Find the value of both quantities and then state the correct relationship. Quantity I- find the distance, if a man goes to a distance with speed of 60km/h reaches 20 min early, and when he goes same distance with speed of 40 km/h he reaches 20 min late? Quantity II- Two buses start travelling in same direction at 6:00 am with the speed of 45km/h and 60km/h. After 4hours faster bus get punctured and takes 3 hours to repair. Find three-fourth of 360% of the distance between these two buses at 6:00 pm if faster bus started after repairing with increasing its speed by 10%

Quantity I: Distance is constant. At 60 km/h, 20 min early; at 40 km/h, 20 min late. Let usual time = t, distance = d. d/60 = t - 1/3, d/40 = t + 1/3. Subtracting: d/40 - d/60 = 2/3, so d/120 = 2/3, d = 80 km. Quantity II: By 6 PM, 12 hours passed. Slower bus: 12 × 45 = 180 km. Faster bus: travels 7 hours at 60 km/h (4 hours + 3 hours after repair) = 420 km. But after 4 hours, gap is 60 km. Faster bus stops for 3 hours, slower covers 135 km more. Gap becomes 195 km. In remaining 5 hours, faster catches 75 km. Final gap = 195 - 75 = 120 km. 360% of 120 = 432. Three-fourth of 432 = 324 km. Since 324 > 80, Quantity II > Quantity I.