This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here are the solutions to the problems: Q1. (a) The sum of angles in a triangle is 180^. i) Find x 72^ + x + 2x = 180^ 72^ + 3x = 180^ 3x = 180^ - 72^ 3x = 108^ x = (108^)/(3) x = 36^ ii) Find (x+m) From the diagram, m and 2x form a linear pair (angles on a straight line), so their sum is 180^. m + 2x = 180^ Substitute the value of x = 36^: m + 2(36^) = 180^ m + 72^ = 180^ m = 180^ - 72^ m = 108^ Now, calculate x+m: x+m = 36^ + 108^ x+m = 144^ Q1. (b) Given: Total pupils, |U| = 25 Pupils with rulers, |R| = 13 Pupils with pencils, |P| = 19 Pupils with neither ruler nor pencil, |(R P)'| = 6 Step 1: Find the number of pupils who have at least one item (ruler or pencil). |R P| = |U| - |(R P)'| = 25 - 6 = 19 Step 2: Find the number of pupils who have both rulers and pencils. Using the formula |R P| = |R| + |P| - |R P|: 19 = 13 + 19 - |R P| 19 = 32 - |R P| |R P| = 32 - 19 |R P| = 13 Step 3: Find the number of pupils who have rulers only and pencils only. Rulers only: |R P| = |R| - |R P| = 13 - 13 = 0 Pencils only: |P R| = |P| - |R P| = 19 - 13 = 6 i) Draw a Venn diagram to illustrate the information: (A Venn diagram cannot be drawn in text, but the values for each region are provided below.) • Pupils with rulers only: 0 • Pupils with pencils only: 6 • Pupils with both rulers and pencils: 13 • Pupils with neither: 6 ii) Find the number of pupils who have both rulers and pencils. The number of pupils who have both rulers and pencils is |R P|. |R P| = 13 iii) Find the number of pupils who have pencil only. The number of pupils who have pencil only is |P R|. |P R| = 6 iv) What percentage of the pupil have both ruler and pencil? Percentage = Number with bothTotal pupils × 100\% Percentage = (13)/(25) × 100\% Percentage = 0.52 × 100\% Percentage = 52\% Q2. (a) Let the marked price be M. The discount is 10% of the marked price, which is 0.10M. The selling price (amount paid) is GHS 810. Selling Price = Marked Price - Discount 810 = M - 0.10M 810 = 0.90M M = (810)/(0.90) M = GHS 900 Q2. (b) i) How many girls like Mathematics? Total pupils in the school = 360. Pupils who like Mathematics = (1)/(10) × 360 = 36. Half of those pupils who like Mathematics are girls. Girls who like Mathematics = (1)/(2) × 36 = 18 ii) Find the truth set of 3x - 4 < 4x - 5. 3x - 4 < 4x - 5 Subtract 3x from both sides: -4 < 4x - 3x - 5 -4 < x - 5 Add 5 to both sides: -4 + 5 < x 1 < x So, x > 1. The truth set is \x R x > 1\ iii) Make p the subject of the relation 3t - pq = 2(P+L). 3t - pq = 2(P+L) Expand the right side: 3t - pq = 2P + 2L Isolate the term containing p: -pq = 2P + 2L - 3t Multiply both sides by -1: pq = -(2P + 2L - 3t) pq = 3t - 2P - 2L Divide by q (assuming q ≠ 0): p = (3t - 2P - 2L)/(q)