Q10. 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 11 km/h more than that of the passenger train, find the average speed of two trains.

Let average speed of passenger train = x km/h

Let average speed of express train = (x + 11) km/h

Time taken by passenger train to cover 132 km = \({{132} \over {x}}\) hours

Time taken by express train to cover 132 km = \({{132} \over {x + 11}}\) hours

According to the given condition,

\({{132} \over {x}} = {{132} \over {x + 11}} + 1\)

=>\(132({{1} \over {x}} - {{1} \over {x + 11}}) = 1\)

=>\(132({{x + 11 - x} \over {x(x + 11)}}) = 1\)

=> \(132 (11) = x (x + 11)\)

=>\(1452 = x^2 + 11x\)

=>\(x^2 + 11x -1452 = 0\)

Comparing equation \(x^2 + 11x -1452 = 0\) with general quadratic equation \(ax^2 + bx + c = 0\), we get a = 1, b = 11 and c = ?1452

Applying quadratic formula = \(x = {{-b ± \sqrt{b^2 - 4ac}} \over {2a}}\)

=>\(x = {{-11 ± \sqrt{(11)^2 - 4(1)(-1452)}} \over {(2)(1)}}\)

=>\(x = {{-11 ± \sqrt{121 + 5808}} \over {2}} = {{-11 ± \sqrt{5929}} \over {2}}\)

=>\(x = {{-11 + 77} \over {2}} , {{-11 - 77} \over {2}}\)

=>\(x = 33, -44\)

As speed cannot be in negative. Therefore, speed of passenger train = 33 km/h

And, speed of express train = x + 11 = 33 + 11 = 44 km/h