1. **Evaluate** $\sqrt{\frac{11}{12}} - \frac{1}{3} + \frac{1}{2}$ without a calculator. Step 1: Simplify $\sqrt{\frac{11}{12}} = \frac{\sqrt{11}}{\sqrt{12}} = \frac{\sqrt{11}}{2\sqrt{3}}$. Step 2: Rationalize the denominator: $$\frac{\sqrt{11}}{2\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{33}}{2 \times 3} = \frac{\sqrt{33}}{6}$$ Step 3: Combine the fractions: $$\frac{\sqrt{33}}{6} - \frac{1}{3} + \frac{1}{2}$$ Step 4: Find common denominator 6: $$\frac{\sqrt{33}}{6} - \frac{2}{6} + \frac{3}{6} = \frac{\sqrt{33} - 2 + 3}{6} = \frac{\sqrt{33} + 1}{6}$$ **Final answer:** $\boxed{\frac{\sqrt{33} + 1}{6}}$ 2. **Baraka's new earnings:** Step 1: Original hourly rate = 210. Step 2: New rate changes in ratio $8:7$, so new rate = $210 \times \frac{7}{8} = 183.75$. Step 3: Hours worked = $10 \frac{1}{2} = 10.5$ hours. Step 4: Total earnings = $183.75 \times 10.5 = 1928.4375$. **Final answer:** $\boxed{1928.44}$ (rounded to 2 decimal places) 3. **Solve for $x$ in:** $$4^{3x} \times 8 = \left(\frac{1}{32}\right)^{2x - 3}$$ Step 1: Express all bases as powers of 2: $$4 = 2^2, \quad 8 = 2^3, \quad 32 = 2^5$$ Step 2: Rewrite equation: $$ (2^2)^{3x} \times 2^3 = (2^{-5})^{2x - 3}$$ Step 3: Simplify exponents: $$2^{6x} \times 2^3 = 2^{-5(2x - 3)}$$ Step 4: Combine left side: $$2^{6x + 3} = 2^{-10x + 15}$$ Step 5: Equate exponents: $$6x + 3 = -10x + 15$$ Step 6: Solve for $x$: $$6x + 10x = 15 - 3$$ $$16x = 12$$ $$x = \frac{12}{16} = \frac{3}{4}$$ **Final answer:** $\boxed{\frac{3}{4}}$ 4. **Solve combined inequality:** $$-1 \leq \frac{5 - 2x}{3} < 2x - 1$$ Step 1: Multiply all parts by 3 (positive, so inequality signs unchanged): $$-3 \leq 5 - 2x < 3(2x - 1)$$ Step 2: Simplify right side: $$-3 \leq 5 - 2x < 6x - 3$$ Step 3: Solve left inequality: $$-3 \leq 5 - 2x$$ $$-3 - 5 \leq -2x$$ $$-8 \leq -2x$$ Divide by -2 (reverse inequality): $$4 \geq x$$ Step 4: Solve right inequality: $$5 - 2x < 6x - 3$$ $$5 + 3 < 6x + 2x$$ $$8 < 8x$$ $$1 < x$$ Step 5: Combine: $$1 < x \leq 4$$ **Final answer:** $\boxed{1 < x \leq 4}$ 5. **Estimate non-flooded area of farm PQRS:** Step 1: Total area = length $\times$ width = $210 \times 120 = 25200$ m$^2$. Step 2: Count flooded squares on grid (assumed from image, say approx 40 squares). Step 3: Each square represents $1 \text{ cm}^2$ on grid, corresponding to $\frac{210}{\text{grid length}} \times \frac{120}{\text{grid width}}$ m$^2$ per square. Step 4: Assuming 1 cm = 1 m (from problem statement), flooded area = number of flooded squares $\times 1$ m$^2$. Step 5: Non-flooded area = total area - flooded area. **Final answer:** $\boxed{\text{Non-flooded area} = 25200 - \text{flooded area in m}^2}$ (exact value depends on counted flooded squares). 6. **Draw labelled net of prism ABCDEFGH:** Step 1: The prism has trapezium cross-section with sides BC = CF = 5 cm, AD = HE = 2 cm, AB = DC = 2.5 cm. Step 2: The net consists of two trapezium faces (top and bottom) and four rectangular faces connecting corresponding sides. Step 3: Label trapezium faces ABCD and EFGH with given side lengths. Step 4: Rectangular faces correspond to edges AB-EF, BC-FG, CD-GH, DA-HE. **Final answer:** A labelled net showing two trapeziums ABCD and EFGH parallel, connected by rectangles ABFE, BCGF, CDHG, DAHE with correct side lengths. 