This economics question tests your understanding of economic models and analysis. The step-by-step answer below applies the relevant framework and explains the reasoning.
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You're on a roll — here are the solutions to the questions: Question 17: The shape below is a rectangle. Study it and answer the questions that follow. a) Find the value of x. In a rectangle, opposite sides are equal. Therefore, the two expressions for the length must be equal. (x+25) cm = (3x+5) cm Subtract x from both sides: 25 = 2x + 5 Subtract 5 from both sides: 20 = 2x Divide by 2: x = (20)/(2) x = 10 b) Find the length. Substitute x=10 into the expression for the length: Length = (x+25) cm = (10+25) cm = 35 cm (Alternatively, using the other expression: (3x+5) cm = (3(10)+5) cm = (30+5) cm = 35 cm) c) Find the width. Substitute x=10 into the expression for the width: Width = 2x cm = 2(10) cm = 20 cm Question 18: A family got 900,000 FRW and had the following projects they wanted to do: (i) Paying school fees worth 120,000 FRW (ii) Painting a house at the cost of 100,000 FRW (iii) Buying food for 200,000 FRW (iv) Buying clothes for 150,000 FRW a) Order the family needs according to priority. The needs are ordered from most essential to least essential: 1. (iii) Buying food for 200,000 FRW (Basic survival need) 2. (i) Paying school fees worth 120,000 FRW (Education is a high priority) 3. (iv) Buying clothes for 150,000 FRW (Basic necessity) 4. (ii) Painting a house at the cost of 100,000 FRW (Maintenance, less urgent than basic needs) b) i) How much does a family require to meet with their budget? This refers to the total cost of all the projects. Total required amount = 120,000 + 100,000 + 200,000 + 150,000 Total required amount = 570,000 FRW b) ii) Do they have the required amount? The family's budget is 900,000 FRW. The total required amount is 570,000 FRW. Since 900,000 FRW > 570,000 FRW, the family has the required amount. b) iii) State the item that could be done later. Based on the priority order, the item that could be done later is (ii) Painting a house. Question 19: The following marks were obtained by a P6 class in a mathematics examination. Marks: 95, 90, 85, 80, 75, 70, 60 Number of learners: 2, 3, 8, 1, 6, 3, 2 a) Complete this frequency table. To complete the table, we calculate the product of marks (x) and the number of learners (f) for each row (fx), and then find the sums f and fx. | Marks (x) | Number of learners (f) | (fx) | | :-------- | :--------------------- | :--- | | 95 | 2 | 95 × 2 = 190 | | 90 | 3 | 90 × 3 = 270 | | 85 | 8 | 85 × 8 = 680 | | 80 | 1 | 80 × 1 = 80 | | 75 | 6 | 75 × 6 = 450 | | 70 | 3 | 70 × 3 = 210 | | 60 | 2 | 60 × 2 = 120 | | | f = 25 | fx = 2000 | b) How many learners did the exam? The total number of learners is the sum of the frequencies ( f). f = 2+3+8+1+6+3+2 = 25 learners c) What is the modal mark? The modal mark is the mark that appears most frequently. From the table, the highest frequency is 8, which corresponds to the mark of 85. Modal mark = 85 d) What is the mean of P6 class? The mean is calculated by dividing the sum of (fx) by the sum of frequencies ( f). Mean = ( fx)/( f) Mean = (2000)/(25) Mean = 80 Got more? Send 'em!