Mathematics 2023 Jamb Examination

Let the height of the first building be \(h_1 = 24\text{ m}\). Let the distance between the two buildings be \(d\). From the top of the first building, the angle of depression to the bottom of the second building is 60°: \[\tan(60^\circ) = \frac{24}{d} \implies \sqrt{3} = \frac{24}{d} \implies d = \frac{24}{\sqrt{3}} = 8\sqrt{3}\text{ m}\] The angle of elevation to the top of the second building is 30°. Let the height of the second building above the 24 m level be \(h\): \[\tan(30^\circ) = \frac{h}{d} \implies \frac{1}{\sqrt{3}} = \frac{h}{8\sqrt{3}} \implies h = 8\text{ m}\] Total height of the second building = \(24 + 8 = 32\text{ m}\).

\(\dfrac {\sin A}{a} = \dfrac {\sin B}{b} = \dfrac {\sin C}{c}\)

\(\implies \dfrac {\sin 82^0}{43.2} = \dfrac {\sin 56^0}{AB}\)

\(\implies AB \times \sin 82^0 = 43.2 \times \sin 56^0\)

\(\therefore AB = \dfrac {43.2 \times \sin 56^0}{\sin 82^0}\) = 36.2cm (to 3 s.f)

First, find the Cost Price (CP). Selling Price (SP) = CP + Profit: \[115\% \text{ of } CP = 230 \implies CP = \frac{230}{1.15} = 200\] If the article is sold for ₦180, there is a loss of \(200 - 180 = 20\). \[\text{Loss } \% = \left( \frac{20}{200} \times 100 \right) = 10\%\]

Consider the right–angled triangle \( \triangle XOB \). Applying the Pythagoras theorem gives\[ 13^2 = 12^2 + h^2\] \[ 169 = 144 + h^2\] \[ h^2 = 169 - 144\] \[ h^2 = 25\] \[ h = \sqrt{25} = 5\text{ cm}\] Using the definition of the tangent ratio,\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}}\] \[ \tan \theta = \frac{12}{5} = 2.4\] \[ \theta = \tan^{-1}(2.4)\] \[ \theta = 67.38^\circ\] Since the central angle \( \angle AOB \) is twice \( \theta \),\[ \angle AOB = 2\theta = 2 \times 67.38^\circ = 134.76^\circ\] The length of the corresponding arc is obtained using the arc length formula\[ L = \frac{\theta}{360^\circ} \times 2\pi r\] \[ L = \frac{134.76}{360} \times 2 \times \frac{22}{7} \times 13\] \[ L = \frac{77082.72}{2520}\] \[ L \approx 30.6\text{ cm}\] \[ \therefore L = 30.6\text{ cm} \text{ (correct to 3 significant figures)}\]

Measurements corrected to the nearest meter have an error of \(\pm 0.5\text{ m}\). Range for 38 m: \([37.5, 38.5)\). Range for 52 m: \([51.5, 52.5)\). Minimum Area = \(37.5 \times 51.5 = 1931.25\text{ m}^2\). Maximum Area = \(38.5 \times 52.5 = 2021.25\text{ m}^2\). The range is \(1931.25\text{ m}^2 \le A < 2021.25\text{ m}^2\).

Let the total number of items be \(x\).
Brand A = \(\dfrac{1}{9}x\).
Remainder = \(x - \dfrac{1}{9}x = \dfrac{8}{9}x\).
Brand B = \(\dfrac{5}{8} \times \dfrac{8}{9}x = \dfrac{5}{9}x\).
Brand C = Remainder \(-\) Brand B = \(\dfrac{8}{9}x - \dfrac{5}{9}x = \dfrac{3}{9}x = \dfrac{1}{3}x\).
Given Brand C = 81, then \(\dfrac{1}{3}x = 81 \implies x = 243\).
Brand B = \(\dfrac{5}{9} \times 243 = 135\). Difference = \(135 - 81 = 54\).

The sum of the ratio parts is \(2 + 3 + 4 = 9\). Sides are: \(a = \frac{2}{9} \times 180 = 40\text{ cm}\), \(b = \frac{3}{9} \times 180 = 60\text{ cm}\), \(c = \frac{4}{9} \times 180 = 80\text{ cm}\). Semi-perimeter \(s = \frac{180}{2} = 90\text{ cm}\). Using Heron's formula: \[\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}\] \[\text{Area} = \sqrt{90(90-40)(90-60)(90-80)} = \sqrt{90 \times 50 \times 30 \times 10}\] \[\text{Area} = \sqrt{1,350,000} \approx 1161.895\text{ cm}^2\] To the nearest \(\text{cm}^2\), the area is 1162.

Using the formula \(A = P(1 + \frac{r}{100})^n\): \[A = 15700(1 + 0.08)^2 = 15700(1.08)^2\] \[A = 15700 \times 1.1664 = 18312.48\] \(\text{Compound Interest} = A - P\): \[CI = 18312.48 - 15700 = 2612.48\]

\[ \text{Volume of the cylinder} = \frac{\theta}{360} \times \pi r^2 h\] \[ \theta = 360^\circ - 90^\circ = 270^\circ\] \[ \text{Volume} = \frac{270}{360} \times \frac{22}{7} \times 14^2 \times 32 \]\[= \frac{37,255,680}{2520} = 14,784\ \text{cm}^3\]

\(U_3 = a + 2d = 6\) and \(U_5 = a + 4d = 12\).
Subtracting the first from the second: \(2d = 6 \implies d = 3\).
Substitute \(d = 3\) into \(a + 2(3) = 6 \implies a = 0\).
Sum of 12 terms \(S_{12} = \frac{n}{2}[2a + (n-1)d]\): \[S_{12} = \frac{12}{2}[2(0) + (11)3] = 6[33] = 198\]