7. **Road problem:** (a)(i) Let time after 11:50 am when vehicles meet be $t$ hours. Distance lorry travels: $45t$ km. Distance car travels: $75t$ km. Total distance: $45t + 75t = 120t$ km. Given total distance between A and B is 160 km, so: $$120t = 160 \Rightarrow t = \frac{160}{120} = \frac{4}{3} = 1 \text{ hour } 20 \text{ minutes}$$ Meeting time = 11:50 am + 1h 20m = 1:10 pm. (a)(ii) Distance from A to C = $45 \times \frac{4}{3} = 60$ km. (b) Car stops at C for 1h 40m = $\frac{5}{3}$ hours. Lorry continues at 45 km/h. Time for lorry from C to B: $$\text{Distance} = 160 - 60 = 100 \text{ km}$$ $$\text{Time} = \frac{100}{45} = \frac{20}{9} \text{ hours} \approx 2.22 \text{ hours}$$ Total time for lorry from meeting to B = $\frac{20}{9}$ hours. Car leaves C after $\frac{5}{3}$ hours stop and arrives at A at same time as lorry arrives at B. Time car takes from C to A = $\frac{20}{9} - \frac{5}{3} = \frac{20}{9} - \frac{15}{9} = \frac{5}{9}$ hours. Distance from C to A = 60 km. Average speed of car from C to A: $$\frac{60}{\frac{5}{9}} = 60 \times \frac{9}{5} = 108 \text{ km/h}$$ **Final answers:** (i) 1:10 pm (ii) 60 km (iii) Average speed = 108 km/h 8. **Simplify:** $$\frac{x^2 - 4y^2}{x^2 + 4xy + 4y^2}$$ Step 1: Factor numerator: $$x^2 - 4y^2 = (x - 2y)(x + 2y)$$ Step 2: Factor denominator: $$x^2 + 4xy + 4y^2 = (x + 2y)^2$$ Step 3: Simplify: $$\frac{(x - 2y)(x + 2y)}{(x + 2y)^2} = \frac{x - 2y}{x + 2y}$$ **Final answer:** $\boxed{\frac{x - 2y}{x + 2y}}$ 9. **Area of sector and cone height:** Given sector area $A = 550$ cm$^2$, radius of cone base $r = 7$ cm. Step 1: Sector arc length $l = 2\pi r_{sector} \times \frac{\theta}{360}$, but here sector forms cone base circumference: $$l = 2\pi \times 7 = 14\pi$$ Step 2: Sector radius $R$ satisfies: $$\text{Sector area} = \frac{1}{2} R l = 550$$ Step 3: Solve for $R$: $$\frac{1}{2} R \times 14\pi = 550$$ $$7\pi R = 550$$ $$R = \frac{550}{7\pi}$$ Step 4: Height $h$ of cone from Pythagoras: $$h = \sqrt{R^2 - r^2} = \sqrt{\left(\frac{550}{7\pi}\right)^2 - 7^2}$$ Calculate numerically with $\pi = 3.142$: $$R \approx \frac{550}{7 \times 3.142} = \frac{550}{21.994} \approx 25$$ $$h = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24$$ **Final answer:** $\boxed{24 \text{ cm}}$ 10. **Clock losing 18 seconds per hour:** Step 1: Total time from Monday 8:00 am to Saturday 11:20 am: - Monday 8:00 am to Saturday 8:00 am = 5 days = 120 hours - Plus 3 hours 20 minutes = 3.333 hours - Total = 123.333 hours Step 2: Total seconds lost: $$18 \times 123.333 = 2220 \text{ seconds} = 37 \text{ minutes}$$ Step 3: Clock time at Saturday 11:20 am real time: $$11:20 - 0:37 = 10:43 \text{ am}$$ **Final answer:** $\boxed{10:43 \text{ am}}$ 11. **Shopkeeper's cost price:** Let cost price = $C$, profit = $3x$, loss = $2x$. Step 1: Selling price with profit: $$C + 3x = 2740$$ Step 2: Selling price with loss: $$C - 2x = 2340$$ Step 3: Subtract second from first: $$(C + 3x) - (C - 2x) = 2740 - 2340$$ $$5x = 400 \Rightarrow x = 80$$ Step 4: Substitute $x$ back: $$C + 3(80) = 2740 \Rightarrow C = 2740 - 240 = 2500$$ **Final answer:** $\boxed{2500}$ 12. **Least number of needy families:** Step 1: Maize = 240 kg, Beans = 150 kg. Step 2: Each family receives equal mass of either maize or beans. Step 3: Find greatest common divisor (GCD) of 240 and 150: $$\text{GCD}(240, 150) = 30$$ Step 4: Number of families for maize: $$\frac{240}{30} = 8$$ Step 5: Number of families for beans: $$\frac{150}{30} = 5$$ Step 6: Total families = 8 + 5 = 13. **Final answer:** $\boxed{13}$ 13. **Quadratic curve tangent at P(1, -1):** Given: $$y = 3x^2 - 4x$$ Step 1: Find derivative: $$\frac{dy}{dx} = 6x - 4$$ Step 2: Slope at $x=1$: $$m = 6(1) - 4 = 2$$ Step 3: Equation of tangent line at P(1, -1): $$y - (-1) = 2(x - 1)$$ $$y + 1 = 2x - 2$$ $$y = 2x - 3$$ Step 4: Rearrange to form $ax + by = c$: $$2x - y = 3$$ **Final answer:** $\boxed{2x - y = 3}$