WAEC Mathematics 2017
Questions & Answers

A student might mistakenly select Option B or D by misplacing the decimal position or failing to round the trailing digits properly. To solve this, convert the value into standard scientific notation as 4.07 × 10^{-5}. Inspecting the initial two figures, 4 and 0, notice that the subsequent digit is 7, which requires incrementing the second digit by one. This transforms the expression into 4.1 × 10^{-5}, which equals 0.000041. Common mistake: dropping zeros after the decimal point or forgetting to round up when the following digit is five or greater.

A student might incorrectly choose Option A or C by treating inverse proportionality as direct variation or squaring the variable. Given that x is inversely proportional to y, we can write the equation as x = k / y using a constant k. Since y is directly proportional to z, substituting y = m z with another constant m yields x = k / (m z). Grouping the constants together as (k/m) / z demonstrates that x varies inversely with z. Common mistake: multiplying the relationship constants instead of substituting the dependent variable correctly.

A student might mistakenly pick Option A or C due to arithmetic errors when simplifying fractions or processing the denominator subtraction. Begin by converting mixed numbers in the numerator into improper fractions: 3 1/4 becomes 13/4 and 1 3/5 becomes 8/5, multiplying to give 104/20 or 26/5. For the denominator, convert 11 1/3 to 34/3 and 5 1/3 to 16/3, which subtracts to 18/3 or 6. Dividing the numerator fraction 26/5 by 6 results in 26/30, which reduces to 13/15. Common mistake: incorrectly subtracting the fractional parts before handling whole numbers or failing to simplify the final fraction.

A student might mistakenly choose Option A or D by miscalculating modular arithmetic additions or subtracting instead of adding. The expression (n ⊕ 4) ⊕ 3 ≡ 0 (mod 5) simplifies by combining the constant values to give n + 7 ≡ 0 (mod 5). Reducing 7 modulo 5 leaves n + 2 ≡ 0 (mod 5), which rearranges to n ≡ -2. Adjusting negative residues within modulo 5 yields 3, making n equal to 3. Common mistake: forgetting to reduce intermediate sums modulo 5 before isolating the unknown variable.

A student might mistakenly pick Option A or B by multiplying the ratios directly without accounting for the intermediate variable. Represent Tunde's age as k, making Ola's age equal to 2k based on the first ratio of 1:2. Using the second ratio of 4:5 for Ola to Musa, write 2k / m = 4/5, which expresses Musa's age m as (5/2)(2k) or 5k/2. Forming the final ratio of Tunde to Musa gives k : 5k/2, which simplifies to 2:5. Common mistake: treating the ratio components as absolute values rather than scaling them through a common variable.

A student might mistakenly choose Option B or D by assuming the boundary number 8 is included in the given inequalities. Statement I is false because 8 is excluded from both set M (x < 8) and set N (x > 8). Statement II is also false because 8 fails to belong to either individual set, meaning it cannot be in their union. Statement III is true because the two sets share no overlapping elements, making their intersection empty (∅). Common mistake: confusing strict inequalities with inclusive inequalities when evaluating boundary membership.

A student might mistakenly select Option C or D by incorrectly applying the power rule or subtraction property of logarithms. Utilizing the multiplication property of logarithms, log 14 can be rewritten as log (7 × 2). Expanding this product gives the sum of the individual logarithms, log 7 + log 2. Substituting the given variables a and b directly yields a + b. Common mistake: converting multiplication inside a logarithm into multiplication or subtraction of the separate log terms.

A student might mistakenly pick Option A or C through sign errors when dividing negative numbers or combining terms. First, compute the product xy as (2/3) × (-6), which equals -4. Next, compute the quotient y/x as (-6) / (2/3), which transforms into -6 × (3/2) resulting in -9. Subtracting the second result from the first gives -4 - (-9), which simplifies to -4 + 9 to equal 5. Common mistake: mishandling double negatives when subtracting a negative division result.

A student might mistakenly choose Option A or C by improperly adding fractions without finding a common denominator. Combining the terms on the left side yields (1 + 5)/(5x) = 3, which simplifies to 6/(5x) = 3. Solving for the denominator term gives 5x = 6/3 = 2, resulting in x = 2/5. Common mistake: adding the numerators while incorrectly altering or dropping the denominators during fraction addition.

