A passenger train takes 3 hours less for a journey of 360 km if its speed is increased by 20km/hr from its usual speed. We need to find its usual speed. Represent the problem situation in the form of a quadratic equation.

A passenger train takes 3 hours less for a journey of 360 km if its sp

1.	A passenger train takes 3 hours less for a journey of 360 km if its speed is increased by 20km/hr from its usual speed. We need to find its usual speed. Represent the problem situation in the form of a quadratic equation.
Answer» Let usual speed of train is `x` km/h. Let the time taken by the train to cover `360` km distance with normal speed is `t` hour. Then, `360/x = t->(1)` When speed is increased by `20` km/h, then, `360/(x+20) = t-3->(2)` Subtracting (1) from (2), `360/(x+20) - 360/x = -3` `=>360x -360x-7200 = -3x(x+20)` `=>3x^2+60x-7200 = 0` `=>x^2+20x-2400 = 0`, which is the required quadratic equation. Now, we will solve it to find the speed of the train. `=> x^2+20x-2400 = 0 ` `=>x^2+60x-40x-2400 = 0` `=>(x+60)(x-40) = 0` `=> x = -60 and x = 40` But, `x` can not be negative, so, `x = 40` km/h. `:.` Usual speed of the train is `40` km/h.