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Expenditure on housing = x Expenditure on food = 2x Expenditure on school fees = 90o Expenditure on transport = 45o x + 2x + 90 + 45 = 360o 3x + 135 = 360o 3x = 360 - 135 3x = 225 x = 225/3 = 75o ∴∠ for food = 2x = 2 * 75 = 150 360o = N30,000 1o = ? 1o = 30,000/360 150o = (30000/360) * (150/1) = N12500

dy/dx = x + cos x y = ∫(x + cos x)dx y = 1/2x2 + sin x + C

y = (cos\theta - sin\theta) dy/dx = 2(cos\theta - sin\theta)(-sin\theta - cos\theta) dy/dx = 2(-cos\thetasin\theta - cos2\theta + sin2\theta) dy/dx = 2(- cos\(^2\)\theta + sin\(^2\)\theta) = -2(cos\(^2\)\theta - sin\(^2\)\theta) = -2(1 - 2sin\(^2\)\theta) = -2cos2\theta

Area of rectangle PQRS = \(10 \times 7 = 70cm^2\) Area of semi-circle: \(\frac{\pi r^{2}}{2}\) r = \(\frac{7}{2} cm\) Area of semi-circle = \(\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) = \(\frac{77}{4} = 19.25 cm^{2}\) Area of figure = \((70 + 19.25) cm^{2}\) = 89.25 cm\(^{2} \approxeq\) 89 cm\(^{2}\)

Tan θ = 5/4 x2 = 52 + 42 = 25 + 16 = 41 x = √41 sin2θ - cos2θ = (5/√41)2 - (4/√41)2 = 25/41 - 16/41 = 9/41

Gradient PQ, P(1,q) and Q(3,2) \(=\frac{(2-q)}{(3-1)} = \frac{(2-q)}{2}\) Gradient of RS : R(3,4) and S(5,2q) \(= \frac{(2q-4)}{(5-3)}= \frac{(2q-4)}{2} = \frac{2(q-2)}{2}\) = q-2 Since PQ and RS are parallel, their gradients are equal \(∴ \frac{(2-q)}{2} = q-2\) 2-q = 2(q-2) 2-q = 2q-4 2+4 = 2q+q 6 = 3q q = 2

∠yxz = 180 - (45 + 15) = 180 - 60 = 120o Using sine rule \(\frac{x}{sinx}=\frac{7}{sinz}\) \(\frac{x}{sin 120}=\frac{7}{sin 45}\) x = \(\frac{7 sin 120}{sin45}\) x = \(\frac{7sin(180-120)}{sin 45}\) x = \(\frac{7 sin 60}{sin 45}=\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = \(\left(7\left(\frac{\sqrt{3}}{2}\right)\div \frac{1}{\sqrt{2}}\right)\) x = 7\(\left(\frac{\sqrt{6}}{2}\right)\)

y = 55 (alternate angles) x = x (same reason) 260 + y + x = 360 (∠s at a point) 260 + 55 + x = 360 315 + x = 360 x = 360 - 315 x = 45o

∠PSQ = 90^\circ (∠ in semi circle) ∠PSQ + ∠QSR = 145^\circ 90 + ∠QSR = 145 ∠QSR = 145 - 90 ∠QSR = 55^\circ BUt x^\circ = ∠QSR (∠s in the same segment) ∴x^\circ = 55^\circ

Each ∠of a regular polygon \(=\frac{(n-2)180}{n}\\ =\frac{(6-2)180}{6}\\ =\frac{(4)180}{6}\) = 120o ΔQSR is isosceles ∴ Q = S = 30o Also ΔTSR is isosceles ∴ T = R = 30o ∠TSV + ∠VSR = 120 ∠TSV + 30 = 120 ∠TSV = 120 - 30 ∠TSV = 90o ∴∴VTS + ∴VST + ∴TVS = 180 (sum of ∠s of a Δ) 30 + 90 + ∠TVS = 180 120 + ∠TVS = 180 ∠TVS = 180 - 120 ∠TVS = 60o

Locus of point equidistant from points P and Q is a perpendicular bisector of the straight joining P and Q ∴PN = NQ

In ΔPST; cos 60 = ST/8 ST = 8cos 60 ST = 8 x 1/2 = 4 TR = 10 cm (opp. sides of a rectangle PQRT) SR = ST + TR SR = 4 + 10 SR = 14 cm

Cost of removing a 6m\(^2\) room = N540 Cost of renovating a 1m\(^2\) room = 540 / 6 Cost of removing a 9m\(^2\) room = (540 / 6) * 9 = N810

No expanded working has been attached for this question yet. The verified answer is Option C.

tan y = 0.404; y = tan-1 0.0404(tables);y = 22o cos 2y = cos 2(22o); cos 44o = 0.719