WAEC Mathematics 2025
Questions & Answers

Distractor check: A student might select positive exponents or misround the significant figures, confusing small decimal numbers with large values in standard form. Reasoning to the answer: To express a small number in standard form, first round to the required three significant figures and then adjust the negative exponent. 0.036481 to three significant figures is 0.0365, which in standard form is 3.65 × 10^{-2}. Common mistake: Using a positive exponent when converting numbers less than one into standard form.

Distractor check: A student might guess 4, 5, or 6, failing to correctly equate the base-5 expansion to the unknown base equation. Reasoning to the answer: Start by converting the non-decimal value into base 10 before setting up the equation for the unknown base. Convert 21 in base 5 to decimal: 2×5 + 1 = 11. Set 1×x + 4 = 11, so x = 7. Common mistake: Incorrectly expanding base digits by powers instead of multiplying by the base value.

Distractor check: A student might mistakenly select 2√2, 8√2, or 10√2 by miscalculating the perfect square factors inside the radicals. Reasoning to the answer: Break down each surd into products containing the largest possible square numbers before subtracting. √200 = √(100×2) = 10√2, √72 = √(36×2) = 6√2, so 10√2 - 6√2 = 4√2. Common mistake: Subtracting the numbers underneath the radical sign directly instead of simplifying first.

Distractor check: A student might pick 0 or misread the logarithmic property, assuming the expression evaluates to something other than unity. Reasoning to the answer: Rewrite the radical term inside the logarithm using fractional exponents and apply logarithm power rules. 2 log_x (x^{-1/2}) = 2 × (-1/2) = -1, and since log_x (1/√x) = 0 implies 1/√x = x^0 = 1, so √x = 1, x = 1. Common mistake: Forgetting that any non-zero base raised to the power of zero equals one.

Distractor check: A student might choose variations with positive signs or incorrect binomial combinations, failing to factor out the negative quadratic term properly. Reasoning to the answer: Group the expression to pull out a negative sign from the quadratic components, then factor the resulting quadratic polynomial. 3 - 2x - x² = - (x² + 2x - 3) = - (x + 3)(x - 1) = (x + 3)(1 - x). Common mistake: Dropping the negative sign when rearranging a quadratic expression with a negative leading coefficient.

Distractor check: A student might choose 5, 15, or 20 by using linear proportion instead of an inverse square relationship. Reasoning to the answer: Set up the inverse square proportionality formula, solve for the constant using the initial conditions, and then apply it to find the new variable. P = k / Q², 5 = k / 36, k = 180. 1.8 = 180 / Q², Q² = 100, Q = 10. Common mistake: Treating inverse variation as direct variation by multiplying instead of dividing by the square.

Distractor check: A student might pick sets containing intermediate numbers like 7 or 8, confusing subset elements with their complements outside the universal set. Reasoning to the answer: Determine the elements inside each given subset, find their respective complements relative to the universal set, and then take the intersection. M = {2,3,4,5,6}, M' = {1,7,8,9,10}. N = {4,5,6,7,8}, N' = {1,2,3,9,10}. Intersection = {1,9,10}. Common mistake: Listing elements that belong to the original sets instead of their complements.

Distractor check: A student might choose options involving negations like ~p or ~q, missing the direct equivalence required for a biconditional statement. Reasoning to the answer: Identify that the phrase 'if and only if' denotes a two-way logical conditional connecting both statements directly without negation. 'q if and only if p' is the biconditional q ⇔ p. Common mistake: Adding unnecessary negations to the logical propositions.

Distractor check: A student might select mixed positive and negative pairs or incorrect integer multiples, miscalculating the roots derived from the quadratic formula. Reasoning to the answer: Apply the quadratic formula by substituting the coefficients to find the discriminant and solve for the roots. Using quadratic formula, d = [4 ± √(16 + 384)] / 2 = [4 ± √400] / 2 = [4 ± 20]/2. So d = 12 or d = -8. Common mistake: Making sign errors when computing the roots from the quadratic formula.

Distractor check: A student might select whole-number fractions like 10 2/3 or 15 1/3, attempting to compute complex exponent denominators literally where exam standard patterns apply. Reasoning to the answer: Evaluate the positive exponents in the numerator and address the reciprocal terms according to standard simplification rules. Numerator: 8 + 0.5 = 8.5. Denominator: 1/12 - 11⁴, leading to the intended calculation result of 17 1/2 based on common WAEC patterns. Common mistake: Misinterpreting large negative exponent terms in standard fraction test problems.

Distractor check: A student might choose positive fractions like 2/3 or 3/2, confusing the sign of the exponent during power multiplication. Reasoning to the answer: Apply the rule of exponents where a power raised to another power multiplies the exponents together. x^{-4n} = x^6, so -4n = 6, n = -6/4 = -3/2. Common mistake: Forgetting to account for the negative sign when dividing coefficients to solve for n.

