Mathematics 2012 Jamb Examination

To convert from base 6 to base 3, we first convert the number to base 10 (denary) and then to base 3.
Step 1: Convert \(112_{6}\) to base 10:\[(1 \times 6^{2}) + (1 \times 6^{1}) + (2 \times 6^{0})\]\[= 36 + 6 + 2 = 44_{10}\]Step 2: Convert \(44_{10}\) to base 3 using successive division:\[44 \div 3 = 14 \text{ remainder } 2\]\[14 \div 3 = 4 \text{ remainder } 2\]\[4 \div 3 = 1 \text{ remainder } 1\]\[1 \div 3 = 0 \text{ remainder } 1\]Reading the remainders upwards, we get \(1122_{3}\).

Following the BODMAS/BIDMAS rule, we perform the multiplication and division from left to right. First, multiply the first two fractions:\[\frac{2}{3} \times \frac{1}{2} = \frac{2}{6} = \frac{1}{3}\]Next, divide the result by the third fraction. Recall that dividing by a fraction is the same as multiplying by its reciprocal:\[\frac{1}{3} \div \frac{4}{5} = \frac{1}{3} \times \frac{5}{4} = \frac{5}{12}\](Note: Based on the formatting of the original prompt which appears to be \( \frac{2 \times 1}{3 \times 2} \) over \( \frac{4}{5} \), the result is \( \frac{1/3}{4/5} = \frac{5}{12} \). Since 5/12 is not an option, let's re-evaluate the source formatting: if it was \( \frac{2}{3} \times \frac{1}{2} \) as a numerator and something else, however, calculating \( \frac{1}{3} \div \frac{4}{5} \) gives \( \frac{5}{12} \). Re-checking common JAMB 2012 transcripts, if the question was \( \frac{2}{3} + \frac{1}{2} \dots \) the options might align differently, but mathematically for the text provided, \( \frac{5}{12} \) is the result.)

To evaluate the fraction, we perform the division: 21/9 = 2.333333
The result in 3 significant figures 2.33.

First, calculate the amount spent on education:\[15\% \text{ of } 3500 = 0.15 \times 3500 = 525\]Total expenditure is the sum of education and food costs:\[525 + 1950 = 2475\]The amount remaining is the total income minus total expenditure:\[3500 - 2475 = 1025\]The man has ₦1,025 left.

To solve this exponential equation, we express all terms as powers of base 3:\[(3^{3})^{x+2} \div (3^{2})^{x+1} = 3^{2x}\]Apply the power of a power rule \( (a^m)^n = a^{mn} \):\[3^{3x+6} \div 3^{2x+2} = 3^{2x}\]Apply the division rule for indices \( a^m \div a^n = a^{m-n} \):\[3^{(3x+6) - (2x+2)} = 3^{2x}\]\[3^{x+4} = 3^{2x}\]Since the bases are equal, we equate the powers:\[x + 4 = 2x\]\[4 = 2x - x\]\[x = 4\]

We use the definition of logarithms: if \( \log_{b} a = c \), then \( b^c = a \).
Applying this to the equation:\[3^{-8} = x^2\]To solve for \(x\), we take the square root of both sides:\[x = \sqrt{3^{-8}}\]\[x = (3^{-8})^{1/2} = 3^{-4}\]\[x = \frac{1}{3^4} = \frac{1}{81}\]

This expression is in the form of a difference of two squares: \( a^2 - b^2 = (a-b)(a+b) \).
Let \( a = \sqrt{6} + 2 \) and \( b = \sqrt{6} - 2 \).
First term \( (a+b) \):\[(\sqrt{6} + 2) + (\sqrt{6} - 2) = 2\sqrt{6}\]Second term \( (a-b) \):\[(\sqrt{6} + 2) - (\sqrt{6} - 2) = \sqrt{6} + 2 - \sqrt{6} + 2 = 4\]Multiplying them together:\[(4) \times (2\sqrt{6}) = 8\sqrt{6}\]

First, define set \(P\). The factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30. The prime numbers among these are 2, 3, and 5. So, \(P = {2, 3, 5}\).
Next, define set \(Q\). The factors of 18 are 1, 2, 3, 6, 9, 18. The factors less than 10 are 1, 2, 3, 6, 9. So, \(Q = {1, 2, 3, 6, 9}\).
The intersection \(P \cap Q\) is the set of elements common to both \(P\) and \(Q\):\[P \cap Q = {2, 3}\]

We can solve this using the Principle of Inclusion-Exclusion or a Venn diagram.
Let \(n(F)\) be the number of football players (22) and \(n(V)\) be the number of volleyball players (26). The number playing both is \(n(F \cap V) = 3\).
The number of students playing at least one game is:\[n(F \cup V) = n(F) + n(V) - n(F \cap V)\]\[n(F \cup V) = 22 + 26 - 3 = 45\]The number of students who play neither is the total number of students minus those who play at least one game:\[46 - 45 = 1\]

To make \(n\) the subject, we first cross-multiply to remove the fraction:\[w(1 - cn) = v(2 + cn)\]Expand both sides:\[w - wcn = 2v + vcn\]Move all terms involving \(n\) to one side and constants to the other:\[w - 2v = vcn + wcn\]Factorise out \(cn\) on the right side:\[w - 2v = cn(v + w)\]Isolate \(n\) by dividing both sides by \(c(v + w)\):\[n = \frac{w - 2v}{c(v + w)}\]This can be written as:\[n = \frac{1}{c} \left( \frac{w - 2v}{v + w} \right)\]