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
CBSE Class 10 Sample Paper for 2019 Boards
Question 1
Question 2 (Or 1st)
Question 2 (Or 2nd) Important
Question 3 (Or 1st) Important
Question 3 (Or 2nd)
Question 4
Question 5 Important
Question 6 Important
Question 7 (Or 1st)
Question 7 (Or 2nd) Important
Question 8 (Or 1st)
Question 8 (Or 2nd)
Question 9
Question 10 Important
Question 11 Important
Question 12 Important
Question 13 Important
Question 14
Question 15
Question 16 (Or 1st) Important
Question 16 (Or 2nd)
Question 17 (Or 1st) Important
Question 17 (Or 2nd)
Question 18 Important
Question 19 (Or 1st)
Question 19 (Or 2nd) Important
Question 20
Question 21 (Or 1st) Important
Question 21 (Or 2nd)
Question 22
Question 23 (Or 1st) Important You are here
Question 23 (Or 2nd)
Question 24
Question 25
Question 26
Question 27 (Or 1st)
Question 27 (Or 2nd) Important
Question 28 (Or 1st)
Question 28 (Or 2nd) Important
Question 29
Question 30 Important
CBSE Class 10 Sample Paper for 2019 Boards
Last updated at April 16, 2024 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
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