Rewrite in exponential form. \[ \frac{3x – 7}{5} = 10^1 \] \[ \frac{3x – 7}{5} = 10 \] Step 3: Solve for \(x\). Multiply both sides by 5: \[ 3x – 7 = 50 \] Add 7 to both sides: \[ 3x = 57 \] Divide by 3: \[ x = 19 \] Answer: \(x = 19\).

Question 5: Bearings and Distance Given: – Distance from \(M\) to \(N\): \(6 \, \text{km}\), bearing: \(065^\circ\). – Distance from \(N\) to \(P\): \(13 \, \text{km}\), bearing: \(146^\circ\). Task: 1. (a) Draw a diagram. The diagram will show two paths: – \(M\) to \(N\) at \(065^\circ\). – \(N\) to \(P\) at \(146^\circ\). 2. (b) Calculate the distance from \(M\) to \(P\) and the bearing of \(N\) from \(P\). — Step 1: Resolve into vectors (Law of Cosines). Let \(MN = 6 \, \text{km}\), \(NP = 13 \, \text{km}\), and \(MP = d\). Using the angle between the two paths: \[ \text{Angle between bearings} = 146^\circ – 65^\circ = 81^\circ \] Apply the Law of Cosines: \[ d^2 = MN^2 + NP^2 – 2(MN)(NP)\cos(81^\circ) \] Substitute values: \[ d^2 = 6^2 + 13^2 – 2(6)(13)\cos(81^\circ) \] \[ d^2 = 36 + 169 – 156 \times \cos(81^\circ) \] Using \(\cos(81^\circ) \approx 0.1564\): \[ d^2 = 36 + 169 – 156(0.1564) \] \[ d^2 = 205 – 24.3984 \] \[ d^2 \approx 180.6 \] \[ d \approx \sqrt{180.6} \approx 13.4 \, \text{km} \] Step 2: Find the bearing of \(N\) from \(P\) (Using Law of Sines). Using trigonometry, solve for the angle at \(P\), but this step needs an illustration to show clearly how it proceeds. Answers: 1. (i) Distance from \(M\) to \(P\): 13.4 km 2. (ii) Bearing of \(N\) from \(P\): Calculation involves further geometric interpretation.

Question 6a: Modulo 9 Multiplication Table We are completing the multiplication table under modulo \(9\). Here’s how: For any two numbers \(a \otimes b\): \[ (a \times b) \mod 9 \] Complete the table: | ⊗ | 2 | 3 | 5 | 7 | 8 | |—-|—-|—-|—-|—-|—-| | 2 | 4 | 6 | 10 mod 9 = 1 | 14 mod 9 = 5 | 16 mod 9 = 7 | | 3 | 6 | 9 mod 9 = 0 | 15 mod 9 = 6 | 21 mod 9 = 3 | 24 mod 9 = 6 | | 5 | 10 mod 9 = 1 | 15 mod 9 = 6 | 25 mod 9 = 7 | 35 mod 9 = 8 | 40 mod 9 = 4 | | 7 | 14 mod 9 = 5 | 21 mod 9 = 3 | 35 mod 9 = 8 | 49 mod 9 = 4 | 56 mod 9 = 2 | | 8 | 16 mod 9 = 7 | 24 mod 9 = 6 | 40 mod 9 = 4 | 56 mod 9 = 2 | 64 mod 9 = 1 | — (ii) Evaluate \((3 \otimes 7) \otimes (8 \otimes 7)\). From the completed table: – \(3 \otimes 7 = 3\) – \(8 \otimes 7 = 2\) Now: \[ (3 \otimes 7) \otimes (8 \otimes 7) = 3 \otimes 2 \] From the table: \[ 3 \otimes 2 = 6 \] Answer: \((3 \otimes 7) \otimes (8 \otimes 7) = 6\). (iii) Find the truth sets: – (\(\alpha\)) \(3 \otimes m = 6\) From the table, check where \(3 \otimes m = 6\): \[ m = 3, \, 8 \] So, the truth set is: \[ \{3, 8\} \] – (\(\beta\)) \(p \otimes p = 4\) From the table, check where \(p \otimes p = 4\): \[ p = 5 \] So, the truth set is: \[ \{5\} \]

