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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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