# Thread: Distance, speed and time problem

1. ## Distance, speed and time problem

A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?

2. Originally Posted by saberteeth
A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?
Solve the following two equations simultaneously for x:

$\frac{x}{40} = t$ .... (1)

$\frac{x}{30} = t + \frac{1}{10}$ .... (2)

3. Originally Posted by saberteeth
A train running between two stations arrives at it's destination 10 minutes late when it travels at 40km/hr and 16 minutes late when it travels at 30km/hr. The distance between the two stations is ?
If the distance between the stations is $x$ km then the time of the journey at $40$ km/hr is $x/40$ hr, and at $30$ km/hr is $x/30$ hr.

So:

$\frac{x}{30} - \frac{x}{40} = \frac{1}{10}$

(that is the difference in journey times at the two speeds is $6$ minutes or $1/10$ of an hour).

CB

4. Originally Posted by CaptainBlack

(that is the difference in journey times at the two speeds is $6$ minutes or $1/10$ of an hour).
Thanks!

5. Hello, saberteeth!

A train running between two stations arrives at its destination
10 minutes late when it travels at 40 km/hr
and 16 minutes late when it travels at 30 km/hr.
Find the distance between the two stations.

Let $D$ = distance (in kilometers).
Let $t$ = time for a normal run (in hours).

We will use: . $\text{Distance} \:=\: \text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$

At 40 kph, it is 10 minutes $\left(\tfrac{1}{6}\text{ hr}\right)$ late.
. . $\frac{D}{40} \:=\:t + \frac{1}{6} \quad\Rightarrow\quad D \:=\:40t + \frac{20}{3}\quad [1]$

At 30 kph, it is 16 minutes $\left(\tfrac{4}{15}\text{ hr}\right)$ late.
. . $\frac{D}{30} \:=\:t + \frac{4}{15} \quad\Rightarrow\quad D \:=\:30t + 8\quad[2]$

Equate [1] and [2]: . $40t + \frac{20}{3} \:=\:30t + 8$

Hence: . $10t \:=\:\frac{4}{3} \quad\Rightarrow\quad t \:=\:\frac{2}{15}$

Substitute into [1]: . $D \:=\:40\left(\frac{2}{15}\right) + \frac{20}{3}$

Therefore: . $D \:=\:12\text{ km}$

But Captain Black has an elegant solution!