Question 6b: Sum of Interior Angles of a Polygon Given: \[ \text{Sum of interior angles} = 16 \, \text{right angles} \] Since \(1 \text{ right angle} = 90^\circ\): \[ \text{Sum of interior angles} = 16 \times 90^\circ = 1440^\circ \] The sum of interior angles of a polygon is given by: \[ \text{Sum of angles} = 180^\circ (n – 2) \] Where \(n\) is the number of sides. Solve for \(n\): \[ 180(n – 2) = 1440 \] Divide both sides by 180: \[ n – 2 = 8 \] Add 2 to both sides: \[ n = 10 \] Answer:** The polygon has 10 sides.

Question 8a: Equation of the Line Given: – \(x\)-intercept = 5 – \(y\)-intercept = -7 The equation of a line in intercept form is: \[ \frac{x}{a} + \frac{y}{b} = 1 \] Where: – \(a = x\)-intercept – \(b = y\)-intercept Substitute \(a = 5\) and \(b = -7\): \[ \frac{x}{5} + \frac{y}{-7} = 1 \] Multiply through by \(35\) (LCM of 5 and 7): \[ 7x – 5y = 35 \] Answer: The equation of the line is \(7x – 5y = 35\).

Question 8b: Angle of Depression Problem Given: – Angle of depression to the **shrine** = \(60^\circ\) – Angle of depression to the **church** = \(75^\circ\) – Distance from the pole’s foot to the **church** = \(42 \, \text{m}\) Let: – Height of the pole = \(h\) – Distance from the pole’s foot to the shrine = \(d\) (i) Height of the Pole From the church’s position: \[ \tan(75^\circ) = \frac{h}{42} \] Solve for \(h\): \[ h = 42 \times \tan(75^\circ) \] Using \(\tan(75^\circ) \approx 3.732\): \[ h = 42 \times 3.732 = 156.74 \, \text{m} \] Height of the pole is approximately \(156.74 \, \text{m}\). — (ii) Distance from Shrine to Church From the shrine’s position: \[ \tan(60^\circ) = \frac{h}{d} \] Solve for \(d\): \[ d = \frac{h}{\tan(60^\circ)} \] Using \(\tan(60^\circ) \approx 1.732\): \[ d = \frac{156.74}{1.732} \approx 90.48 \, \text{m} \] The distance from the shrine to the church: \[ \text{Distance (Shrine to Church)} = d – 42 = 90.48 – 42 = 48.48 \, \text{m} \] — Answers: – (i) Height of the pole = 156.74 m – (ii) Distance from shrine to church = 48.48 m

Question 9a: Cycling Problem Let the distance between the two villages be \(d\). – On the journey to the next village, Mabel’s speed = \(12 \, \text{km/h}\). – On the return journey, Mabel’s speed = \(16 \, \text{km/h}\). – Total time for both journeys = \(3.5 \, \text{hours}\). Step 1: Write time equations. Time taken to travel to the next village: \[ t_1 = \frac{d}{12} \] Time taken to return: \[ t_2 = \frac{d}{16} \] Total time: \[ t_1 + t_2 = 3.5 \] Substitute the expressions for \(t_1\) and \(t_2\): \[ \frac{d}{12} + \frac{d}{16} = 3.5 \] Step 2: Solve for \(d\). Find the LCM of 12 and 16, which is 48: \[ \frac{4d}{48} + \frac{3d}{48} = 3.5 \] \[ \frac{7d}{48} = 3.5 \] Multiply both sides by 48: \[ 7d = 3.5 \times 48 \] \[ 7d = 168 \] Divide both sides by 7: \[ d = \frac{168}{7} = 24 \, \text{km} \] Answer: The distance between the two villages is 24 km.

