Chapter 4: Quadratic Equations

Example – 18. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution :- Let the original speed of the train = x km/h and increased speed = x + 5 km/h.

The time taken in two cases are : 360/x and 360/(x + 5)

A/Q                                   360/x – 360/(x + 5) = 1
Or,                     [360(x + 5) – 360x] /x(x + 5) = 1
Or,             [360x + 369 x 5 – 360x]/(x2 + 5x) = 1
Or,                                            x2 + 5x – 1800 = 0

Here, a = 1, b = 5, c = – 1800 ;   and D = b2 – 4ac

= 52 – 4 × 1 × (– 1800) 
= 25 + 7200 = 7225

Hence,       x = [– b ± √ D]/ 2a = [– 5 ± √(7225)]/(2 × 1)

= [– 5± 85]/2 = 80/2 or – 90 /2 = 40 or – 45

As speed cannot be negative, therefore, x = 40 or speed = 40 km/h.

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Subjects Maths (Part-1) by Mr. M. P. Keshari
Chapter 1 Linear Equations in Two Variables
Chapter 2 HCF and LCM
Chapter 3 Rational Expression
Chapter 4 Quadratic Equations
Chapter 5 Arithmetic Progressions
Chapter 6 Instalments
Chapter 7 Income Tax
Chapter 8 Similar Triangles