Mathematics 2023 Jamb Examination

The area of a trapezium is calculated using the formula:\[ \text{Area} = \frac{1}{2}(a + b)h \]Given that the ratio of the parallel sides is 2 : 3, we can represent the lengths of the sides as \(2x\) and \(3x\). The perpendicular distance (height) is 16 cm, and the area is 200 cm\(^{2}\).

Substituting these values into the formula:\[ 200 = \frac{1}{2}(2x + 3x) \times 16 \]\[ 200 = 8(5x) \]\[ 200 = 40x \]\[ x = \frac{200}{40} = 5 \]
Now, find the lengths of the parallel sides:
First side = \(2x = 2(5) = 10\text{ cm}\)
Second side = \(3x = 3(5) = 15\text{ cm}\)

Therefore, the lengths of the parallel sides are 10 cm and 15 cm.

Opposite angles in a cyclic quadrilateral are supplementary.\[ \angle PQR + \angle PSR = 180^\circ\] \[ (5x - y + 10) + (-2x + 3y + 145) = 180\] \[ 5x - y + 10 - 2x + 3y + 145 = 180\] \[ 3x + 2y + 155 = 180\] \[ 3x + 2y = 25 \qquad (i)\] Also,\[ \angle QPS + \angle QRS = 180^\circ\] \[ (-4x - 7y + 150) + (2x + 8y + 105) = 180\] \[ -4x - 7y + 150 + 2x + 8y + 105 = 180\] \[ -2x + y + 255 = 180\] \[ -2x + y = -75 \qquad (ii)\] From equation (ii):\[ y = -75 + 2x \qquad (iii)\] Substitute \(y = -75 + 2x\) into equation (i):\[ 3x + 2(-75 + 2x) = 25\] \[ 3x - 150 + 4x = 25\] \[ 7x = 175\] \[ x = \frac{175}{7} = 25\] Substituting into equation (iii):\[ y = -75 + 2(25)\] \[ y = -75 + 50\] \[ y = -25\] \[ x + y = 25 + (-25) = 0\]

Let the distance between the two airports be \(d\).
Time taken to fly to the airport: \(t_{1} = \frac{d}{120}\)
Time taken to return: \(t_{2} = \frac{d}{150}\)

The total time is given as 3 hours:\[ \frac{d}{120} + \frac{d}{150} = 3 \]To solve for \(d\), find a common denominator for 120 and 150, which is 600:\[ \frac{5d + 4d}{600} = 3 \]\[ \frac{9d}{600} = 3 \]\[ 9d = 1800 \]\[ d = \frac{1800}{9} = 200\text{ km} \]
The distance between the two airports is 200 km.

In an isosceles right-angled triangle, the two legs are equal in length. Let the length of each leg be \(x\). By the Pythagorean theorem, the hypotenuse (the longer side) is \(\sqrt{x^{2} + x^{2}} = x\sqrt{2}\).

The perimeter is the sum of all sides:\[ \text{Perimeter} = x + x + x\sqrt{2} = 2 \]\[ 2x + x\sqrt{2} = 2 \]\[ x(2 + \sqrt{2}) = 2 \]\[ x = \frac{2}{2 + \sqrt{2}} \]
The longer side is the hypotenuse, \(x\sqrt{2}\):\[ \text{Hypotenuse} = \left( \frac{2}{2 + \sqrt{2}} \right) \times \sqrt{2} = \frac{2\sqrt{2}}{2 + \sqrt{2}} \]To simplify, rationalise the denominator by multiplying the top and bottom by \(2 - \sqrt{2}\):\[ \frac{2\sqrt{2}(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{4\sqrt{2} - 2(2)}{4 - 2} = \frac{4\sqrt{2} - 4}{2} = 2\sqrt{2} - 2 \]This can be rewritten as \(-2 + 2\sqrt{2}\) metres.

Both ports are on the same latitude (86\(^{\circ}\)N). The distance between two points on the same latitude is calculated along the small circle arc:\[ \text{Distance} = \frac{\theta}{360} \times 2\pi R \cos \alpha \]Where:
\(\theta\) = Longitude difference = \(64^{\circ} - 56^{\circ} = 8^{\circ}\)
\(\alpha\) = Latitude = 86\(^{\circ}\)
\(R\) = Radius of the Earth = 6370 km

Calculation:\[ \text{Distance} = \frac{8}{360} \times 2 \times 3.142 \times 6370 \times \cos(86^{\circ}) \]\[ \text{Distance} = \frac{1}{45} \times 6.284 \times 6370 \times 0.069756 \]\[ \text{Distance} \approx 0.02222 \times 40029.08 \times 0.069756 \]\[ \text{Distance} \approx 62.05\text{ km} \]Rounding to the nearest km, the distance is 62 km.

