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 = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/(𝑇𝑖𝑚𝑒 𝑜𝑟𝑔𝑖𝑛𝑎𝑙)
x = 300/(𝑇𝑖𝑚𝑒 𝑜𝑟𝑔𝑖𝑛𝑎𝑙)
Time original = 300/𝑥
Speed 5 km/h more
Distance = 300 km
Speed = (x + 5) km/hr
Speed = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒/(𝑇𝑖𝑚𝑒 𝑆𝑝𝑒𝑒𝑑 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑)
x + 5 = 300/(𝑇𝑖𝑚𝑒 𝑆𝑝𝑒𝑒𝑑 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑)
Time speed increased = 300/(𝑥 + 5)
Given that train takes 2 hours less after speed increased
Time Speed increased = Time original – 2 hours
300/(𝑥 + 5) = 300/𝑥 – 2
2 = 300/𝑥 – 300/(𝑥 + 5)
300/𝑥 – 300/(𝑥 + 5) = 2
300(1/𝑥−1/(𝑥 + 5)) = 2
300(((𝑥 + 5) − 𝑥)/(𝑥(𝑥 + 5))) = 2
300(5/(𝑥(𝑥 + 5))) = 2
300 × 5/2 = x (x + 5)
150 × 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 × 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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