Here are the solutions to your questions: Q1. a) Find the total marks Step 1: Let the total marks be $T$. We are given that 45 marks is 60% of the total marks. $$0.60 \times T = 45$$ Step 2: Divide by 0.60 to find $T$. $$T = \frac{45}{0.60}$$ $$T = 75$$ The total marks are 75. $$\boxed{\text{75 marks}}$$ b) How many more marks are needed to reach 80%? Step 1: Calculate 80% of the total marks. $$80\% \text{ of } 75 = 0.80 \times 75$$ $$0.80 \times 75 = 60$$ Step 2: Find the difference between 80% of the total marks and the student's current marks. $$\text{Marks needed} = 60 - 45$$ $$\text{Marks needed} = 15$$ The student needs 15 more marks to reach 80%. $$\boxed{\text{15 marks}}$$ Q2. a) Kofi’s share b) Ama’s share c) Yaw’s share Step 1: Find the total number of ratio parts. $$\text{Total ratio parts} = 2 + 3 + 4 = 9$$ Step 2: Find the value of one ratio part. $$\text{Value per part} = \frac{\text{GH₵}360}{9} = \text{GH₵}40$$ Step 3: Calculate each person's share. a) Kofi's share: $$\text{Kofi's share} = 2 \times \text{GH₵}40 = \text{GH₵}80$$ $$\boxed{\text{GH₵80}}$$ b) Ama's share: $$\text{Ama's share} = 3 \times \text{GH₵}40 = \text{GH₵}120$$ $$\boxed{\text{GH₵120}}$$ c) Yaw's share: $$\text{Yaw's share} = 4 \times \text{GH₵}40 = \text{GH₵}160$$ $$\boxed{\text{GH₵160}}$$ Q3. a) Find the perimeter Step 1: The formula for the perimeter of a rectangle is $P = 2(L+W)$. $$P = 2(20 \text{ m} + 15 \text{ m})$$ $$P = 2(35 \text{ m})$$ $$P = 70 \text{ m}$$ The perimeter is 70 m. $$\boxed{\text{70 m}}$$ b) Find the area Step 1: The formula for the area of a rectangle is $A = L \times W$. $$A = 20 \text{ m} \times 15 \text{ m}$$ $$A = 300 \text{ m}^2$$ The area is 300 m$^2$. $$\boxed{\text{300 m}^2}$$ c) Find the cost of fencing it at GH₵5 per metre Step 1: Multiply the perimeter by the cost per metre. $$\text{Cost} = 70 \text{ m} \times \text{GH₵}5/\text{m}$$ $$\text{Cost} = \text{GH₵}350$$ The cost of fencing is GH₵350. $$\boxed{\text{GH₵350}}$$ Q4. a) Find x Step 1: Angles on a straight line sum to 180°. $$(x+10)^\circ + (3x-20)^\circ = 180^\circ$$ Step 2: Combine like terms. $$4x - 10 = 180$$ Step 3: Add 10 to both sides. $$4x = 190$$ Step 4: Divide by 4. $$x = \frac{190}{4}$$ $$x = 47.5$$ The value of $x$ is 47.5. $$\boxed{x = 47.5}$$ b) If one of the angles is part of a triangle, find the third angle if the other two are 50° and that angle. Step 1: Calculate the values of the two angles on the straight line. First angle: $x+10 = 47.5+10 = 57.5^\circ$ Second angle: $3x-20 = 3(47.5)-20 = 142.5-20 = 122.5^\circ$ Let's use the first angle, $57.5^\circ$, as "that angle" for the triangle. Step 2: The sum of angles in a triangle is 180°. $$\text{Third angle} = 180^\circ - 50^\circ - 57.5^\circ$$ $$\text{Third angle} = 130^\circ - 57.5^\circ$$ $$\text{Third angle} = 72.5^\circ$$ The third angle is 72.5°. $$\boxed{\text{72.5}^\circ}$$ Q5. a) What was the original price? Step 1: A 10% discount means the shirt was sold for $100\% - 10\% = 90\%$ of its original price. Let the original price be $P$. $$0.90 \times P = \text{GH₵}180$$ Step 2: Divide by 0.90 to find $P$. $$P = \frac{\text{GH₵}180}{0.90}$$ $$P = \text{GH₵}200$$ The original price was GH₵200. $$\boxed{\text{GH₵200}}$$ b) A shopkeeper bought goods for GH₵450 and sold them for GH₵540. Find the profit percentage. Step 1: Calculate the profit. $$\text{Profit} = \text{Selling Price} - \text{Cost Price}$$ $$\text{Profit} = \text{GH₵}540 - \text{GH₵}450$$ $$\text{Profit} = \text{GH₵}90$$ Step 2: Calculate the profit percentage. $$\text{Profit Percentage} = \frac{\text{Profit}}{\text{Cost Price}} \times 100\%$$ $$\text{Profit Percentage} = \frac{\text{GH₵}90}{\text{GH₵}450} \times 100\%$$ $$\text{Profit Percentage} = 0.2 \times 100\%$$ $$\text{Profit Percentage} = 20\%$$ The profit percentage is 20%. $$\boxed{20\%}$$ Q6. i. Find the mode Step 1: List the marks and count their frequencies: 42 (1), 55 (3), 50 (2), 48 (2), 62 (1), 38 (1), 60 (1), 45 (1) Step 2: The mode is the mark that appears most frequently. The mark 55 appears 3 times, which is more than any other mark. The mode is 55. $$\boxed{55}$$ ii. Find the range Step 1: Identify the highest and lowest marks. Highest mark = 62 Lowest mark = 38 Step 2: Calculate the range. $$\text{Range} = \text{Highest mark} - \text{Lowest mark}$$ $$\text{Range} = 62 - 38$$ $$\text{Range} = 24$$ The range is 24. $$\boxed{24}$$ b) A bag contains 4 red, 3 blue and 3 green balls. One ball is drawn at random. Step 1: Find the total number of balls. $$\text{Total balls} = 4 (\text{red}) + 3 (\text{blue}) + 3 (\text{green}) = 10$$ i. Find the probability that it is: Blue Step 2: The probability of drawing a blue ball is the number of blue balls divided by the total number of balls. $$P(\text{Blue}) = \frac{\text{Number of blue balls}}{\text{Total balls}}$$ $$P(\text{Blue}) = \frac{3}{10}$$ The probability of drawing a blue ball is $\frac{3}{10}$. $$\boxed{\frac{3}{10}}$$ ii. Not green Step 2: The number of balls that are not green is the sum of red and blue balls. $$\text{Not green balls} = 4 (\text{red}) + 3 (\text{blue}) = 7$$ Step 3: The probability of drawing a ball that is not green is the number of non-green balls divided by the total number of balls. $$P(\text{Not green}) = \frac{\text{Number of not green balls}}{\text{Total balls}}$$ $$P(\text{Not green}) = \frac{7}{10}$$ The probability of drawing a ball that is not green is $\frac{7}{10}$. $$\boxed{\frac{7}{10}}$$ Q7. a) A taxi company charges a fixed fee of GH₵10 plus GH₵1.20 per kilometer. How much will a passenger pay for a trip of 15 km? Step 1: Calculate the variable cost for 15 km. $$\text{Variable cost} = 15 \text{ km} \times \text{GH₵}1.20/\text{km}$$ $$\text{Variable cost} = \text{GH₵}18.00$$ Step 2: Add the fixed fee to the variable cost. $$\text{Total cost} = \text{Fixed fee} + \text{Variable cost}$$ $$\text{Total cost} = \text{GH₵}10.00 + \text{GH₵}18.00$$ $$\text{Total cost} = \text{GH₵}28.00$$ A passenger will pay GH₵28.00 for a 15 km trip. $$\boxed{\text{GH₵28.00}}$$ b) A rectangular field is 85 m long and 60 m wide. A farmer wants to put a fence around the field and also divide the field into 5 equal smaller plots by lines parallel to the shorter side. i. What is the perimeter of the field? Step 1: The formula for the perimeter of a rectangle is $P = 2(L+W)$. $$P = 2(85 \text{ m} + 60 \text{ m})$$ $$P = 2(145 \text{ m})$$ $$P = 290 \text{ m}$$ The perimeter of the field is 290 m. $$\boxed{\text{290 m}}$$ ii. How many metres of fencing will be needed altogether? Step 1: The fencing around the field is its perimeter, which is 290 m. Step 2: To divide the field into 5 equal smaller plots by lines parallel to the shorter side (60 m), 4 dividing lines are needed. Each dividing line will be 60 m long. $$\text{Length of dividing lines} = 4 \times 60 \text{ m} = 240 \text{ m}$$ Step 3: Add the perimeter and the length of the dividing lines to find the total fencing needed. $$\text{Total fencing}