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

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.