# Train journey problem

• Feb 25th 2010, 05:09 AM
Natasha1
Train journey problem
I can't do this at all, please explain clearly. Many many thanks!

A train does the first part of a journey at 30mph, then the second part at 60 mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?
• Feb 25th 2010, 07:02 AM
u2_wa
Quote:

Originally Posted by Natasha1
I can't do this at all, please explain clearly. Many many thanks!

A train does the first part of a journey at 30mph, then the second part at 60 mph, and the average speed for the whole journey is 42 6/17mph (this is made of a whole number 42 and a fraction 6/17). What fraction of the distance of the whole journey is the first (30mph) part?

I got $5/12$ and this is how I arrived at my answer:

T=time for journey with speed 30Kmph
t=time for journey with speed 60Kmph

Think of two trains: train 1 is moving at avg. speed and Train 2 is the train given in the question.

During first part the difference in the distance covered by these trains would be:
(42 6/17 - 30)T

During second part the difference in the distance covered by these trains would be:
(60 - 42 6/17)t

As we know both trains cover the same distance finally so net effect of the differences would be zero.

(42 6/17 - 30)T-(60 - 42 6/17)t $=0$

$T/t=10/7$
$D/d=(10*30)/(7*60)$ (I multiplied time with speed to calculate distance)

$D/(D+d)=5/12$
• Feb 26th 2010, 10:20 PM
pflo
Quote:

Originally Posted by u2_wa
I got $5/12$

I concur, using a different method.

$d_{1}$ is the distance of the first part
$t_{1}$ is the time of the first part

$d_{2}$ is the distance of the second part
$t_{2}$ is the time of the second part

$D$ is the total distance and $D=d_1+d_2$
$T$ is the total time and $T=t_1+t_2$

The question is asking for $\frac{d_1}{D}$ and we know $\frac{d_1}{D}+\frac{d_2}_D=1$.

Simply using $d=rt$, you can come up with the following formula: $30t_1+60t_2=42\frac{6}{7}T$. You can solve this equation for [tex]\frac{t_2}{T}[tex]:

$30t_1+30t_2+30t_2=42\frac{6}{7}T$
$30(t_1+t_2)+30t_2=42\frac{6}{7}T$
$30T+30t_2=42\frac{6}{7}T$
$30t_2=12\frac{6}{7}T$
$\frac{t_2}{T}=\frac{7}{17}$

So, $\frac{t_1}{T}=\frac{10}{17}$

Again using $d=rt$, you can perform the following calculation:
$\frac{d_1}{D}=\frac{30*10}{42\frac{6}{17}*17}=\fra c{5}{12}$.