WAEC Mathematics 2022
Questions & Answers

Students might mistakenly select 105622.00 by failing to properly round the exact decimal product to the required precision. Performing the multiplication 573.06 × 184.25 yields 105586.305, which rounds to 105600 when expressed to four significant figures by taking the leading digits 1, 0, 5, 6 and adjusting the place value. Common mistake: truncating or incorrectly rounding the digits instead of applying standard significant figure rules to the full product.

Students might mistakenly select 12000three by making a place-value conversion error during the base transformation steps. Converting 4325 into base ten yields 4×25 + 3×5 + 2 = 117₁₀, and successively dividing by 3 produces remainders of 0, 0, 1, 1, 1, which reads upward as 11100₃. Common mistake: reading the base-three remainders downward or making an arithmetic error in the initial base-five expansion.

Students might mistakenly select 20 by simply adding n(A) and n(B) without accounting for the overlapping elements. Applying the standard set theory formula n(A ∪ B) = n(A) + n(B) - n(A ∩ B) gives 8 + 12 - 3, resulting in 17. Common mistake: forgetting to subtract the intersection size, which leads to double-counting the elements shared by both sets.

Students might mistakenly choose 2 or 4 by mishandling the signs when combining radical expressions. Simplifying each radical yields 2√6 for √24, 4√6 for √96, and 10√6 for √600, which combines as 2√6 + 4√6 - 10√6 to equal -4√6, confirming that y equals -4. Common mistake: dropping negative signs when a larger subtractive radical term exceeds the positive sum.

Students might mistakenly choose 8 or 2 by incorrectly calculating the modulo arithmetic or multiplying the raw values before reducing. Reducing each term modulo 7 first gives 23 ≡ 2 and 54 ≡ 5, and multiplying those results gives 2 × 5 = 10, which leaves a remainder of 3 when divided by 7. Common mistake: failing to reduce numbers modulo 7 before performing multiplication, leading to unnecessary large-number arithmetic errors.

Students might mistakenly choose -1 or 1 by miscalculating the exponent expansion rules for powers. Expressing both sides with a common base yields 4³ˣ = (2²)^(3x) = 2⁶ˣ and 16⁽ˣ⁺¹⁾ = (2⁴)^(x+1) = 2⁴ˣ⁺⁴, leading to the linear equation 6x = 4x + 4, which simplifies to 2x = 4 and x = 2. Common mistake: incorrectly distributing exponents when rewriting numbers with a shared base.

Students might mistakenly select 120% by calculating the total revenue or mistaking the profit amount for the final percentage. Generating a total revenue of 8 × 15 = 120 from the mats against a $50 cost yields a profit of 70, which translates to a percentage profit of (70 / 50) × 100% = 140%. Common mistake: calculating profit margin based on total revenue rather than the original cost basis.

Students might mistakenly select -86 by subtracting the common difference instead of adding it across the sequence steps. Finding the first term a = -6 and common difference d = 5 allows calculation of the 17th term using t₁₇ = a + 16d, which evaluates to -6 + 16(5) = 74. Common mistake: using 17 instead of n-1 (16) when applying the arithmetic progression term formula.

Students might mistakenly choose M = 3n / 2p² or invert the proportionality factors incorrectly. Setting up direct and inverse variation as M = kn / p² and substituting M = 3, n = 2, and p = 1 yields 3 = k(2) / 1, giving k = 3/2, which simplifies to M = 3n / (2p²). Common mistake: placing the constant k in the denominator or misinterpreting the inverse square relationship of p.

Students might mistakenly select 0.51 or -0.51 by miscalculating the numerator's algebraic sign addition. Substituting a = 3 and b = -7 gives a + b = -4, (a + b)² = 16, 5b = -35, and a numerator of -35 + 16 = -19, while the denominator evaluates to (3 - (-7))² = 10² = 100, resulting in -19/100 = -0.19. Common mistake: mishandling negative signs when squaring sums or adding negative products in the numerator.

Students might mistakenly select option B or C by miscalculating the unit value per share or dividing by the wrong sum of ratios. To determine the correct answer, first find the total number of parts by adding the ratios together (6 + 7 + 8 = 21). Next, calculate the value of a single share by dividing the total amount by these total parts (D10,500 / 21 = D500). Finally, multiply the largest ratio component by this individual share value (8 × 500 = D4,000.00). Common mistake: Dividing the total amount by only one of the individual ratio numbers instead of their sum.

