# Two Train speeds?

## Two trains stations, A and B, are 300km apart. Two trains leave A and B simultaneously and proceed at constant speeds to other station. The train from A reaches station B nine hours after trains have met, and the train from B reaches the station A four hours after the trains have met. Find the speed of each train?

Aug 10, 2018

One solution is ${v}_{A} = 20 k m {h}^{-} 1$ and ${v}_{B} = 30 k m {h}^{-} 1$

#### Explanation:

Let the speeds of the train be $= {v}_{A}$ and $= {v}_{B}$

The time taken by train $A$ to reach $B$ is

${t}_{A} = \frac{300}{v} _ A$

The time taken by train $B$ to reach $A$ is

${t}_{B} = \frac{300}{v} _ B$

Let the meeting point be $x$ km from $A$

The time taken by train $A$ to reach the meeting point with $B$ is

${t}_{1 A} = \frac{x}{v} _ A$

The time taken by train $B$ to reach the meeting point with $A$ is

${t}_{1 B} = \frac{300 - x}{v} _ B$

Then

${t}_{A} = {t}_{1 A} + 9 = \frac{x}{v} _ A + 9$

and

${t}_{B} = {t}_{1 B} + 4 = \frac{300 - x}{v} _ B + 4$

$\frac{300}{v} _ A = \frac{x}{v} _ A + 9$.......................$\left(1\right)$

$\frac{300}{v} _ B = \frac{300 - x}{v} _ B + 4$..........................$\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$

$300 = x + 9 {v}_{A}$

$x = 300 - 9 {v}_{A}$

$\frac{300}{v} _ B = \frac{300 - \left(300 - 9 {v}_{A}\right)}{v} _ B + 4$

$300 = 9 {v}_{A} + 4 {v}_{B}$

This is a Diophantine equation

One solution is

${v}_{A} = 20 k m {h}^{-} 1$

${v}_{B} = 30 k m {h}^{-} 1$

graph{(9x+4y-300)=0 [-104.4, 106.45, -27.4, 78.1]}