From the scale of the graph:

On the \(y\)-axis, four equal boxes represent the interval from \(0\) to \(1\). Hence the value represented by each box is\[ \frac{1-0}{4} = \frac{1}{4} = 0.25 \text{ units}\] On the \(x\)-axis, six equal boxes cover the interval from \(0^\circ\) to \(90^\circ\). Therefore each box corresponds to\[ \frac{90^\circ-0}{6} = \frac{90^\circ}{6} = 15^\circ\] For the first angle, counting \(2.5\) boxes to the right of \(180^\circ\):\[ \theta_1 = 180^\circ + (2.5 \times 15^\circ)\] \[ \theta_1 = 180^\circ + 37.5^\circ\] \[ \theta_1 = 217.5^\circ \approx 218^\circ\] For the second angle, \(3.5\) boxes are measured to the right of \(270^\circ\):\[ \theta_2 = 270^\circ + (3.5 \times 15^\circ)\] \[ \theta_2 = 270^\circ + 52.5^\circ\] \[ \theta_2 = 322.5^\circ \approx 323^\circ\] \[ \therefore \theta = 218^\circ,\; 323^\circ\]

When a perpendicular line is drawn from the centre of a circle to a chord, it bisects the chord. This forms a right-angled triangle where:
The hypotenuse = Radius (r) = 13 cm
The perpendicular distance = 12 cm
The base = Half the chord (x)

Using the Pythagorean theorem:\[ r^{2} = 12^{2} + x^{2} \]\[ 13^{2} = 12^{2} + x^{2} \]\[ 169 = 144 + x^{2} \]\[ x^{2} = 169 - 144 = 25 \]\[ x = \sqrt{25} = 5\text{ cm} \]
Since \(x\) is only half of the chord, the full length of the chord is:\[ \text{Chord length} = 2x = 2(5) = 10\text{ cm} \]

[Image explaining the rules for significant figures including zeros between non-zero digits and trailing zeros]
To determine the number of significant figures, we apply the following rules:
1. All non-zero digits are significant (5, 2, 7).
2. Zeros between non-zero digits are significant (the zeros between 5 and 2, and between 2 and 7).
3. Trailing zeros in a whole number without a decimal point are generally not considered significant (they are place-holders).

In the number 5,020,700:
- The digits '5', '0', '2', '0', '7' are significant because the zeros are trapped between non-zero digits.
- The final two zeros ('00') are trailing zeros at the end of a whole number and are not significant.

Thus, there are 5 significant figures.

[Image illustrating simultaneous equations solving by substitution or elimination]
Let \(v\) be the number of VIP tickets and \(r\) be the number of regular tickets.
From the total tickets sold:\[ v + r = 200 \quad \text{--- (Equation 1)} \]From the total revenue:\[ 1200v + 700r = 180000 \quad \text{--- (Equation 2)} \]
Multiply Equation 1 by 700:\[ 700v + 700r = 140000 \quad \text{--- (Equation 3)} \]Subtract Equation 3 from Equation 2:\[ (1200v - 700v) + (700r - 700r) = 180000 - 140000 \]\[ 500v = 40000 \]\[ v = \frac{40000}{500} = 80 \]
Now, find \(r\) using Equation 1:\[ 80 + r = 200 \]\[ r = 200 - 80 = 120 \]
Therefore, 80 VIP tickets and 120 regular tickets were sold.

The amount of work done is proportional to the number of men (M), the number of hours worked per day (H), and the number of days (D). For the same piece of work, we use the relation:\[ M_{1} \times H_{1} \times D_{1} = M_{2} \times H_{2} \times D_{2} \]
Given:
\(M_{1} = 12\), \(H_{1} = 8\), \(D_{1} = 4\)
\(M_{2} = 4\), \(H_{2} = 16\), \(D_{2} = ?\)

Calculation:\[ 12 \times 8 \times 4 = 4 \times 16 \times D_{2} \]\[ 384 = 64 \times D_{2} \]\[ D_{2} = \frac{384}{64} = 6 \]
It will take 6 days for the 4 men to complete the work.