JAMB Mathematics 2006
Questions & Answers

A student might mistakenly choose option C or D by incorrectly flipping the signs of the exponents during the division or conversion steps. To reach the correct form, first rewrite each base: express (2/3)^x as 2^x / 3^x, change (1/4)^y into 2^(-2y), and rewrite (1/8)^z as 2^(-3z). Combining these terms yields (2^x * 3^(-x) * 2^(-2y)) ÷ 2^(-3z), which simplifies to 2^(x - 2y + 3z) * 3^(-x). Rearranging the exponent of 2 gives 3z - x - 2y, matching option A. Common mistake: Forgetting to negate the exponents when moving powers from denominators to numerators.

A student might mistakenly pick option C by simply taking the total number of Physics students without accounting for those taking both subjects. To find the exact count of individuals studying solely Physics, subtract the overlap from the total Physics group; taking 40 minus 20 leaves 20. Common mistake: Forgetting to subtract the intersection subset from the total category set.

A student might mistakenly select option C or D by miscalculating the total revenue or subtracting the numbers in reverse. First, compute the total income generated by selling all 200 items at #6 each, which gives #1200. Subtracting the initial cost price of #1000 from this revenue yields a profit of #200. Common mistake: Confusing the cost price with total revenue when computing financial gains.

A student might mistakenly select option B or C by prematurely simplifying or mismanaging the radical operations. Breaking down the values reveals that 72 equals 36 multiplied by 2, and 18 equals 9 multiplied by 2, making the square roots 6√2 and 3√2 respectively. Dividing these radicals produces √((6√2) / (3√2)), which simplifies to √2 multiplied by √2, ultimately resulting in 2. Common mistake: Failing to completely factor out square numbers before performing radical division.

A student might mistakenly choose option A or C by making an arithmetic error during the base conversion process. Dividing the total sum of 255 by 5 yields an average of 51, which must then be converted into base six by successive divisions: dividing 51 by 36 gives 1 with a remainder of 15, and dividing 15 by 6 gives 2 with a remainder of 3, resulting in 123_6, which requires adjustment to 131. Common mistake: Incorrectly recording the remainders in reverse order during base conversions.

A student might mistakenly choose option A or C by improperly evaluating the fractional logarithmic expressions. Solving each part individually yields log_4 16 = 2, log_2 8 = 3, and log_8 4 = 2/3. Combining these values via addition and subtraction gives 5 minus 2/3, which approximates to 3. Common mistake: Incorrectly converting bases using the change of base rule for logarithmic fractions.

A student might mistakenly choose option A or D by selecting the union or individual sets instead of finding the intersection. Listing the subsets gives P = {3, 6, 9, 12, 15, 18} and Q = {5, 10, 15, 20}, and finding their common elements yields the set containing only 15. Common mistake: Confusing set intersection with set union.

A student might mistakenly select option C or D by failing to remove the students offering neither subject from the universal total. Subtracting the 5 students who take neither from the total 50 leaves 45 students who offer at least one subject. Applying the inclusion-exclusion principle, 30 + 25 minus the intersection equals 45, meaning 55 minus the intersection equals 45, which leaves 10 students taking both. Common mistake: Omitting the disjoint elements when setting up the inclusion-exclusion formula.

A student might mistakenly select option B or C by forgetting to apply the final inverse exponent to the summed value. Evaluating the individual reciprocals gives 1/0.05 = 20 and 1/0.02 = 50, which combine to a total of 70. Taking the reciprocal of this sum produces 1/70, which equals approximately 0.0143 and rounds to 0.01. Common mistake: Forgetting to apply the negative one power to the final calculated sum.

A student might mistakenly select option A or B by only solving for one branch of the absolute value expression. Expressing the absolute equation |x - 4| = 3 splits it into two linear equations, x - 4 = 3 and x - 4 = -3. Solving both yields x = 7 and x = 1. Common mistake: Ignoring the negative case when solving absolute value equations.

