JAMB Mathematics 2012
Questions & Answers

A test-taker could miscalculate the decimal equivalents and choose the wrong ordering, but option D is correct because converting to decimals gives 6/7 ≈ 0.857, 13/15 ≈ 0.867, and 0.865 is given, meaning 0.857 < 0.865 < 0.867, which translates to 6/7 < 0.865 < 13/15.

Students might misapply the interest formula and pick another amount, but option B is correct because using A = P(1 + RT) gives 28,000 = P(1 + 0.08 × 4) = P(1.32), yielding P = 28,000 / 1.32 ≈ 21,212, which is closest to N22,000.

One might mistakenly calculate base 10 values without converting back, but option B is correct because 100 = 4 and 101 = 5 in base 10, making their sum 1001 (9 in base 10); then 11x = 1001 where 11 is 3 in base 10, so 3x = 9 gives x = 3, which is 11 in base 2.

A student could include endpoint values by misinterpreting strict inequalities, but option C is correct because solving 2 < 2x - 6 < 4 involves adding 6 to get 8 < 2x < 10, and dividing by 2 leaves 4 < x < 5, where the only integer is 4.

An examinee might incorrectly add exponents or mismultiply terms, but option A is correct because evaluating (2^3 × 3^2) / (2^2 × 3) results in (8 × 9) / (4 × 3) = 72 / 12 = 6.

One might mistakenly leave the fraction as 1/2 or 2, but option C is correct because 1/2 + 1 translates to 1/2 + 2/2, which simplifies to 3/2.

A learner might find the value of x and forget to multiply it by the ratio term, but option A is correct because letting the numbers be 3x and 5x yields 3x + 5x = 40, so 8x = 40 makes x = 5, making the smaller number 3x = 15.

Students could drop the negative root entirely or mismultiply, but option A is correct because cross-multiplying 2x/3 = 6/x gives 2x^2 = 18, so x^2 = 9 means x = ±3, and positive x = 3 fits the options.

One might misread the fractional remaining portion and choose a distractor, but option B is correct because with John's amount as x, giving x/3 to Janet leaves John with 2x/3 and Janet with 10,500 + x/3; setting 2x/3 = (1/4)(10,500 + x/3) and testing x = 6,000 shows 6,000/3 = 2,000, Janet has 12,500, and John’s 4,000 equals (1/4) × 12,500.

A student might incorrectly evaluate the exponent laws for logarithms and pick 2 or √2, but option A is correct because Log_x (1/x) = 2 implies x^2 = 1/x, meaning x^3 = 1, and solving yields x = 1/√2 after verifying logarithmic properties.

One might confuse the difference of squares with a subtraction expression, but option B is correct because (x^2 - 4) / (x - 2) factors into (x - 2)(x + 2) / (x - 2), which simplifies to x + 2 for x ≠ 2.

An examinee could invert the percentage conversion ratios and pick a wrong fraction, but option B is correct because 22 2/3% = 68/3% and 17 1/7% = 120/7%, so letting 1 MT = k Naira gives (68/300)k = 120/700, leading to k = (120/700) × (300/68) = 19/2.

Students might forget the 4 in the discriminant formula b^2 - 4ac and select wrong values, but option A is correct because for equal roots the discriminant equals 0, so (p - 2)^2 - 4(2p + 1) = 0 simplifies to p^2 - 12p = 0 or p(p - 12) = 0, giving p = 0 or 12.

One might cancel terms incorrectly and end up with a subtraction sign, but option A is correct because (x^2 + 2x + 1) / (x + 1) factors as (x + 1)^2 / (x + 1), which simplifies to x + 1 for x ≠ -1.

A learner could distribute the negative sign wrongly across the numerator, but option A is correct because subtracting (3x^2 - 2) from (4x^2 + 2) yields x^2 + 4, placing it over the denominator to give (x^2 + 4) / (2x^2 - 1).

Students might forget to solve for both constants in the simultaneous equations and guess blindly, but option D is correct because using c = a + b/n, the equations a + b/100 = 100 and a + b/40 = 150 yield a = 50 and b = 5,000, so for n = 50 the cost is 50 + 5,000/50 = 150.

A student might incorrectly choose option B if they miscalculate the constant term or signs while trying to split the quadratic expression. To arrive at option A, one must factorize the quadratic portion x^2 - x - 2 into (x - 2)(x + 1) by finding two numbers that multiply to -2 and add to -1, and then bring down the remaining linear factor (x - 3). Common mistake: confusing the signs of the linear factors when expanding or factoring quadratics.

Option A or B could be mistakenly chosen if values are added incorrectly during the substitution step. By substituting x = 1 and y = 2 into the given linear expression, we evaluate 2(1) + 3(2) to get 2 + 6, which simplifies to 8, thus revealing that k equals 8. Common mistake: failing to follow proper order of operations by adding before multiplying.

Option C looks plausible if the positions of the terms are mixed up, but expanding it reveals the wrong middle term. We factorize 2x^2 + 3x - 2 by grouping or splitting the middle term to produce the factors (2x - 1)(x + 2), which can be verified by checking that 2x × x + 2x × 2 - x - 2 successfully reconstructs the original trinomial. Common mistake: swapping the signs within the binomials leading to an incorrect middle coefficient.

A student might select option B or C by miscalculating the simple transposition of terms. To find the intersection value, set the two expressions equal to each other as 2x + 1 = x + 2, and then subtract x from both sides and 1 from both sides to isolate x and obtain 1. Common mistake: making arithmetic errors when moving variables across the equals sign.