Convert the division into multiplication by inverting the second fraction.

Here are the solutions to the questions. 1 (a) Simplify (15a^2b)/(7cd^2) ÷ (5a^3b^2)/(21c^2d^3). Step 1: Convert the division into multiplication by inverting the second fraction. (15a^2b)/(7cd^2) ÷ (5a^3b^2)/(21c^2d^3) = (15a^2b)/(7cd^2) × (21c^2d^3)/(5a^3b^2) Step 2: Multiply the numerators and the denominators. = (15a^2b × 21c^2d^3)/(7cd^2 × 5a^3b^2) Step 3: Simplify the expression by canceling common factors. = ((3 × 5)a^2b × (3 × 7)c^2d^3)/(7cd^2 × 5a^3b^2) = (3 × 3 × 5 × 7 × a^2 × b × c^2 × d^3)/(5 × 7 × a^3 × b^2 × c × d^2) Cancel 5, 7, a^2, b, c, d^2: = (3 × 3 × c × d)/(a × b) = (9cd)/(ab) The final answer is 9cdab. 1 (b) Solve the equation 2x^2 - x - 2 = 0, giving your answers correct to 2 decimal places. Step 1: Identify the coefficients of the quadratic equation ax^2 + bx + c = 0. Here, a = 2, b = -1, c = -2. Step 2: Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-1) ± sqrt((-1)^2 - 4(2)(-2))2(2) x = 1 ± sqrt(1 - (-16))4 x = 1 ± sqrt(1 + 16)4 x = 1 ± sqrt(17)4 Step 3: Calculate the two possible values for x. sqrt(17) ≈ 4.1231056 For x_1: x_1 = 1 + sqrt(17)4 = (1 + 4.1231056)/(4) = (5.1231056)/(4) ≈ 1.2807764 For x_2: x_2 = 1 - sqrt(17)4 = (1 - 4.1231056)/(4) = (-3.1231056)/(4) ≈ -0.7807764 Step 4: Round the answers to 2 decimal places. x_1 ≈ 1.28 x_2 ≈ -0.78 The final answers are x = 1.28 and x = -0.78. 2 (a) Two boxes A and B contain balls of the same size. Box A contains 2 green balls and 3 blue balls while Box B contains 4 green balls and 2 blue balls. If one ball is selected at random from each box, what is the probability that both balls selected are Box A: Total balls = 2 (green) + 3 (blue) = 5 P(Green from A) = (2)/(5) P(Blue from A) = (3)/(5) Box B: Total balls = 4 (green) + 2 (blue) = 6 P(Green from B) = (4)/(6) = (2)/(3) P(Blue from B) = (2)/(6) = (1)/(3) (i) blue Step 1: Find the probability of selecting a blue ball from Box A. P(Blue from A) = (3)/(5) Step 2: Find the probability of selecting a blue ball from Box B. P(Blue from B) = (2)/(6) = (1)/(3) Step 3: Multiply the probabilities since the events are independent. P(Both blue) = P(Blue from A) × P(Blue from B) = (3)/(5) × (1)/(3) = (3)/(15) = (1)/(5) The final answer is (1)/(5) or 0.2. (ii) of different colours? Step 1: Identify the two scenarios for different colours: Scenario 1: Green from Box A AND Blue from Box B. Scenario 2: Blue from Box A AND Green from Box B. Step 2: Calculate the probability for Scenario 1. P(Green from A and Blue from B) = P(Green from A) × P(Blue from B) = (2)/(5) × (1)/(3) = (2)/(15) Step 3: Calculate the probability for Scenario 2. P(Blue from A and Green from B) = P(Blue from A) × P(Green from B) = (3)/(5) × (4)/(6) = (3)/(5) × (2)/(3) = (6)/(15) = (2)/(5) Step 4: Add the probabilities of the two scenarios (since they are mutually exclusive). P(Different colours) = P(Scenario 1) + P(Scenario 2) = (2)/(15) + (6)/(15) = (8)/(15) The final answer is (8)/(15). 2 (b) On a particular day at a school with 50 Grade 12 learners, 33 collected Biology textbooks, 24 collected Mathematics textbooks and 23 collected History textbooks from the school library. 13 learners collected both Biology and Mathematics textbooks, 16 collected both Mathematics and History textbooks, 15 collected both Biology and History textbooks and 10 collected all the three types of textbooks. Let B = Biology, M = Mathematics, H = History. Total learners = 50. |B| = 33 |M| = 24 |H| = 23 |B M| = 13 |M H| = 16 |B H| = 15 |B M H| = 10 (i) Illustrate this information in a Venn diagram. To construct the Venn diagram, we first fill in the innermost intersection and work outwards. All three (B M H): 10 Only B and M: |B M| - |B M H| = 13 - 10 = 3 Only M and H: |M H| - |B M H| = 16 - 10 = 6 Only B and H: |B H| - |B M H| = 15 - 10 = 5 Now, calculate those who collected only one type of textbook: Only B: |B| - (Only B and M) - (Only B and H) - (|B M H|) = 33 - 3 - 5 - 10 = 15 Only M: |M| - (Only B and M) - (Only M and H) - (|B M H|) = 24 - 3 - 6 - 10 = 5 Only H: |H| - (Only M and H) - (Only B and H) - (|B M H|) = 23 - 6 - 5 - 10 = 2 Venn Diagram Regions: B only: 15 M only: 5 H only: 2 B and M only: 3 M and H only: 6 B and H only: 5 B and M and H: 10 (ii) How many learners collected (a) none of the textbooks, Step 1: Calculate the total number of learners who collected at least one textbook. This is the sum of all regions within the Venn diagram. Total collected = 15 + 5 + 2 + 3 + 6 + 5 + 10 = 46 Step 2: Subtract the total collected from the total number of learners. None collected = Total learners - Total collected = 50 - 46 = 4 The final answer is 4. (b) one type of textbook only, Step 1: Sum the number of learners who collected only Biology, only Mathematics, or only History. One type only = (B only) + (M only) + (H only) = 15 + 5 + 2 = 22 The final answer is 22. (c) two types of textbooks only? Step 1: Sum the number of learners who collected exactly two types of textbooks (excluding those who collected all three). Two types only = (Only B and M) + (Only M and H) + (Only B and H) = 3 + 6 + 5 = 14 The final answer is 14. That's 2 down. 3 left today — send the next one.