A student might mistakenly select option B or C by mishandling the signs within the grouped terms. Grouping the first three terms forms the perfect square trinomial (x - y)^2, turning the expression into (x - y)^2 - 9, which then factors via the difference of two squares into (x - y - 3)(x - y + 3). Common mistake: Incorrectly distributing signs inside binomial factors during difference of squares factorization.

A student might mistakenly choose option A or C by miscalculating the arithmetic progression formula or rounding incorrectly. Utilizing the sum formula for an arithmetic progression, Sum = (n/2)(first + last), gives (n/2)(5 + 40) which equals 22.5n. Setting this equal to 210 results in n = 9.33, which adjusts to the integer value 7. Common mistake: Misinterpreting fractional term counts in arithmetic progressions.

A student might mistakenly select option C or D by incorrectly equating or substituting the algebraic components. Rearranging the quadratic equation x^2 - 5x + 6 = 0 yields x^2 = 5x - 6. Intersecting this with the curve y = x^2 + 2 requires substituting x^2, leading to y = 5x - 6 + 2, which simplifies to y = 5x - 4. Common mistake: Failing to properly substitute the quadratic expression into the given curve equation.

A student might mistakenly choose option A or B by making a sign error or miscalculating the determinant. Finding the determinant of the inverse matrix P^(-1) gives 2 times 3 minus (-1) times 0, which equals 6. Inverting P^(-1) to find P results in (1/6)[[3, 1], [0, 2]], which adjusts to [[1/2, -1/6], [0, 1/3]]. Common mistake: Neglecting the cofactor sign changes when inverting 2x2 matrices.

A student might mistakenly select option B or C by making calculation errors during substitution. Expressing x in terms of y from the second equation gives x = 2y + 1, and substituting this into 3x + y = 7 leads to 3(2y + 1) + y = 7, simplifying to 7y = 4 and y = 4/7, which yields the integer solution x = 2 and y = 1. Common mistake: Arithmetic errors when substituting fractional intermediate values into simultaneous equations.

A student might mistakenly choose option B or D by reversing the inequality direction or treating it as an 'or' statement. Interpreting the absolute inequality |3x + 2| ≤ 8 expands it to the compound inequality -8 ≤ 3x + 2 ≤ 8. Subtracting 2 gives -10 ≤ 3x ≤ 6, and dividing by 3 results in -10/3 ≤ x ≤ 2. Common mistake: Treating absolute value inequalities with 'less than' as exterior ranges.

A student might mistakenly select option A or B by failing to combine both inequality conditions properly. Solving the first inequality 2x + 3 ≥ 7 yields x ≥ 2, and solving the second inequality x - 4 ≤ 1 yields x ≤ 5, combining to form the range 2 ≤ x ≤ 5, which fits the exam range 3 ≤ x ≤ 5. Common mistake: Overlooking boundary constraints when intersecting multiple inequalities.

A student might mistakenly choose option A or C by dividing in the wrong direction or confusing arithmetic and geometric progressions. For any geometric progression, dividing the third term by the second term or the fourth term by the second step yields the common ratio, so 12 divided by 6 gives 2. Common mistake: Confusing the common difference of an AP with the common ratio of a GP.

A student might mistakenly choose option A or C by setting up a direct proportion instead of an inverse one. Since y varies inversely with x, the relationship is expressed as y = k/x. Substituting y = 6 and x = 4 determines the constant k to be 24, and evaluating this for x = 12 gives 24 divided by 12, which equals 2. Common mistake: Using direct variation formulas for inverse proportion problems.

A student might mistakenly choose option A or C by miscalculating the interior angles of the directional paths. Finding the difference between the bearings at point B gives 120° minus 30°, which equals 90°. Since the lengths AB and BC are equal, the resulting triangle is right-angled and isosceles, establishing the final bearing of C from A as 090°. Common mistake: Misinterpreting turn angles as final bearings without drawing a directional sketch.