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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. - Mathematics

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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.

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Solution

Let the speed of the passenger train be x km/hr. Then,

Speed of the express train = (x + 11)km/hr

Time taken by the passenger train to cover 132 km between Mysore to Bangalore `132/x`hr

Time taken by the express train to cover 132 km between Mysore to Bangalore `=132/(x+11)`hr

Therefore,

`132/x-132/(x+11)=1`

`(132(x+11)-132x)/(x(x+11))=1`

`(132x+1452-132x)/(x^2+11)=1`

`1452/(x^2+11)=1`

1452 = x2 + 11

x2 + 11 - 1452 = 0

x2 - 33x + 44x - 1452 = 0

x(x - 33) + 44(x - 33) = 0

(x - 33)(x + 44) = 0

So, either 

x - 33 = 0

x = 33

Or

x + 44 = 0

x = -44

But, the speed of the train can never be negative.

Thus, when x = 33 then speed of express train

= x + 11

= 33 + 11

= 44

Hence, the speed of the passenger train is x = 33 km/hr

and the speed of the express train is x = 44 km/hr respectively.

Concept: Solutions of Quadratic Equations by Factorization
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APPEARS IN

NCERT Class 10 Maths
Chapter 4 Quadratic Equations
Exercise 4.3 | Q 10 | Page 88
RD Sharma Class 10 Maths
Chapter 4 Quadratic Equations
Exercise 4.8 | Q 11 | Page 59
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