Question 9b: Standard Deviation Given ages: \[ 25, 20, 29, 13, 15, 10, 11, 11, 6, 10 \] Mean (\(\mu\)) = 15. Step 1: Calculate the deviations from the mean. For each value \(x_i\), compute \((x_i – \mu)^2\): | \(x_i\) | \(x_i – 15\) | \((x_i – 15)^2\) | |——–|————–|——————| | 25 | 10 | 100 | | 20 | 5 | 25 | | 29 | 14 | 196 | | 13 | -2 | 4 | | 15 | 0 | 0 | | 10 | -5 | 25 | | 11 | -4 | 16 | | 11 | -4 | 16 | | 6 | -9 | 81 | | 10 | -5 | 25 | Sum of \((x_i – \mu)^2\): \[ 100 + 25 + 196 + 4 + 0 + 25 + 16 + 16 + 81 + 25 = 488 \] Step 2: Compute the variance. Variance (\(\sigma^2\)) is: \[ \sigma^2 = \frac{\sum (x_i – \mu)^2}{N} \] \[ \sigma^2 = \frac{488}{10} = 48.8 \] Step 3: Compute the standard deviation. \[ \sigma = \sqrt{48.8} \approx 7 \] Answer:The standard deviation is 7 years.

Question 10: Seamstress Charges Problem Given: – 4 hours: ₦9,750 – 7 hours: ₦15,000 Step 1: Let the cost equation be: \[ C = a + bh \] Where: – \(a\) = fixed price – \(b\) = charge per hour – \(h\) = hours From the given data: 1. \(9,750 = a + 4b\) 2. \(15,000 = a + 7b\) Step 2: Solve the system of equations. Subtract the first equation from the second: \[ (15,000 – 9,750) = (a + 7b) – (a + 4b) \] \[ 5,250 = 3b \] Solve for \(b\): \[ b = \frac{5,250}{3} = 1,750 \, \text{(charge per hour)} \] Substitute \(b = 1,750\) into the first equation: \[ 9,750 = a + 4(1,750) \] \[ 9,750 = a + 7,000 \] Solve for \(a\): \[ a = 9,750 – 7,000 = 2,750 \] Step 3: Solve the tasks. (a) Fixed price \(a = 2,750\). (b) Seamstress charges per hour \(b = 1,750\). (c) Cost of a dress for 11 hours: \[ C = a + 11b \] \[ C = 2,750 + 11(1,750) = 2,750 + 19,250 = 22,000 \] (d) Find the time \(h\) for a dress costing ₦16,750: \[ 16,750 = 2,750 + 1,750h \] Subtract 2,750: \[ 14,000 = 1,750h \] \[ h = \frac{14,000}{1,750} = 8 \, \text{hours} \] — Answers: – (a) Fixed price: ₦2,750 – (b) Charge per hour: ₦1,750 – (c) Cost for 11 hours: ₦22,000 – (d) Time for ₦16,750: 8 hours

Step-by-Step Solutions — Question 11a: Volume of a Cuboid Given: – Dimensions of the container: \(300 \, \text{cm} \times 90 \, \text{cm} \times 84 \, \text{cm}\) – Dimensions of each packet of sugar: \(36 \, \text{cm} \times 30 \, \text{cm} \times 20 \, \text{cm}\) We are to find how many packets of sugar can fit in the container. Step 1: Calculate the volume of the container. The volume of a cuboid is given by: \[ \text{Volume} = \text{Length} \times \text{Width} \times \text{Height} \] For the container: \[ \text{Volume of container} = 300 \times 90 \times 84 \] \[ \text{Volume of container} = 2,268,000 \, \text{cm}^3 \] Step 2: Calculate the volume of one packet of sugar. \[ \text{Volume of one packet} = 36 \times 30 \times 20 \] \[ \text{Volume of one packet} = 21,600 \, \text{cm}^3 \] Step 3: Calculate the number of packets. \[ \text{Number of packets} = \frac{\text{Volume of container}}{\text{Volume of one packet}} \] \[ \text{Number of packets} = \frac{2,268,000}{21,600} \approx 105 \] Answer: The container can hold 105 packets of sugar.