A student might mistakenly pick Option A or B by distributing the total amount incorrectly among the participants. Let each girl's share be represented by g, making each boy's share g + 20. Setting up the total sum equation as 5(g + 20) + 4g = 18100 expands to 9g + 100 = 18100, which simplifies to 9g = 18000 and gives g = 2000. Adding the ₦20 difference gives the boy's share as ₦2,020.00. Common mistake: applying the monetary difference to the total sum instead of scaling it across the specific group of boys.

A student might mistakenly select Option A or D by guessing incorrect factor signs or miscalculating the product terms. To factor 7x² + 33x - 10, identify two numbers that multiply to 7 × (-10) = -70 and add to 33, which are 35 and -2. Rewrite the middle term to get 7x² + 35x - 2x - 10, then factor by grouping as 7x(x + 5) - 2(x + 5), yielding the factors (7x - 2)(x + 5). Common mistake: selecting factor pairs that add up to the correct coefficient but fail to produce the required product.

A student might mistakenly pick Option B or D by mishandling inequality signs during division. Divide the compound inequality by the positive fraction 3/4, giving -1/3 < 3x - 2 < 1/3. Add 2 across all parts to get 5/3 < 3x < 7/3, and finally divide by 3 to arrive at 5/9 < x < 7/9. Common mistake: forgetting to apply the addition and division steps uniformly across all three parts of the compound inequality.

A student might mistakenly choose Option B or D by distributing negative signs incorrectly across the parentheses. Expanding the expression 3x - (p - x) - (r - p) yields 3x - p + x - r + p. Combining the like terms for x gives 4x, while the -p and +p terms cancel out completely, leaving -r. Common mistake: failing to distribute the negative sign to every term inside the second set of parentheses.

A student might mistakenly pick Option A or C by using degree formulas instead of radian relationships or miscalculating the conversion factor. Using the arc length formula s = r θ, substitute the given values to find 7.5 = 7.5 θ, which means θ equals 1 radian. Convert this angle to degrees by multiplying by 180/π, giving 180 × (7/22) ≈ 57.27°, which rounds to the nearest degree as 57°. Common mistake: mixing up radian and degree measures when applying the arc length formula.

A student might mistakenly select Option B or C by miscalculating the cylinder base area or multiplying the flow rate incorrectly. Determine the total volume after 4 seconds by multiplying the flow rate 40π by 4, yielding 160π cm³. Compute the base area using πr² with a radius of 4, which gives 16π cm². Dividing the total volume by the base area (160π / 16π) results in a height of 10 cm. Common mistake: squaring the radius incorrectly when calculating the circular base area of the container.

A student might mistakenly pick Option B or D by forgetting to divide the total volume in half or converting cubic centimeters to litres incorrectly. Calculate the full volume of the tank as 13 × 10 × 70 = 9100 cm³, and halve it to account for the half-filled state, resulting in 4550 cm³. Convert this volume to litres by dividing by 1000, yielding 4.55 litres. Common mistake: misapplying the conversion factor between cubic centimeters and litres.

A student might mistakenly choose Option B or D by using incorrect formulas for hemispherical surface area or volume. Set the curved surface area of a solid hemisphere, given by 2πr², equal to its volume formula, (2/3)πr³. Dividing both sides by 2πr² provides 1 = (1/3)r, which solves to r = 3.0 cm. Common mistake: including the flat circular base area in the surface equation when the question specifies only the curved surface area.

A student might mistakenly select Option B or D by confusing supplementary angles with complementary or reflex measurements. In any parallelogram, consecutive interior angles are supplementary, meaning their measures sum to 180°, while opposite angles are equal. Option A correctly states this geometric property. Common mistake: confusing the supplementary nature of adjacent angles with the equality of opposite angles.

A student might mistakenly pick Option A or D by misapplying trigonometric ratios or special triangle side proportions. In a 30-60-90 right-angled triangle, the side lengths follow a ratio of 1 : √3 : 2. Since the side opposite the 30° angle is half the length of the hypotenuse, multiplying 5 cm by 2 gives the hypotenuse length as 10 cm. Common mistake: multiplying the short side by √3 instead of 2 when attempting to find the hypotenuse.

A student might mistakenly choose Option B or D by confusing interior angle sums with exterior angle properties. The sum of all exterior angles in any convex polygon is always 360°. For a regular hexagon with six equal sides, dividing 360° by 6 yields an individual exterior angle of 60°. Common mistake: calculating the interior angle instead of dividing the total 360 degrees by the number of sides.