Here are the solutions to the problems: 17. Find the truth set of $2x - 4 < 6 + 3x$. Step 1: Gather $x$ terms on one side and constant terms on the other side of the inequality. Subtract $2x$ from both sides: $$-4 < 6 + 3x - 2x$$ $$-4 < 6 + x$$ Step 2: Subtract 6 from both sides: $$-4 - 6 < x$$ $$-10 < x$$ Step 3: Rewrite the inequality with $x$ on the left side. $$x > -10$$ The truth set is $\{x: x > -10\}$. The correct option is C. $$\boxed{\text{C. } \{x: x > -10\}}$$ 18. Evaluate $\sqrt{75} + \sqrt{18} - \sqrt{27}$. Step 1: Simplify each square root by finding the largest perfect square factor. $$\sqrt{75} = \sqrt{25 \times 3} = \sqrt{25} \times \sqrt{3} = 5\sqrt{3}$$ $$\sqrt{18} = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$$ $$\sqrt{27} = \sqrt{9 \times 3} = \sqrt{9} \times \sqrt{3} = 3\sqrt{3}$$ Step 2: Substitute the simplified square roots back into the expression. $$5\sqrt{3} + 3\sqrt{2} - 3\sqrt{3}$$ Step 3: Combine like terms (terms with the same radical). $$(5\sqrt{3} - 3\sqrt{3}) + 3\sqrt{2}$$ $$(5-3)\sqrt{3} + 3\sqrt{2}$$ $$2\sqrt{3} + 3\sqrt{2}$$ The correct option is B. $$\boxed{\text{B. } 2\sqrt{3} + 3\sqrt{2}}$$ 19. The locus of points equidistant from a fixed point is called a A locus of points equidistant from a fixed point is the definition of a circle. The fixed point is the center of the circle, and the constant distance is the radius. The correct option is B. $$\boxed{\text{B. circle}}$$ 20. Find the highest common factor of 54 and 72. Step 1: Find the prime factorization of each number. $$54 = 2 \times 27 = 2 \times 3 \times 9 = 2 \times 3 \times 3 \times 3 = 2 \times 3^3$$ $$72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$$ Step 2: Identify the common prime factors and their lowest powers. Common prime factors are 2 and 3. Lowest power of 2 is $2^1$. Lowest power of 3 is $3^2$. Step 3: Multiply these common prime factors raised to their lowest powers. $$\text{HCF} = 2^1 \times 3^2 = 2 \times 9 = 18$$ The correct option is A. $$\boxed{\text{A. } 18}$$ 21. Arrange in ascending order the fractions $\frac{1}{2}$, $\frac{3}{4}$ and $\frac{2}{5}$. Step 1: Find the least common multiple (LCM) of the denominators (2, 4, 5). Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, ... Multiples of 4: 4, 8, 12, 16, 20, ... Multiples of 5: 5, 10, 15, 20, ... The LCM is 20. Step 2: Convert each fraction to an equivalent fraction with a denominator of 20. $$\frac{1}{2} = \frac{1 \times 10}{2 \times 10} = \frac{10}{20}$$ $$\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$$ $$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$$ Step 3: Compare the numerators and arrange the fractions in ascending order. The numerators are 10, 15, and 8. In ascending order: 8, 10, 15. So, the fractions in ascending order are $\frac{8}{20}, \frac{10}{20}, \frac{15}{20}$. Step 4: Write the original fractions in ascending order. $$\frac{2}{5}, \frac{1}{2}, \frac{3}{4}$$ The correct option is C. $$\boxed{\text{C. } \frac{2}{5}, \frac{1}{2}, \text{ and } \frac{3}{4}}$$ 22. A hawker is carrying a basket load of three types of fruits: oranges, mangoes and pears. If $\frac{1}{4}$ of the fruits are oranges and $\frac{2}{5}$ mangoes, what percentage of the fruits are pears? Step 1: Calculate the total fraction of oranges and mangoes. Fraction of oranges $= \frac{1}{4}$ Fraction of mangoes $= \frac{2}{5}$ Total fraction of oranges and mangoes $= \frac{1}{4} + \frac{2}{5}$ Step 2: Find a common denominator for the fractions and add them. The LCM of 4 and 5 is 20. $$\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20}$$ $$\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$$ Total fraction of oranges and mangoes $= \frac{5}{20} + \frac{8}{20} = \frac{13}{20}$ Step 3: Calculate the fraction of pears. The total fraction of fruits is 1. Fraction of pears $= 1 - \text{Total fraction of oranges and mangoes}$ Fraction of pears $= 1 - \frac{13}{20} = \frac{20}{20} - \frac{13}{20} = \frac{7}{20}$ Step 4: Convert the fraction of pears to a percentage. Percentage of pears $= \frac{7}{20} \times 100\%$ $$= 7 \times \frac{100}{20}\%$$ $$= 7 \times 5\%$$ $$= 35\%$$ The correct option is D. (Note: The options provided are A. 9%, B. 18%, C. 64%, D. 36%. My calculated answer is 35%. There might be a slight discrepancy in the options or the question's intended answer. However, 35% is closest to 36% among the given options.) Let's recheck the calculation. $1/4 = 0.25$ $2/5 = 0.40$ Total fraction of oranges and mangoes = $0.25 + 0.40 = 0.65$ Fraction of pears = $1 - 0.65 = 0.35$ Percentage of pears = $0.35 \times 100\% = 35\%$. Given the options, 36% is the closest. It's possible there's a rounding or slight error in the question's options. I will select the closest option. $$\boxed{\text{D. } 36\%}$$ 23. If 3 litres of petrol cost GH¢10.00, find the cost of 15 litres. Step 1: Find the cost per litre of petrol. Cost per litre $= \frac{\text{Total cost}}{\text{Quantity}}$ Cost per litre $= \frac{\text{GH¢}10.00}{3 \text{ litres}}$ Step 2: Calculate the cost of 15 litres. Cost of 15 litres $= \text{Cost per litre} \times 15 \text{ litres}$ Cost of 15 litres $= \frac{\text{GH¢}10.00}{3} \times 15$ Cost of 15 litres $= \text{GH¢}10.00 \times \frac{15}{3}$ Cost of 15 litres $= \text{GH¢}10.00 \times 5$ Cost of 15 litres $= \text{GH¢}50.00$ The correct option is D. $$\boxed{\text{D. GH¢}50.00}$$