Try This If You Are Bored Waiting For The Train

This is a very nice question. Here, we approach the problem with a set of formulae involving relative speed, although I believe a solution involving graphs is also possible. Let the velocity of the train be $v$ & that of the two girls be $c$ . For Shina, who walks in the same direction as that of the train, relative speed is $v-c$ . Let the time in which the train passes her be $t_1$ . It is given $ct_1 = 45 \Rightarrow t_1=\frac{45}{c}$ . Similarly for Belle, relative speed is $v+c$ & the time in which train passes her is $\frac{30}{c}$ . We know

$(v-c)\frac{45}{c} = (v+c)\frac{30}{c} = l$ . Here we evaluate $\frac{v}{c}$ from the two equations, & equate them, getting, $l= \boxed{180}$ .

Let x be speed of the girls. Since 30 m of train will be behind the point from where they started. When last wagon reaches belle shina wouldh Have travelled 30 m on the other Side. So the train travels 75 m in the same as Shina covers 15 m. So train, s speed is 5x. Solving from here we get length of train as 180 m

$(V_{train}$ - $V_{girl}$ )* $\frac{45}{V_{girl}}$ =L

$(V_{train}$ + $V_{girl}$ )* $\frac{30}{V_{girl}}$ =L

where L is the length of the train.

Solve the equations to get $\frac{V_{train}}{V_{girl}}$ =5 and L=180

Let the speed at which Shina and Belle frolick around the train station in their respective directions be $x$ m/s. Let the speed of the train by $y$ m/s. Let the length of the train be $l$ metres.

Since Belle walks in the opposite direction to that of the train, by the time Belle walks $30$ metres, the train will have moved $l-30$ metres. The time taken for both is the same, and $\large time=\frac{distance}{speed}$ so we obtain the following equation: $\large\frac{30}{x}=\frac{l-30}{y}$ Rearranging gives us: $\large\frac{y}{x}=\frac{l-30}{30}$

Since Shina walks in the same direction as that of the train, by the time Shina walks $45$ metres, the train will have moved $l+45$ metres. The time taken for both is the same, so we obtain the following equation: $\large\frac{45}{x}=\frac{l+45}{y}$ Rearranging gives us: $\large\frac{y}{x}=\frac{l+45}{45}$

Equating the two equations gives us: $\large\frac{l-30}{30}=\frac{l+45}{45}$

Cross-multiplying gives us: $45l-45\times30=30l+45\times30$

Rearranging gives us: $45l-30l=2\times45\times30$

$15l=2\times45\times30$

$\therefore l=\large 180$

A easy and nice question.. let the speed of shina and belle be x and speed of train be y distance of train be D Firstly speed of belle in respect to train will be (x+y) and, speed of shina in respect to train will be (y-x)

there fore eqn. will be :-

dx/(y+x) = 30 .................................1

dx/(y-x) = 45 ..................................2

by solving these equations we get D = 180