Question

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Answer 1

let the speed of the train be x
when it is increased the speed becomes=x+5
360/x-360/x+5=1
360(1/x-1/x+5)=1
360(x+5-x/x²+5x)=1
360(5/x²+5x)=1
1800=x²+5x
x²+5x-1800=0
-5+-√25+7200/2
-5+-√7225/2
-5+-85/2
-90/2and80/2
therefore answer=40

Answer 2

Appropriate Answer:

• A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let the original speed of the train be x km/hr.

Total Speed = ( x + 5 )km/h

$\bf{Time \: (T_1) \: taken \: to \: cover \: 360 \: km \: t \: original \: speed = \frac{360}{x}hr }$

$\bf{\therefore \: Time \: (T_2) \: taken \: to \: cover \: 360 \: km \: at \: new \: speed = \frac{360}{x + 5} hr}$

According to the Statement given,

$\tt{T_1 = T_2+1} \\ : \implies \tt{T_1-T_2=1}$

$: \implies \sf{ \frac{360}{x} - \frac{360}{x + 5} = 1 } \\ \\: \implies \sf{ \frac{360(x + 5) - 360x}{x(x + 5)} = 1} \\ \\ : \implies \sf{360x + 1800 - 360x = {x}^{2} + 5x} \\ \\ : \implies \sf{1800 = {x}^{2} + 5x} \\ \\ : \implies \sf{ {x}^{2} + 5x - 1800 = 0 } \\ \\ : \implies \sf{ {x}^{2} + 45x - 40x - 1800 = 0} \\ \\ : \implies \sf{x(x + 45) - 40(x + 45) = 0} \\ \\ : \implies \sf{(x - 40)(x + 45) = 0} \\ \\ : \implies \sf{x - 40 = 0 \: \: \: \: \: } \tt{or} \: \: \: \: \: \sf{x + 45 = 0} \\ \\ : \implies \sf{x = 40 \: \: \: } \tt{or} \: \: \: \sf{x = - 45\bigg(Rejected\bigg)}$

$\therefore \: \: \: \: \: \fbox{ Speed \: of \: the \: train = 40km/h.}$