Arithmetic

Time, Speed & Distance

Medium

Train $T$ leaves station $X$ for station $Y$ at $3 \mathrm{pm}$ . Train $S$ , traveling at three quarters of the speed of $T$ , leaves $Y$ for $X$ at $4 \mathrm{pm}$ . The two trains pass each other at a station $Z$ , where the distance between $X$ and $Z$ is three-fifths of that between $X$ and $Y$ . How many hours does train $T$ take for its journey from $X$ to $Y$ ?

Entered answer:

✅ Correct Answer: 15

We have two trains traveling toward each other on the same route. When they meet at point Z, they've been traveling for different amounts of time due to the 1-hour head start.

Given Information:

Train T leaves X at 3 pm, heading to Y

Train S leaves Y at 4 pm, heading to X

Speed of S = $\tfrac{3}{4}$ × Speed of T

They meet at station Z

Distance XZ = $\tfrac{3}{5}$ × Distance XY

Let's define:

Speed of train T = $v_T$

Speed of train S = $v_S = \tfrac{3}{4}v_T$

Time taken by S to reach Z = $t$ hours

Time taken by T to reach Z = $(t+1)$ hours

Since T starts 1 hour before S, when they meet, T has been traveling for 1 hour more than S.

Since XZ is three-fifths of XY, we can write:

XZ = $\tfrac{3}{5}$ × XY

This means YZ = XY - XZ = XY - $\tfrac{3}{5}$ XY = $\tfrac{2}{5}$ XY

Key Insight: The ratio XZ : YZ = 3 : 2

For the meeting point Z:

Distance XZ = $v_T(t+1)$ (distance covered by train T)

Distance YZ = $v_S \times t = \tfrac{3}{4}v_T \times t$ (distance covered by train S)

Since XZ : YZ = 3 : 2, we can write:

$\dfrac{XZ}{YZ} = \dfrac{3}{2}$

Substituting our expressions:

$\dfrac{v_T(t+1)}{\tfrac{3}{4}v_T \times t} = \dfrac{3}{2}$

Simplifying the left side:

$\dfrac{v_T(t+1)}{\tfrac{3}{4}v_T \times t} = \dfrac{4(t+1)}{3t}$

$\dfrac{4(t+1)}{3t} = \dfrac{3}{2}$

Cross-multiplying:

$4(t+1) \times 2 = 3t \times 3$

$8(t+1) = 9t$

$8t + 8 = 9t$

$8 = t$

Therefore: $t = 8$ hours

We now know:

Train S takes 8 hours to travel from Y to Z

Train T takes 9 hours to travel from X to Z

Finding the relationship:

Distance XZ = $v_T \times 9$

Since XZ = $\tfrac{3}{5}$ × XY, we have: $v_T \times 9 = \tfrac{3}{5} \times XY$

Solving for XY:

$XY = \dfrac{9v_T \times 5}{3} = 15v_T$

Time for T's complete journey:

$\text{Time} = \dfrac{XY}{v_T} = \dfrac{15v_T}{v_T} = 15 \text{ hours}$

Train T takes 15 hours for its journey from X to Y.

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