Question

A train travel 360 km at a uniform speed. if the speed had been 5km/h more , it would have taken 1hours less for the same journo. find the speed of the train.​

Answer 1

$\begin{gathered}\Large{\bold{\purple{\underline{CaLcUlAtIoN\::}}}} \\ \end{gathered}$

Case :- 1

☆ Let speed of the train be x km/ hour.

☆ Distance travel = 360 km

☆ So, time taken to travel 360 km with speed of 'x' km/hr is given by

$\tt \:  \longrightarrow \boxed{ \tt \purple{\: t_1 \: = \: \dfrac{360}{x} \: hours}}$

Case :- 2

☆ Let speed of the train be (x + 5) km/ hour.

☆ Distance travel = 360 km

☆ So, time taken to travel 360 km with speed of 'x + 5' km/hr is given by

$\tt \:  \longrightarrow \boxed{ \purple{ \tt\: t_2 \: = \: \dfrac{360}{x + 5} \: hours}}$

$\begin{gathered}\bf\red{According \: to \: statement}\end{gathered}$

$\tt \:  \longrightarrow \: t_1 \: - \: t_2 \: = 1$

$\tt \:  \longrightarrow \: \dfrac{360}{x} - \dfrac{360}{x + 5} = 1$

$\tt \:  \longrightarrow \: \dfrac{360x + 1800 - 360x}{x(x + 5)} = 1$

$\tt \:  \longrightarrow \: {x}^{2} + 5x = 1800$

$\tt \:  \longrightarrow \: {x}^{2} + 5x - 1800 = 0$

$\tt \:  \longrightarrow \: {x}^{2} + 45x - 40x - 1800 = 0$

$\tt \:  \longrightarrow \: x(x + 45) - 40(x + 45) = 0$

$\tt \:  \longrightarrow \: (x + 45)(x - 40) = 0$

$\tt \:  \longrightarrow \: x = 40 \: or \: x \: = - 45 \: (rejected)$

$\tt\implies \:speed \: of \: train \: = \: 40 \: km \: per \: hour$