Betina has 42 Kg of rice costing K640 per Kg. She wants to add rice costing K820 per Kg to make a mixture that costs K760 per Kg. How many kilograms of the K820 rice should she add?

You're on a roll — 13. To find how many kilograms of rice Betina should add: Let x be the number of kilograms of rice costing K820 per Kg that Betina adds. Step 1: Calculate the total cost of the initial rice. Initial rice: 42 Kg at K640 per Kg. Cost_1 = 42 Kg × K640/Kg = K26880 Step 2: Calculate the total cost of the added rice. Added rice: x Kg at K820 per Kg. Cost_2 = x Kg × K820/Kg = K820x Step 3: Calculate the total mass of the mixture. Total mass = (42 + x) Kg Step 4: Calculate the total cost of the mixture. The mixture should cost K760 per Kg. Total cost of mixture = (42 + x) Kg × K760/Kg = K760(42 + x) Step 5: Set up the equation where the sum of individual costs equals the total cost of the mixture. Cost_1 + Cost_2 = Total cost of mixture 26880 + 820x = 760(42 + x) Step 6: Solve the equation for x. 26880 + 820x = 760 × 42 + 760x 26880 + 820x = 31920 + 760x Subtract 760x from both sides: 26880 + 820x - 760x = 31920 26880 + 60x = 31920 Subtract 26880 from both sides: 60x = 31920 - 26880 60x = 5040 Divide by 60: x = (5040)/(60) x = 84 Betina must add 84 Kg of rice. The final answer is 84 Kg. 14. In one Mathematics continuous assessment, the following marks were obtained by top ten performing students: 37, 36, 37, 35, 37, 35, 37, 36, 38, 36. i) Draw a bar graph to illustrate the given information. Step 1: Create a frequency table for the given marks. • Mark 35: Appears 2 times • Mark 36: Appears 3 times • Mark 37: Appears 4 times • Mark 38: Appears 1 time Step 2: Draw the axes. • Draw a horizontal axis (x-axis) and label it "Marks". • Draw a vertical axis (y-axis) and label it "Number of Students" or "Frequency". Step 3: Choose a suitable scale for the y-axis. • The maximum frequency is 4, so the y-axis should go up to at least 4 or 5. A scale of 1 unit per grid line is appropriate. Step 4: Draw the bars. • For each mark, draw a bar whose height corresponds to its frequency. • Ensure the bars have equal width and are equally spaced from each other. Here is a description of the bar graph: • X-axis (Marks): Label points at 35, 36, 37, 38. • Y-axis (Number of Students): Label points at 0, 1, 2, 3, 4. • Bar for Mark 35: Height up to 2. • Bar for Mark 36: Height up to 3. • Bar for Mark 37: Height up to 4. • Bar for Mark 38: Height up to 1. ii) Find the median for the given marks. Step 1: Arrange the marks in ascending order. 35, 35, 36, 36, 36, 37, 37, 37, 37, 38 Step 2: Determine the number of data points (n). There are 10 marks, so n = 10. Step 3: Find the position of the median. Since n is an even number, the median is the average of the two middle values. These are the ((n)/(2))^th and ((n)/(2) + 1)^th values. • (10)/(2) = 5^th value • (10)/(2) + 1 = 6^th value Step 4: Identify the 5th and 6th values from the ordered list. The 5th value is 36. The 6th value is 37. Step 5: Calculate the median. Median = 5^th value + 6^th value2 = (36 + 37)/(2) Median = (73)/(2) = 36.5 The median mark is 36.5. Got more? Send 'em!