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(a) Let the cost of the car to the dealer be x. Sales of car to a man with 15% profit : (115% of x = 1.15 x\) Sale of the car to the woman by the man at 5% loss : (95% of 1.15x = N120,175.00) \(0.95 \times 1.15x = N120,175.00\) \(1.0925x = N120,175.00\) \(x = \frac{120,175.00}{1.0925}\) \(x = N110,000\) The original cost of the car to the dealer = N110,000.00. (b) Circumference of a circle = \(2\pi r\) Distance covered by the wheel = 1.05 km Radius of the wheel = 18cm Circumference of the wheel = \(2 \times \frac{22}{7} \times 18\) = \(113.14 cm\) Note : 1.05 km = 105,000 cm Number of revolution = \(\frac{105,000}{113.14}\) = \(928.05 revs\) \(\approxeq 928 revs\)

\(\stackrel\frown{AOB} = \stackrel\frown{BOC} = \stackrel\frown{COD} = \stackrel\frown{DOE} = 60^\circ\) In \(\Delta AOB\), Area of sector = \(\frac{60}{360} \times \frac{22}{7} \times 7^{2}\) = \(25.67 cm^{2}\) Area of triangle AOB = \(\frac{1}{2} \times \7 \times 7 \times \sin 60\) = \(24.5 \times 0.866\) = \(21.22 cm^{2}\) Area of the 4 sectors = \(4 \times 25.67 = 102.68 cm^{2}\) Area of the 4 triangles = \(4 \times 21.22 = 84.88 cm^{2}\) Area of the shaded portion = \((102.68 - 84.88)cm^{2}\) = \(17.80 cm^{2}\) \(\approxeq 18 cm^{2}\)

(a) (b) \(x - 5 + 2x + 2x + 2x + 7 + 2x + 2x - 1 = 144\) \(11x + 1 = 144\) \(11x = 143\) \(x = \frac{143}{11} = 13cm\)

(a) Note that triangles AXC and BXD are similar. \(\therefore \frac{6}{|XD|} = \frac{3}{4}\) \(|XD| = \frac{6 \times 4}{3}\) = \(8 cm\) (b)(i) From \(\Delta ZYX\), \(\stackrel\frown{YZX} = 90^\circ - 22^\circ = 68^\circ\) \(\stackrel\frown{YXZ} = 312^\circ - 270^\circ = 42^\circ\) \(\stackrel\frown{ZYX} = 180^\circ - (68^\circ + 42^\circ)\) = \(180^\circ - 110^\circ\) = \(70^\circ\) \(a^{2} = b^{2} + c^{2} - 2bc \cos A\) \(|XZ|^{2} = 50^{2} + 32^{2} - 2(50)(32) \cos 70\) = \(2500 +1024 - 3200 \times 0.342\) = \(3524 - 1094.46\) \(|XZ|^{2} = 2429.54\) \(|XZ| = \sqrt{2429.54} = 49.29 km\) \(\approxeq 49.3 km\) (to 1 d.p) (ii) \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) \(\frac{50}{\sin \theta} = \frac{49.3}{\sin 70}\) \(\sin \theta = \frac{50 \times \sin 70}{49.3}\) \(\sin \theta = 0.9530\) \(\theta = \sin^{-1} (0.9530)\) = \(72.37^\circ\) \(\therefore\) The bearing of Z from X = \(312^\circ - 72.37^\circ\) = \(239.63^\circ \approxeq 239.6^\circ\)

To find the interest on \( N250.00 \) for \( 6 \) months at the same rate, we follow these steps: Calculate the Interest Rate Given: P = N150.00, T = 2\(\frac{1}{2}\) years and I = N4.50 Using the formula for simple interest: \(I = P \times r \times T\) Rearranging to find the interest rate \( r \): \(r = \frac{I}{P \times T}\) Substituting the values: \(r = \frac{4.50}{150 \times 2.5}\) Calculating \( P \times T \): \(P \times T = 150 \times 2.5 = 375\) Now, substituting back to find \( r \): \(r = \frac{4.50}{375} = 0.012\) Calculate the Interest for N250.00 \) for 6 Months We need to calculate the interest for: P = N250.00, T = 6 months = 0.5yrs, r = 0.012 Using the interest formula again: \(I = P \times r \times T\) Substituting the values: \(I = 250 \times 0.012 \times 0.5\) \(I = 250 \times 0.012 = 3\) Then: \(I = 3 \times 0.5 = 1.50\) The interest on \( N250.00 \) for \( 6 \) months at the same rate is N1.5

∠GOK = 85o (vertical opposite angle ∠s) ∠EOG + ∠GOK + ∠KOF = 180 (∠s on a straight line) ∠EOg + 85o + 52o = 180o ∠EOG + 137o = 180o ∠EOG = 43o

Sum of interior ∠s = 1800o ∴(n - 2) 180o = 1800o 180n -360o = 1800o 180n = 1800o + 360o 180n = 2160o n = 2160o/180o n = 12 sides Each exterior ∠ = 360o/n = 360o/12 = 30o

Area of a sector = \(\frac{\theta}{360}\times \pi r^2\) 55 = \(\frac{\theta}{360}\times \frac{22}{7}\times \frac{10\times 10}{1}\) \(\theta = \frac{360 \times 55 \times 7}{22 \times 10 \times 10}\) \(\theta = 63^{\circ}\)

The locus of points equidistant from lines PQ and ST lies on the angle bisector of the angles formed by the intersection of those lines.

Chord Length (L) = \(2 \times Radius \times (sin \frac{angle}{2})\) = \(2 \times 14 \times (sin \frac{60}{2})\) = \( 28 \times (sin 30)\) = \(28 \times \frac{1}{2}\) = \(14cm\)

sin θ = -1/2 = -0.5 θ = sin-1 (0.5) θ = 30o Since θ is negative θ = 180 + 30 = 210o θ = 360 - 30 = 330o ∴θ = 210o and 330o

y = sin (2x\(^3\) + 3x - 4) let u = 2x\(^3\) + 3x - 4 ∴du/dx = 6x\(^2\) y = sin u dy/du = cos u dy/dx = du/dx * dy/du ∴dy/dx = (6x\(^2\) + 3) cos u = (6x\(^2\) + 3)cos(2x\(^3\) + 3x - 4)

Increase in radius dr/dt = 0.5 Area of the circular disc = πr2 Increase in area: dA/dr = 2πr Rate of increase in area: dA/dt = dA/dr * dr/dt when r = 6cm = 2πr * 0.5 = 2 * π * 6 * 0.5 2 * π * 6 * 1/2 = 6 π cm2/sec

\(\int_0^{\frac{\pi}{2}}\)sin 2x dx = [-1/2cos 2x + C]\(_0^{\frac{\pi}{2}}\) =[-1/2 cos 2 * \pi/2 + C] - [-1/2 cos 2 * 0] = [-1/2 cos \pi] - [-1/2 cos 0] = [-1/2x - 1] - [-1/2 * 1] = 1/2 -(-1/2) = 1/2 + 1/2 = 1

Angle of Excellence = 360 - (120+80+90) = 360 - 290 = 70o If 360o represents 36 students 1º will represent \(\frac{36}{360}\) 50^\circ will represent \(\frac{36}{360}\) * \(\frac{70}{1}\) = 7