## 1 Answer

Let the distance between station $X$ and $Y$ be $`d\text{’} \; \text{km}.$

We can draw the diagram for clear visualization.

Let $`k\text{’} \; \text{km/hr}$ be the speed of train $\text{T}.$

Then the speed of train $\text{S} = \frac{3}{4}k \; \text{km/hr}$

The distance between $X$ and $Z = \frac{3}{5}d \; \text{km}$

Then the distance between $Y$ and $Z= d – \frac{3d}{5} = \frac{2d}{5} \; \text{km}$

Let $`t\text{’} \; \text{hours}$ be the time taken by train $\text{T}$ to reach station $\text{Z}.$

Then, time taken by train $\text{S}$ to reach station $\text{Z} = (t-1) \; \text{hours}$

We know that, $ \boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

The speed of train $\text{T}: k = \frac{ \frac{3d}{5}}{t} $

$ \Rightarrow \boxed {k = \frac{3}{5} \left( \frac{d}{t} \right)} \quad \longrightarrow (1) $

The speed of train $\text{S}: \frac{3K}{4} = \frac{\frac{2d}{5}}{(t-1)} $

$ \Rightarrow \frac{3k}{4} = \frac{2}{5} \left[ \frac{d}{(t-1)} \right] $

Put the value of $`k\text{’}$ from the equation $(1),$ we get

$ \frac{3}{4} \left[ \frac{3}{5} \left( \frac{d}{t} \right) \right] = \frac{2}{5} \left [ \frac{d}{(t-1)} \right] $

$ \Rightarrow \frac{9}{4t} = \frac{2}{(t-1)}$

$ \Rightarrow 9(t-1) = 8t$

$ \Rightarrow 9t – 9 = 8t $

$ \Rightarrow \boxed{t = 9 \; \text {hours}} $

Put the value of $`t\text{’}$ in the equation $(1),$ we get

The speed of train $\text{S} : k = \frac{3d}{5 \times 9} \; \text{km/hr}$

$\Rightarrow k = \frac {d}{15} \; \text{km/hr}$

The time taken by train $\text{T}$ to reach $Y$ from $X = \frac{d}{\frac{d}{15}} = 15 \; \text{hours.}$

$\therefore$ Time taken by train $\text{T}$ to complete journey from $X$ to $Y\;\text{is}\; 15 \; \text{hours.}$

Correct Answer $: 15 $