WAEC Mathematics 2023
Questions & Answers

A student might mistakenly choose option A or C by dropping the negative sign when distributing across the parentheses. By properly distributing the negative sign across the second parenthesis, we get (2x + 3y) - x + 2y, which rearranges to 2x - x + 3y + 2y. Combining these like terms gives x + 5y, making option B correct.

One might pick option B or C by making an arithmetic error when isolating x. Expanding the right side of the equation yields 3x - 7 = 2x + 2; subtracting 2x from both sides gives x - 7 = 2; and finally adding 7 to both sides results in x = 9, matching option A.

A test-taker might select option A or C by confusing the exponent with a coefficient or miscalculating powers of ten. By definition, log10(1000) asks for the exponent to which 10 must be raised to equal 1000, and since 10^3 = 1000, option B is correct.

A student might erroneously choose option C or D by picking the wrong signs for the factors. To factor the quadratic, we search for two numbers that multiply to 6 and add up to -5, which are -2 and -3, yielding (x - 2)(x - 3) and making option A correct.

An unwary student might pick option B by forgetting to divide by 2 in the triangle area formula. Applying the formula for the area of a triangle, which is (1/2) multiplied by the base and height, we compute (1/2) × 10 × 5, resulting in 25 cm² and proving option A is correct.

One might choose option B or C by failing to recognize the structure of a perfect square. Recognizing the expression as a perfect square trinomial allows us to rewrite it as (x - 2)^2 = 0, which yields a double root at x = 2, making option A correct.

A student might mistakenly pick option A by confusing sine with cosine at 90 degrees. On the unit circle, sin(90°) corresponds directly to the y-coordinate at the top of the circle, which is 1, pointing to option B.

An unwary candidate might select option B or C due to an error in substitution or elimination. Adding the two given equations together eliminates y: (2x + y) + (x - y) = 5 + 1 simplifies to 3x = 6, which yields x = 2 and confirms option A.

A student might choose option D by accidentally finding the sum instead of the mean. The mean requires dividing the sum of the numbers by their count, so (2 + 4 + 6 + 8) / 4 equals 20 / 4, which gives 5, making option A correct.

One might mistakenly select option B or C by dropping the middle cross-terms. Using the binomial theorem or the FOIL method on (a + b)(a + b), we get a^2 + ab + ab + b^2, which simplifies to a^2 + 2ab + b^2, making option A correct.

A student might pick option B by forgetting to multiply after subtracting. Subtracting 2 from both sides of y/3 + 2 = 5 gives y/3 = 3, and multiplying both sides by 3 results in y = 9, validating option A.

An unwary test-taker might select option B by attempting to factor into distinct integers without checking the discriminant. Recognizing this expression as a perfect square yields (x - 3)^2 = 0, meaning the discriminant is zero and gives a repeated root at x = 3, matching option A.

A student might choose option B by confusing cosine with sine at zero degrees. On the unit circle, cos(0°) represents the x-coordinate at the rightmost point, which equals 1, making option A correct.

One might erroneously pick option B by multiplying the side length by 3 instead of cubing it. The volume of a cube is calculated by cubing the side length, giving 3 × 3 × 3 = 27 cm³, which means option A is correct.

A student might select option B by treating log x = 2 as multiplying by 10 instead of exponentiating. Assuming the common logarithm base 10, log10 x = 2 implies x = 10^2, which evaluates to 100 and makes option A correct.

An unwary candidate might select option D by stopping halfway through the simplification. Simplifying inside the radical first gives 16 / 4 = 4, and taking the square root of 4 results in 2, confirming option A.

Students might mistakenly choose B or D if they confuse the slope-intercept equation components or misidentify negative signs. To find the correct answer, evaluate the linear expression y = mx + c where m represents the slope or gradient and corresponds directly to the numerical coefficient attached to x, yielding 2. Common mistake: Confusing the y-intercept constant c with the gradient coefficient m.

Options like B or C might be tempting if trigonometric values are guessed without recalling exact coordinate ratios. Working through the trigonometric identity reveals that tan theta equals sin theta divided by cos theta; evaluating this at an angle of 45 degrees yields equal values of the square root of two over two for both sine and cosine, making their quotient equal to 1. Common mistake: Forgetting that sin(45°) and cos(45°) are identical, which leads to miscalculating their division.

A student could mistakenly select B by picking an arbitrary number or failing to sequence the dataset properly. Finding the correct middle value requires organizing the sequence in ascending order first: 1, 3, 5, 7, and 9; since there is an odd count of observations, the exact midpoint is the third entry, which evaluates to 5. Common mistake: Selecting the first number in the sequence instead of sorting and locating the actual midpoint.

Options like C or B might be selected if exponential rules are misapplied or indices are ignored entirely. To solve this, express both sides of the equation under a common base of 2, recognizing that 4 equals 2 squared and 16 equals 2 to the fourth power, so that (2^2)^x = 2^(2x) = 2^4 lets you equate the exponents 2x = 4 to isolate x as 2. Common mistake: Failing to convert both sides of the equation to share the same base before equating the exponents.