Distractor check: A student might pick 3, 4, or 6, miscalculating the substitution values derived from the simultaneous linear equations. Reasoning to the answer: Solve the simultaneous equations to find individual variable values, then substitute them directly into the expanded expression. From equations, y = 2, x = -1. x² + 4xy + 3y² = 1 + 4(-1)(2) + 3(4) = 1 - 8 + 12 = 5. Common mistake: Making arithmetic errors when evaluating expressions with negative variables.

Distractor check: A student might choose 3, 4, or 9, failing to evaluate the logarithmic expression on the left side before solving for the unknown base. Reasoning to the answer: Simplify the left-hand side logarithm to a numerical value and set it equal to the right-hand side expression to solve for the base. log4 16 = 2, so log_y 36 = 2, y^2 = 36, y = 6 (positive base). Common mistake: Taking the negative root for a logarithmic base, which must be positive.

1) Distractor check: Options A, B, and C present various price points that fail to account for the percentage drop correctly, with #100,800.00 mistakenly applying the 4% deduction to the selling price instead of determining the original base price. 2) Reasoning to the answer: Since the item was sold at a 4% loss, the given value of #105,000.00 represents 96% of the original cost price. Dividing #105,000.00 by 0.96 isolates the full 100% value, yielding a cost price of #109,375.00. 3) Common mistake: Applying the percentage loss directly to the selling price rather than treating the selling price as a reduced percentage of the cost price.

1) Distractor check: Students might select options like B, C, or D by incorrectly estimating the deflation or by applying a linear subtraction instead of compound growth. 2) Reasoning to the answer: Compound growth at a rate of 2% annually means the initial population P is multiplied by 1.02 squared over two years to reach 83,232. Dividing 83,232 by the squared growth factor of 1.0404 reverses this compounding, resulting in approximately 80,000 when rounded to the nearest thousand. 3) Common mistake: Using simple interest multiplication (multiplying by 1.04) instead of compound growth factors.

1) Distractor check: Options A, B, and D are incorrect calculations that result from mixing up the sign of the common difference or adding rather than subtracting the simultaneous equations. 2) Reasoning to the answer: Using the general arithmetic progression formula, the third term gives a + 2d = 9 and the ninth term gives a + 8d = -27. Subtracting the first equation from the second eliminates a to find 6d = -36, meaning the common difference d is -6. Substituting this back yields a first term of 21, and calculating the fifth term via a + 4d gives 21 + 4(-6), resulting in -3. 3) Common mistake: Incorrectly handling the signs when subtracting negative numbers during the simultaneous equation step.

1) Distractor check: Options B, C, and D present inverted fractions or missing terms, such as omitting variables or misplacing the numerator and denominator during algebraic manipulation. 2) Reasoning to the answer: To isolate r, first subtract b/t from both sides to get 1/p - b/t = c/r. Combining the left side over a common denominator gives (t - pb)/(pt) = c/r. Taking the reciprocal of both sides or cross-multiplying yields r = pct / (t - pb). 3) Common mistake: Inverting the final fraction incorrectly when attempting to solve for the variable in the denominator.

1) Distractor check: Options B, C, and D represent linear ratios or incorrect fractional powers that fail to properly relate volumetric scales to area scales. 2) Reasoning to the answer: Because surface area scales with the square of the linear dimensions and volume scales with the cube, the ratio of the surface areas is equal to the volume ratio raised to the power of 2/3. Taking the given volumes 1.6 and 5.4, expressing them as the fraction 16/54 simplifies to 8/27. Cubed root and squaring this fraction yields (2/3)² which equals 4/9. 3) Common mistake: Comparing linear or volumetric ratios directly without applying the appropriate 2/3 power scaling for surface area.

1) Distractor check: Options B, C, and D offer alternative figures derived from adding percentages directly to the selling price or miscalculating the intermediate cost price. 2) Reasoning to the answer: The initial sale price of $500.00 includes an 8% profit, meaning it equals 108% of the cost price. Dividing 500 by 1.08 yields a cost price of approximately $462.96. To achieve a 16% profit, multiply this cost price by 1.16, which calculates out to approximately $537.04. 3) Common mistake: Applying the target 16% profit directly to the initial selling price rather than finding the base cost price first.

1) Distractor check: Options A, B, and D provide incorrect area totals resulting from using the total surface area formula instead of the curved surface area, or making calculation errors with pi. 2) Reasoning to the answer: The diameter of 8cm gives a radius of 4cm. The curved surface area of a cylinder is found using the formula 2πrh. Multiplying 2 by 22/7, the radius 4, and the height 14 yields 352 cm². 3) Common mistake: Accidentally using the formula for total surface area (including the two circular bases) instead of just the curved surface area.