Question 12a: Compound Interest Given: – Principal (\(P\)) = ₦500,000 – Annual interest rate (\(r\)) = 20% = 0.20 – Compounding frequency = quarterly (4 times a year) – Time (\(t\)) = 1 year 6 months = 1.5 years The formula for compound interest is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] Where: – \(A\) = total amount after interest – \(P\) = principal – \(r\) = annual interest rate – \(n\) = number of compounding periods per year – \(t\) = time in years Step 1: Substitute values. \[ A = 500,000 \left(1 + \frac{0.20}{4}\right)^{4 \times 1.5} \] Simplify: \[ A = 500,000 \left(1 + 0.05\right)^6 \] \[ A = 500,000 (1.05)^6 \] Using \( (1.05)^6 \approx 1.3401 \): \[ A = 500,000 \times 1.3401 = 670,050 \] Step 2: Calculate the compound interest. \[ \text{Compound Interest} = A – P = 670,050 – 500,000 = 170,050 \] Answer: The compound interest is ₦170,050.

Question 12b: Probability and Ratios Given: – Initial ratio: \(6:5:4\) (Blue: Yellow: Red) – Red balls = 12, so total balls = \(6x + 5x + 4x = 15x\). From \(4x = 12\), solve for \(x\): \[ x = 3 \] Initial counts: – Blue balls = \(6x = 18\) – Yellow balls = \(5x = 15\) – Red balls = \(4x = 12\) – Total = \(18 + 15 + 12 = 45\) (i) New Ratio After Removal – Removed 8 blue balls: \(18 – 8 = 10\) – Removed 7 yellow balls: \(15 – 7 = 8\) New total: \[ 10 + 8 + 12 = 30 \] New ratio of balls: \[ 10:8:12 = 5:4:6 \] Answer: The new ratio is 5:4:6. (ii) Probability of Selecting Two Yellow Balls Without Replacement Yellow balls = 8, total balls = 30. First draw: \[ P(\text{Yellow on first draw}) = \frac{8}{30} \] Second draw: \[ P(\text{Yellow on second draw}) = \frac{7}{29} \] Combined probability: \[ P(\text{Both yellow}) = \frac{8}{30} \times \frac{7}{29} = \frac{56}{870} = \frac{28}{435} \] Answer: The probability is \(\frac{28}{435}\).

Question 13a: Ticket Problem Given: – Total attendees = 429 – Adult ticket = $10.00 – Child ticket = $7.50 – Total proceeds = $3,725.00 Let the number of adults be \(a\) and children be \(c\). From the total attendees: \[ a + c = 429 \quad \text{(Equation 1)} \] From the total proceeds: \[ 10a + 7.5c = 3,725 \quad \text{(Equation 2)} \] Step 1: Solve Equation 1 for \(c\). \[ c = 429 – a \] Step 2: Substitute into Equation 2. \[ 10a + 7.5(429 – a) = 3,725 \] Simplify: \[ 10a + 3,217.5 – 7.5a = 3,725 \] \[ 2.5a = 507.5 \] Solve for \(a\): \[ a = \frac{507.5}{2.5} = 203 \] From Equation 1: \[ c = 429 – 203 = 226 \] Step 3: Calculate the percentage of adults. \[ \text{Percentage of adults} = \frac{203}{429} \times 100 \approx 47.3\% \] Answer: Approximately 47% of attendees were adults.

