JAMB Mathematics 2012 Past Questions & Answers

To convert 112 in base 6 to base 3, you first need to convert it to base 10. So, 112₆ = (1 x 6²) + (1 x 6¹) + (2 x 6⁰) = 36 + 6 + 2 = 44₁₀. Next, convert 44 in base 10 to base 3 by repeatedly dividing by 3 and taking the remainders. Dividing 44 by 3 gives 14 remainder 2, dividing 14 by 3 gives 4 remainder 2, dividing 4 by 3 gives 1 remainder 1, and dividing 1 by 3 gives 0 remainder 1. Reading the remainders from the bottom up gives the final answer, which is 1122₃.

Common mistake: Students often make the mistake of reading the remainders from top to bottom after the division, which would incorrectly lead them to choose 2211.

First, convert all the mixed numbers into improper fractions: $2(2/3)$ becomes $8/3$, $1(1/2)$ becomes $3/2$, and $4(4/5)$ becomes $24/5$. Next, solve the expression in the bracket: $(8/3)$ × $(3/2)$ = $24/6$, which simplifies to 4. Finally, divide this result by the denominator: 4 ÷ $(24/5)$ is the same as 4 × $(5/24)$, which gives $20/24$. Simplifying this fraction by dividing both the numerator and denominator by 4 gives the final answer, $5/6$.

Common mistake: Students may incorrectly handle the final division, perhaps by dividing the whole number 4 by the whole number part of the denominator (also 4) and ignoring the fractional part, leading them to choose $4/5$.

First, divide 21 by 9 to get the decimal 2.3333... To approximate this to 3 significant figures, we count the first three important digits, which are 2, 3, and 3. We then look at the fourth digit, which is also 3, to decide if we should round up. Since 3 is less than 5, we do not round up the last digit, so the number remains 2.33.

Common mistake: A common mistake is to incorrectly round to two significant figures (2.3) and choose option A (2.30), forgetting that the third figure must be included before rounding.

First, calculate the amount spent on education, which is 15% of his N3,500 salary. This is $(15/100)$ * 3500 = N525. Next, add the additional amount spent on food to get the total expenses: N525 + N1,950 = N2,475. Finally, subtract the total expenses from his salary to find how much he has left: N3,500 - N2,475 = N1,025.

Common mistake: A common mistake is to only calculate the 15% spent on education (N525) and select that as the final answer, forgetting to account for the other expenses.

To solve this problem, first express all numbers in base 3, so 27 is $3^3$ and 9 is $3^2$. The equation becomes $(3^3)^(x+2)$ ÷ $(3^2)^(x+1)$ = $3^(2x)$. Applying the laws of indices, this simplifies to $3^(3x+6)$ ÷ $3^(2x+2)$ = $3^(2x)$, and then $3^((3x+6) - (2x+2))$ = $3^(2x)$. This gives $3^(x+4)$ = $3^(2x)$, and by equating the powers, we get x + 4 = 2x, which solves to x = 4.

Common mistake: A common mistake is incorrectly expanding the bracket 2(x+1) as 2x+1 instead of 2x+2, which leads to an incorrect final equation of x+5 = 2x and the wrong answer x=5.

To solve this, first apply the power rule of logarithms, which brings the power down, so log₃ x² = -8 becomes 2 log₃ x = -8. Next, divide both sides by 2 to isolate the logarithm, giving you log₃ x = -4. To find x, convert this logarithmic equation to its index form, which is x = 3⁻⁴. A negative power means 'one over', so x = $1/3⁴$, which simplifies to $1/81$.

Common mistake: A common mistake is to correctly get to log₃ x = -4 but then incorrectly calculate x as 3⁻² $(which is 1/9)$, perhaps by confusing the original x² and taking a square root of the power -4.

This problem is best solved using the difference of two squares formula: a² − b² = (a + b)(a − b). Here, let a = $(√6 + 2)$ and b = $(√6 − 2)$. First, calculate (a + b) = $(√6 + 2)$ + $(√6 − 2)$ = $2√6$. Then, calculate (a − b) = $(√6 + 2)$ − $(√6 − 2)$ = $√6$ + 2 − $√6$ + 2 = 4. Finally, multiply these two results: $(2√6)$ × 4 = $8√6$.

Common mistake: A common mistake is to only calculate the middle term of one of the expansions, 2 × $√6$ × 2 = $4√6$, and incorrectly select it as the answer instead of completing the full simplification.

First, we list the elements of each set. Set P contains all the prime factors of 30, so P = 2, 3, 5. Set Q contains all factors of 18 that are less than 10, so Q = 1, 2, 3, 6, 9. The intersection P ∩ Q means we find the elements that are present in both set P and set Q. Comparing the two sets, the common elements are 2 and 3, therefore P ∩ Q = 2, 3.

Common mistake: Students may correctly identify the elements of set P as 2,3,5 and mistakenly choose it as the answer, forgetting to find the intersection with set Q.

To find the number of students who play at least one game, we add the number of football players and volleyball players, then subtract those who play both. So, the number of students playing either football or volleyball is 22 + 26 - 3 = 45. Since there are 46 students in the class altogether, the number of students who play neither game is the total number of students minus those who play at least one game. Therefore, the answer is 46 - 45 = 1.

Common mistake: A common mistake is to simply add the number of football and volleyball players $(22 + 26 = 48)$ without subtracting the 3 students who play both games, leading to an incorrect calculation.

To make 'n' the subject, first multiply both sides by the denominator (1−cn) to get w(1−cn) = v(2+cn). Expand the brackets on both sides, which gives w − wcn = 2v + vcn. Next, collect all terms with 'n' on one side and the rest on the other: w − 2v = vcn + wcn. Factor out 'n' to get w − 2v = n(vc + wc), then divide to find n = (w − 2v) / (vc + wc), which can be written as (1/c) * (w - 2v)/(v + w).

Common mistake: Students often make sign errors when rearranging terms, for example, incorrectly calculating 'vcn - wcn' instead of 'vcn + wcn' after moving terms across the equals sign, which leads to a wrong denominator.