Last updated at Feb. 25, 2017 by Teachoo

Transcript

Ex 4.3 ,10 An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains. Let the average speed of passenger train = x km/h Given that , express train taken 1 hour less ∴ Time taken by passenger train – Time taken by express = 1 train 132/𝑥 – 132/(𝑥 + 11)=1 (132(𝑥 + 11) − 132𝑥)/(𝑥(𝑥 + 11))=1 (132𝑥 + 132(11) − 132𝑥)/(𝑥(𝑥 + 11))=1 (132𝑥 − 132𝑥 + 1452)/(𝑥(𝑥 + 11))=1 1452/(𝑥(𝑥 + 11))=1 1452 = x(x + 11) 1452 = x2 + 11x 0 = x2 + 11x – 1452 x2 + 11x – 1452 = 0 We solve this by quadratic method Comparing equation with ax2 + bx + c = 0, Here a = 1, b = 11, c = –1452 We know that D = b2 – 4ac D = (11)2 – 4 × 1 × (–1452) D = 121 + 5808 D = 5929 So, the roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (−11 ± √5929)/(2 × 1) x = (−11 ± √5929)/2 x = (−11 ± 77)/2 Solving So, x = 33 & x = – 44 So, x = 33 & x = – 44 is the roots of the equation Since x is the average speed of the passenger train, it cannot be negative So, x = 33 Average speed of passenger train = x = 33 km/hr So, average speed of express train = x + 11 = 33 + 11 = 44 km/hr

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

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