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

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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·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’/(π‘‡π‘–π‘šπ‘’ π‘œπ‘Ÿπ‘”π‘–π‘›π‘Žπ‘™) x = 300/(π‘‡π‘–π‘šπ‘’ π‘œπ‘Ÿπ‘”π‘–π‘›π‘Žπ‘™) Time original = 300/π‘₯ Speed 5 km/h more Distance = 300 km Speed = (x + 5) km/hr Speed = π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’/(π‘‡π‘–π‘šπ‘’ 𝑆𝑝𝑒𝑒𝑑 πΌπ‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘) x + 5 = 300/(π‘‡π‘–π‘šπ‘’ 𝑆𝑝𝑒𝑒𝑑 πΌπ‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘‘) Time speed increased = 300/(π‘₯ + 5) Given that train takes 2 hours less after speed increased Time Speed increased = Time original – 2 hours 300/(π‘₯ + 5) = 300/π‘₯ – 2 2 = 300/π‘₯ – 300/(π‘₯ + 5) 300/π‘₯ – 300/(π‘₯ + 5) = 2 300(1/π‘₯βˆ’1/(π‘₯ + 5)) = 2 300(((π‘₯ + 5) βˆ’ 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 – 750 = 0 We factorize by splitting the middle term method x2 + 30x – 25 – 750 = 0 x (x + 30) – 25(x + 30) = 0 (x + 30) (x – 25) = 0 So, x = –30, x = 25 Splitting the middle term method We need to find two numbers whose Sum = 5 Product = – 750 Γ— 1 = – 750 We know that Speed of train = x So, x cannot be negative ∴ 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 ∴ x = 25 is the solution So, Speed of train = x = 25 km/hr

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.