Question 23 (OR 1 st question)
A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train
Last updated at Oct. 1, 2019 by Teachoo
Question 23 (OR 1 st question)
A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train
Transcript
Question 23 (OR 1st question) A train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/h from its usual speed. Find the usual speed of the train Let the speed of train be x km/hr Normal speed Distance = 300 km Speed = x km/hr Speed = π·ππ π‘ππππ/(ππππ πππππππ) x = 300/(ππππ πππππππ) Time original = 300/π₯ Speed 5 km/h more Distance = 300 km Speed = (x + 5) km/hr Speed = π·ππ π‘ππππ/(ππππ πππππ πΌππππππ ππ) x + 5 = 300/(ππππ πππππ πΌππππππ ππ) Time speed increased = 300/(π₯ + 5) Given that train takes 2 hours less after speed increased Time Speed increased = Time original β 2 hours 300/(π₯ + 5) = 300/π₯ β 2 2 = 300/π₯ β 300/(π₯ + 5) 300/π₯ β 300/(π₯ + 5) = 2 300(1/π₯β1/(π₯ + 5)) = 2 300(((π₯ + 5) β π₯)/(π₯(π₯ + 5))) = 2 300(5/(π₯(π₯ + 5))) = 2 300 Γ 5/2 = x (x + 5) 150 Γ 5 = x(x + 5) 750 = x(x + 5) 750 = x2 + 5x x2 + 5x = 750 x2 + 5x β 750 = 0 We factorize by splitting the middle term method x2 + 30x β 25 β 750 = 0 x (x + 30) β 25(x + 30) = 0 (x + 30) (x β 25) = 0 So, x = β30, x = 25 Splitting the middle term method We need to find two numbers whose Sum = 5 Product = β 750 Γ 1 = β 750 We know that Speed of train = x So, x cannot be negative β΄ x = 25 is the solution So, Speed of train = x = 25 km/hr We know that Speed of train = x So, x cannot be negative β΄ x = 25 is the solution So, Speed of train = x = 25 km/hr
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