(d) Leonard travels a distance of 240 km to visit his parents. He calculates that, if he increases his average speed by \( 20 \mathrm{~km} / \mathrm{h} \), the journey will take him 2 hours less. Determine his original speed.

Let Leonard’s original speed be v km/h. The time taken at his original speed is 240/v hours, and if he increases his speed by 20 km/h (to v + 20 km/h), the time taken becomes 240/(v + 20) hours. According to the problem, the journey time is reduced by 2 hours when the speed is increased. This gives us the equation:   240/v – 240/(v + 20) = 2 Multiply through by v(v + 20) to eliminate the fractions:   240(v + 20) – 240v = 2v(v + 20) Simplify the left-hand side:   240v + 4800 – 240v = 4800 So the equation becomes:   4800 = 2v² + 40v Divide the whole equation by 2:   2400 = v² + 20v Rearrange into standard quadratic form:   v² + 20v – 2400 = 0 Next, use the quadratic formula:   v = (–b ± √(b² – 4ac)) / (2a), where a = 1, b = 20, and c = –2400. Compute the discriminant:   Δ = 20² – 4(1)(–2400) = 400 + 9600 = 10000 Taking the square root:   √Δ = 100 Now, the solutions for v are:   v = (–20 ± 100) / 2 We discard the negative root, so:   v = (–20 + 100) / 2 = 80 / 2 = 40 Thus, Leonard's original speed is 40 km/h.