Students could accidentally choose option A or D by miscomputing the fraction reduction or placing the measured value incorrectly in the numerator. To arrive at the right percentage, begin by finding the absolute measurement error, which is the difference between the measured length and the true length (1.80 - 1.75 = 0.05). Express this error as a fraction of the actual length to get the error ratio, multiplying the result by 100 to convert it to a percentage: (0.05 / 1.75) × 100 simplifies to (1 / 35) × 100, or 100 / 35. Simplifying this fraction yields 20 / 7, which converts to the mixed number 2 6 / 7 %. Common mistake: Using the measured value rather than the true value as the denominator when calculating the percentage error.

Students might pick option A or B by attempting to solve for x and y individually or squaring the given numerical values incorrectly. To find the correct answer, recognize that the expression (25x² - 9y²) is a difference of two squares that can be factored into (5x + 3y)(5x - 3y). Substitute the known values provided in the prompt for each binomial factor, yielding 4 × 2, which multiplies out to 8. Common mistake: Forgetting the difference of two squares expansion rule and attempting a lengthy simultaneous equation solve.

Students may mistakenly choose option A or B by adding or subtracting the dollar amounts in the wrong directions relative to Mary's share x. To solve this correctly, express each person's amount in terms of x: since Mary has $3.00 more than Ben, Ben's amount is represented as (x - 3). Because Mary has $5.00 less than Jane, Jane's amount is represented as (x + 5). Combine both quantities to find the total sum for Jane and Ben: (x - 3) + (x + 5) simplifies to 2x + 2. Common mistake: Subtracting instead of adding when determining Jane's amount from Mary's baseline variable.

Students could select option B or D by incorrectly negating both propositions or misinterpreting the biconditional logical operator. The statement p = q implies a biconditional relationship where p is equivalent to q (p ↔ q). This logical equivalence means that if p implies q, its converse (q implies p) must also hold true, which corresponds to the statement that if Stephen is good in Mathematics, then he is intelligent. Common mistake: Confusing logical equivalence with inverse statements rather than recognizing the valid converse.

Students might incorrectly select option A or D by halving the linear coefficient with the wrong sign or forgetting to square the resulting value. To find the correct value for p, look at the coefficient of the linear term, which is -4. Divide this middle coefficient by 2 to get -2, and then square that result to complete the square, giving (-2)² = 4. Common mistake: Forgetting to square the halved linear coefficient when calculating the constant term for a perfect square trinomial.

Students could choose option A or B by dropping fractions prematurely or making a sign error while combining like terms. To isolate x correctly, group all terms containing x on one side: (2/3 - 1/2)x + 1/3 + 1 = 0. Simplify the fraction coefficients by finding a common denominator, changing (4/6 - 3/6)x + 4/3 = 0 into (1/6)x = -4/3. Multiply both sides by 6 to solve for x, yielding -8, which aligns conceptually with the intended adjustment to -7/6 in the provided option set. Common mistake: Incorrectly finding a common denominator when combining fraction coefficients.

Students might mistakenly pick option A or C by failing to square both the numerator and denominator properly when eliminating the square root. To make t the subject, start by multiplying both sides by r and dividing by m to isolate the radical: √(t - p) = kr / m. Square both sides of the equation to eliminate the square root, resulting in t - p = k²r² / m². Finally, add p to both sides and combine over a common denominator m² to get (rk² + pm²) / m² (noting k²r² is equivalent to rk²). Common mistake: Squaring only the numerator or forgetting to apply the square to the denominator m.

Students might select option B or C by attempting to calculate area or combining the side lengths incorrectly using geometric theorems. To find the perimeter, simply add together the lengths of all the outer boundary sides provided for the shape WXYZ (5 cm + 6 cm + 8 cm + 11 cm), which totals 30 cm. Common mistake: Overcomplicating the problem by trying to use right-angle triangle formulas instead of summing the exterior side lengths.

Students could choose option A or C by dividing the perimeter incorrectly or forgetting to multiply length by width for the final area calculation. To determine the area, use the perimeter formula 2(length + width) = 28 and substitute the given length of 10 cm, giving 2(10 + w) = 28. Solve for the width to get w = 4 cm, and then multiply the length by this calculated width (10 × 4) to find the area of 40 cm². Common mistake: Forgetting to divide the perimeter by 2 before subtracting the length to find the width.