Question
While going to a destination after travelling 75km, the
driver noticed some problem in the clutch of the bus due to which he had to proceed at 2/5 of its former rate and reached the destination 5 hours late. Had the problem in the clutch happened 15 km further on, it would have arrived 30 minutes sooner. Find the speed of the bus and total distance covered by the bus?Solution
Let A is the starting point and B is actual point where bus met with an accident C is the assumed point bus met with an accident Given BC = 15km and Difference of is 30 minutes basically due to the speed covering this 15km. Let original speed = x  km/hr  Speed after clutch problem = (2x)/5 km/hr ⇒ 15/((2x)/5)-15/x=30/60 ⇒ 75/(2x)-15/x=1/2 ⇒ (75-30)/(2x)=1/2 ⇒ 45/(2x)=1/2 ⇒ x = (45 × 2)/2 x = 45 km/hr Let the distance be ‘D’ km  Time taken at original speed =D/45 hr……….(i) Time taken after clutch problem took place at 75 km distance = (75/45)+(D-75)/(45×(2/5)` = 75/45+(5(D-75))/(45×2) = 5/3+((D-75))/18 = (30+D-75)/18 = (D-45)/18 ………………………..(ii) As per the question, (ii) – (i) = 5 hours (D-45)/18-D/45=5  = (5D-225-2D)/90=5  = (3D-225)/90=5  3D – 225 = 450  3D = 675     D = 225 km
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