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An Express Train Takes 1 Hour Less than a Passenger Train to Travel 132 Km Between Mysore and Bangalore. If the Average Speed of the Express Train is 1 1 Km/Hr More than that of the Passenger Train, Form the Quadratic Equation to Find the Average Speed of Express Train. - Mathematics

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 1 1 km/hr more than that of the passenger train, form the quadratic equation to find the average speed of express train.

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Solution

Let the arrange speed of express train be denoted by x km/hr

Given that average speed of express train is 11km/hr more than that of the passenger train

⇒ Average speed of passenger train = (x - 11)km/hr

Total distance travelled by the train = 132 km

We know that,

`"Time taken to travel" = "Distance travelled"/"Average speed"=132/x hr`

`rArr"Time taken by express train" ="Distance travelled"/"Average speed of express train"=132/(x-11) hr`

Given that time taken by express train is 1 hour less than that of passenger train.

⇒ Time taken by passenger train - Time taken by express train = 1 hour

`rArr132/(x-11)-132/x=1`

`rArr132(1/(x-11)-1/x)=1`

`rArr132((x-(x-11))/(x(x-11)))=1`

⇒ 132(x - x + 11) = x(x - 11)

⇒ 132(11) = x2 - 11x

⇒ 1452 = x2 - 11x

⇒ x2 - 11x - 1452 = 0

The required quadratic is x2 - 11x - 1452 = 0

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 4 Quadratic Equations
Exercise 4.2 | Q 5 | Page 8
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