Mathematics 2020 Jamb Examination

If the chord ST is equal to the radius \(r\), then the triangle formed by the center of the circle (O) and the points S and T is an equilateral triangle. In an equilateral triangle, all internal angles are 60^o. Therefore, the angle subtended by the arc at the center, \(\theta\), is 60^o.
The formula for arc length is:\[L = \frac{\theta}{360} \times 2\pi r\]Substituting \(\theta = 60^o\):\[L = \frac{60}{360} \times 2\pi r = \frac{1}{6} \times 2\pi r = \frac{2\pi r}{6} = \frac{\pi r}{3}\]

When a sector is folded into a cone, the arc length of the sector becomes the circumference of the base of the cone (\(2\pi r_{base}\)), and the radius of the sector becomes the slant height of the cone (\(l\)).
Arc length of sector = \(\frac{\theta}{360} \times 2\pi R_{sector}\)\[2\pi r_{base} = \frac{300}{360} \times 2\pi (7.2)\]Dividing both sides by \(2\pi\):\[r_{base} = \frac{5}{6} \times 7.2\]\[r_{base} = 5 \times 1.2 = 6\text{cm}\]

The capacity of the tank refers to its volume. The volume \(V\) of a cylinder is given by:\[V = \pi r^2 h\]Given the diameter is 14m, the radius \(r = \frac{14}{2} = 7\)m. The volume \(V = 3080\)m³ and we use \(\pi \approx \frac{22}{7}\).\[3080 = \frac{22}{7} \times 7^2 \times h\]\[3080 = 22 \times 7 \times h\]\[3080 = 154h\]\[h = \frac{3080}{154} = 20\text{m}\]

First, find the total number of acres for all crops:\[\text{Total} = 2 + 5 + 3 + 11 + 9 = 30 \text{ acres}\]The angle for a specific sector is calculated as:\[\text{Angle} = \frac{\text{Value of crop}}{\text{Total value}} \times 360^o\]For cassava, the value is 3:\[\text{Angle of Cassava} = \frac{3}{30} \times 360^o = \frac{1}{10} \times 360^o = 36^o\]

In a geometric progression, the common ratio \(r\) is found by dividing a term by its preceding term. Therefore:\[r = \frac{n}{n - 2} = \frac{n + 3}{n}\]Cross-multiplying gives:\[n^2 = (n - 2)(n + 3)\]\[n^2 = n^2 + 3n - 2n - 6\]\[n^2 = n^2 + n - 6\]Subtract \(n^2\) from both sides:\[0 = n - 6 \implies n = 6\]Now substitute \(n = 6\) back into the common ratio expression:\[r = \frac{n}{n - 2} = \frac{6}{6 - 2} = \frac{6}{4} = \frac{3}{2}\]

Let \(n(M)\) be the students offering Mathematics, \(n(P)\) those offering Physics, and \(x\) the students offering both.
The total number of students who offer at least one subject is:\[40 - 4 = 36\]Using the set theory formula:\[n(M \cup P) = n(M) + n(P) - n(M \cap P)\]\[36 = 32 + 24 - x\]\[36 = 56 - x\]\[x = 56 - 36 = 20\]

The sum of the interior angles of a polygon with \(n\) sides is given by \((n - 2) \times 180^o\). For a pentagon, \(n = 5\):\[\text{Sum} = (5 - 2) \times 180^o = 3 \times 180^o = 540^o\]We are given the sum as \(6x + 6y\), so:\[6x + 6y = 540\]Divide the entire equation by 6 to simplify:\[x + y = 90\]To find y in terms of x:\[y = 90 - x\]

Let the number of students be \(n\). The sum of the ages of the students is \(15n\).
When the teacher's age (45) is added, the new total number of people is \(n + 1\) and the new sum of ages is \(15n + 45\).
The new mean is given as 18:\[\frac{15n + 45}{n + 1} = 18\]Cross-multiply:\[15n + 45 = 18(n + 1)\]\[15n + 45 = 18n + 18\]Collect like terms:\[45 - 18 = 18n - 15n\]\[27 = 3n\]\[n = \frac{27}{3} = 9\]

The distance walked up the hill (500m) represents the hypotenuse of a right-angled triangle. The vertical height (h) is the side opposite to the angle of 30^o.
Using the sine ratio:\[\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}\]\[\sin 30^o = \frac{h}{500}\]Since \(\sin 30^o = 0.5\):\[0.5 = \frac{h}{500}\]\[h = 500 \times 0.5 = 250\text{m}\]

To find the value of x, we find the determinant of the matrix and set it to zero:\[x \begin{vmatrix} x & 1 \\ 1 & x \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 0 & x \end{vmatrix} + 0 \begin{vmatrix} 1 & x \\ 0 & 1 \end{vmatrix} = 0\]\[x(x^2 - 1) - 1(x - 0) + 0 = 0\]\[x^3 - x - x = 0\]\[x^3 - 2x = 0\]Factorise the equation:\[x(x^2 - 2) = 0\]This gives \(x = 0\) or \(x^2 - 2 = 0\).
Since we are looking for a non-zero positive value:\[x^2 = 2 \implies x = \sqrt{2}\]