Question 13b: Direct and Inverse Proportionality Given: – \(Y\) is directly proportional to \(X\) and inversely proportional to \(Z^2\). \[ Y = k \frac{X}{Z^2} \] Where \(k\) is a constant. (i) Express \(Y\) in terms of \(X\) and \(Z\). Given \(Y = 6\), \(X = 12\), \(Z = 2\): \[ 6 = k \frac{12}{2^2} \] \[ 6 = k \frac{12}{4} \] \[ 6 = 3k \] Solve for \(k\): \[ k = 2 \] Thus, the equation is: \[ Y = 2 \frac{X}{Z^2} \] Answer: \(Y = \frac{2X}{Z^2}\). (ii) Find \(Y\) when \(X = 5\) and \(Z = \frac{1}{3}\). Substitute: \[ Y = 2 \frac{5}{\left(\frac{1}{3}\right)^2} \] \[ Y = 2 \frac{5}{\frac{1}{9}} \] \[ Y = 2 \times 5 \times 9 = 90 \] Answer: \(Y = 90\).

WAEC GCE Mathematics Past Questions and Answers are loading.. keep refreshing…

Candidates who sit for the WAEC GCE Mathematics examination are taken through two papers in mathematics.

Paper 1 (Objective) – Here, a candidate is expected to fill in answers in multiple-choice objective questions based on various topics in mathematics.

Paper 2 (Theory & Essay) – This section comprises short as well as long structured questions in which the candidate is put on test based on their applied mathematical abilities and logical thinking.

As you keep practicing these past questions, you will be exposed to the objective as well as theory questions. Topics addressed included–

Algebra,

Geometry,

Statistics,

Probability,

Trigonometry and,

Calculus.

Effective usage of WAEC GCE Mathematics Past Questions

As with everything in life, there are proper and incorrect ways of approaching past questions practice. To attain the maximum utility provided by past questions abdominal exercises, observe the following:

Study Each Topic Thoroughly: Go through each question in a number of the level in order to make sure that all ) of the syllabus has been in .

Simulate Exam Conditions: Set a timer and try answering the questions within the allocated time. This is meant to aid you in managing yourself within the time frame in the real examination.

Check Your Answers: After attempting the questions, there is always look at the given answers and clear the way of working of each who problems.

Focus on Weak Areas: Determine which areas of the subject are troublesome to you, and practice more on those areas.

Revise Regularly: Try to revise the past questions so that the concepts do not fade away in your mind.

FAQs on WAEC GCE Mathematics Past Questions

1. How often are the WAEC GCE Mathematics questions set reused?

Past or practice questions are very useful in preparing for the exams as the same or related questions may be posed in a different form.

2. Is this PDF available for free?

Indeed, all the WAEC GCE Mathematics past questions and their accompanying answers are very low of cost(#1,000) to download contact us on whatsapp:09061221656 to download.

3. Can past questions assure victory in the examination?

The practice of past questions enhances the performance rate of the candidates sitting for the exam.

4. Is it possible to achieve success in the examination based on the previous questions only?
   Inasmuch as practicing past questions increases the odds of your success, one must also appreciate the need to read the textbook and comprehend the concepts that are covered.

ALSO READ: WAEC GCE Past Questions and Answers PDF Download – All Subjects

Final Remarks

Revising WAEC GCE Mathematics past questions and answers papers that were set in the previous years has proven to be very helpful to one’s performance in the exam. Such materials will coalesce into a deeper appreciation of the structure, time management and readiness towards the outcome at hand. Remember to grab your copy as well and get to practice right away!

Access Your WAEC GCE Mathematics Past Questions PDF Now!

---

Post Tags:

WAEC GCE Mathematics past questions, WAEC GCE Mathematics answers PDF, WAEC Mathematics past papers, WAEC GCE Mathematics download, WAEC GCE past questions PDF, WAEC GCE 2024 Mathematics, WAEC GCE Mathematics exam prep, WAEC Mathematics past questions download, free WAEC Mathematics PDF, WAEC GCE Mathematics past questions and answers