Mathematics 2024 Jamb Examination

To find the perimeter, we calculate the length of each side using the distance formula: \(d = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}\).

Let the vertices be A(3, 2), B(4, 5), and C(6, 2).
Length AB = \(\sqrt{(4-3)^{2} + (5-2)^{2}} = \sqrt{1^{2} + 3^{2}} = \sqrt{1 + 9} = \sqrt{10}\)
Length BC = \(\sqrt{(6-4)^{2} + (2-5)^{2}} = \sqrt{2^{2} + (-3)^{2}} = \sqrt{4 + 9} = \sqrt{13}\)
Length AC = \(\sqrt{(6-3)^{2} + (2-2)^{2}} = \sqrt{3^{2} + 0^{2}} = \sqrt{9} = 3\)

Perimeter = AB + BC + AC = \(\sqrt{10} + \sqrt{13} + 3\).

This is a Geometric Progression (GP) where the first term \(a = -10\).
The common ratio \(r = \frac{50}{-10} = -5\).
The formula for the \(n^{th}\) term is \(T_{n} = ar^{n-1}\).

For the 7\(^{th}\) term (\(n=7\)):\[ T_{7} = -10 \times (-5)^{7-1} \]\[ T_{7} = -10 \times (-5)^{6} \]\[ T_{7} = -10 \times 15625 \]\[ T_{7} = -156250 \]

The mode is the value that appears most frequently in a data set. Let's count the frequency of each score:
1: 1 time
2: 2 times
3: 3 times
4: 1 time
5: 2 times
7: 1 time
8: 1 time

The score 3 occurs the most (3 times), so the mode is 3.

The midpoint \(M(x, y)\) between two points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is given by:\[ M = \left( \frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2} \right) \]We are given the midpoint \(x\)-coordinate is 9, \(x_{1} = x\), and \(x_{2} = 10\).\[ 9 = \frac{x + 10}{2} \]\[ 18 = x + 10 \]\[ x = 18 - 10 = 8 \]

Express all terms in base 4:
\(16 = 4^{2}\), \(4 = 4^{1}\), and \(64 = 4^{3}\).\[ \frac{(4^{2})^{2 + x}}{4^{1}} = (4^{3})^{x} \]\[ \frac{4^{4 + 2x}}{4^{1}} = 4^{3x} \]\[ 4^{4 + 2x - 1} = 4^{3x} \]\[ 4^{3 + 2x} = 4^{3x} \]Since the bases are the same, equate the powers:\[ 3 + 2x = 3x \]\[ 3 = 3x - 2x \]\[ x = 3 \]

Identify the elements of each set:
P (primes 0-10) = {2, 3, 5, 7}
Q (odds 0-10) = {1, 3, 5, 7, 9}

Find elements in P not in Q: {2}
Find elements in Q not in P: {1, 9}
The union of these two results is {1, 2, 9}. This represents the symmetric difference (P Δ Q).

In geometric construction, the first arc usually defines a 60\(^{\circ}\) angle from the baseline. Point T is located where an arc from the 60\(^{\circ}\) mark and an arc from the 0\(^{\circ}\) mark intersect. This process is the standard method for bisecting an angle. Bisecting 60\(^{\circ}\) results in 30\(^{\circ}\).

Convert the integer part (10111):\[ (1 \times 2^{4}) + (0 \times 2^{3}) + (1 \times 2^{2}) + (1 \times 2^{1}) + (1 \times 2^{0}) \]\[ = 16 + 0 + 4 + 2 + 1 = 23 \]Convert the fractional part (.11):\[ (1 \times 2^{-1}) + (1 \times 2^{-2}) \]\[ = \frac{1}{2} + \frac{1}{4} = \frac{2+1}{4} = \frac{3}{4} \]Combining both parts gives 23\(\frac{3}{4}\).

\[ \begin{array}{r} \ \ 20045_{7}\\ -14256_{7}\\ \hline \end{array} \] Since \(5 < 6\), borrow \(1\) from the next column (which equals \(7\) in base \(7\)): \[ (7+5)-6 = 6 \] The \(4\) becomes \(3\). Next column: \[ 3 < 5 \] Borrow from the next column. Since the next digit is \(0\), borrowing proceeds from the first non–zero digit \(2\): \[ 2 \rightarrow 1,\quad 0 \rightarrow 7,\quad 0 \rightarrow 6 \] Now compute: \[ (7+3)-5 = 5 \] Next column: \[ 6-2 = 4 \] Next column: \[ 6-4 = 2 \] Final column: \[ 1-1 = 0 \] \[ \therefore\quad 20045_{7} - 14256_{7} = 2456_{7}. \]

To find the overall average, calculate the total age and divide by the total number of teachers.
Total age of females = \(4 \times 40 = 160\)
Total age of males = \(8 \times 25 = 200\)
Total age of all teachers = \(160 + 200 = 360\)
Total number of teachers = \(4 + 8 = 12\)

Average age = \(\frac{360}{12} = 30\).