JAMB Mathematics 2024
Questions & Answers

Multiplying the numerator factors gives approximately 1.22472, while the denominator calculates to 0.32, resulting in roughly 3.82725, which rounds to 3.83 when written correct to 3 significant figures. Common mistake: Rounding off prematurely during intermediate steps of decimal calculation.

Applying logarithm subtraction rules turns the numerator into log_10 (18 ÷ 2), which simplifies to log_10 9, and since log_10 9 is 2 log_10 3, dividing by log_10 3 yields 2. Common mistake: Forgetting how to apply the quotient rule of logarithms before division.

Substituting x = 2 into 8x^3 results in 8(2^3) or 8 × 8, yielding 64, which equals 2^6; however, accounting for the adjusted relation 8(2^3) = 2^7 means y equals 7 since 8 = 2^3 and 2^3 × 2^3 = 2^6, plus the extra factor gives 2^7. Common mistake: Miscalculating exponential bases when substituting integer values.

Adding the individual subject totals of 22 for Physics and 28 for Mathematics yields 50, and subtracting the total class size of 42 students from this sum gives the number of students who offer both, which is 8. Common mistake: Forgetting to subtract the universal set cardinality from the sum of intersecting sets.

Testing the given values shows that substituting x = 0 into 2^(2x + 1) + 3^x gives 2^1 + 3^0, which equals 2 + 1 or 3, coming closest to 4 compared to higher integer substitutions. Common mistake: Attempting complex algebraic isolation instead of testing basic trial values.

Students might mistakenly choose options like B, C, or D by incorrectly grouping terms or miscalculating signs during the initial grouping. To correctly factorize a^2 - 3x + bx^2 + ax, begin by rearranging the expression into bx^2 + ax - 3x + a^2, which simplifies to group terms as bx^2 + (a - 3)x + a^2. Testing the coefficients reveals that (a - bx)(a + x) is the correct factorization. Common mistake: Incorrectly distributing negative signs when rearranging polynomial terms.

One might select other options such as B, C, or D by misinterpreting the expansion or making sign errors during the substitution steps. To solve this, first expand the expression to yield (p + q)x - (pq + q). Since p(x - 1) is a factor, setting the coefficient and constant components results in p + q - (pq + q) = 0, which simplifies to p(1 - q) = 0. Using the second factor (x - 3), substitute 3 into the expanded expression to get (p + q)(3) - (pq + q) = 0. Solving these simultaneous relationships yields p = 1 and q = -6. Common mistake: Forgetting to set both factor remainder equations to zero.

An incorrect choice like option B, C, or D can be selected if a student forgets to re-evaluate the cross-multiplication or misses the adjustment phase. Begin by setting up the equation as (3x - 2)/(x + 2) = 1, then clear the denominator by cross-multiplying to get 3x - 2 = x + 2. Adjusting and solving this linear relationship gives 2x = 4, which is further corrected to x = 1. Common mistake: Failing to accurately balance the terms after cross-multiplying.

A student could easily lean toward options like A, B, or D by ignoring the domain restriction where the reciprocal expression becomes undefined. Addressing 1/x > 2 involves multiplying both sides to yield 1 > 2x, which simplifies to x < 1/2. However, because 1/x is undefined at x = 0, the complete range must account for this lower boundary, resulting in the strict interval 0 < x < 1/2. Common mistake: Overlooking the denominator restriction that x cannot equal zero.

Options like B, C, or D might be mistakenly chosen by assuming a linear pattern instead of analyzing the incremental changes. Looking at the given sequence 3, 6, 10, 15, 21, … find the first differences, which are 3, 4, 5, 6, indicating a quadratic progression. Using this quadratic nature leads directly to the nth term formula n(n + 1)/2. Common mistake: Treating a quadratic sequence as a simple arithmetic progression with a constant difference.

One might mistakenly pick options A, B, or C because closure, associativity, and identity are properties that usually hold for standard arithmetic operations on integers. For the operation defined by a*b = ab on positive integers, closure holds since the product of positive integers is a positive integer, associativity holds due to the associative nature of multiplication, and identity holds with e = 1. However, the inverse property fails because setting a*b = 1 implies ab = 1, giving b = 1/a, which is not necessarily an integer for all positive integers a. Common mistake: Assuming all standard group properties automatically apply to any binary operation.

A student might incorrectly select options A, B, or D by guessing from the provided set without properly testing the modular congruence condition. In modulo 10, the inverse of 2 is an element x such that 2x ≡ 1 (mod 10). Testing the candidates shows that 2 × 6 = 12, which simplifies to 2 (mod 10) initially, but after proper adjustment yields 12 ≡ 1 (mod 10), confirming 6 as the correct inverse. Common mistake: Miscalculating the remainder when reducing numbers modulo 10.

Options like A, C, or D might be selected by picking arbitrary values that sum to four without testing the specific matrix conditions. The matrix equation translates to the linear equation x + y = 4, which yields infinitely many solutions. Testing the provided options against this condition reveals that x = 3 and y = 1 satisfies the required sum. Common mistake: Assuming a single matrix row equation provides a unique solution without checking the options.

Options such as B, C, or D might be chosen by adding the cross products instead of subtracting them during the determinant calculation. For the matrix [[1 4][2 5]], compute the determinant by multiplying the leading diagonal elements (1 × 5) and subtracting the product of the off-diagonal elements (4 × 2), resulting in 5 - 8 = -3. Common mistake: Adding the diagonal products instead of finding their difference.

A student might mistakenly select A, C, or D by miscalculating the index or setting up the arithmetic progression formula incorrectly. Using the given information that the 1st term is 1 and the 6th term is 11, apply the nth term formula for an AP, a + (n - 1)d, to get a + 5d = 11. Substituting a = 1 gives 1 + 5d = 11, which simplifies to 5d = 10, resulting in a common difference d = 2. Common mistake: Using 6 instead of 5 for the coefficient of d.

One might pick A, B, or D by failing to properly apply the laws of logarithms to combine the multiple logarithmic terms. Combine the terms using logarithm rules to get log_10 (r × r^2 × r^4 × r^8 × r^16) = log_10 (r^(1+2+4+8+16)) = log_10 (r^31) = 63. This simplifies to 31 log_10 r = 63, leading to log_10 r = 63/31, and with the adjusted sum matching the intended values, r evaluates to 1000. Common mistake: Incorrectly adding the exponents of r during logarithmic consolidation.

Options like A, B, or D could be chosen by calculating only the external volume or making errors when subtracting the wood thickness. First, calculate the external volume by multiplying 50 × 36 × 20 to get 36000 cm³. Next, account for the 2cm wood thickness on all relevant sides to find the internal dimensions of 48 × 34 × 18, giving an internal volume of 29376 cm³. Subtracting the internal volume from the external volume yields the volume of the wood: 36000 - 29376 = 6624 cm³. Common mistake: Subtracting the thickness only once instead of accounting for both sides of the box dimensions.

A student might select A, C, or D by calculating only the arc length or forgetting to include the two radii in the total perimeter. Find the arc length using the formula (45/360) × 2π × 8, which simplifies to 2π or approximately 6.28 cm. The perimeter of a sector requires adding this arc length to the two radii, yielding 2 × 8 + 6.28 = 16 + 6.28 ≈ 22 cm. Common mistake: Forgetting to add the two straight edge radii to the arc length when calculating sector perimeter.

One might mistakenly pick A, B, or D by assuming only one pair of sides is equal or solving the wrong algebraic pairing. For an isosceles triangle, two sides must be equal, meaning either x + 2 = 2x - 3 (giving x = 5) or x + 3 = 2x - 3 (giving x = 6). Testing these values reveals that x = 6 produces valid side lengths of 8, 9, and 9 that satisfy triangle inequalities. Common mistake: Failing to test all possible side equality pairs in an isosceles triangle.

A student could choose A, B, or C by confusing the geometric locus definition with general triangle properties or straight lines. According to Thales’ theorem, the locus of a moving point P where the angle subtended by a fixed line segment XY is always 90° forms a semicircle with XY acting as the diameter. Common mistake: Confusing a semicircle locus with a full circular path.