A train travelling from town $A$ to town $B$ meets with an accident after 1 hour. The train is stopped for 30 minutes, after which....(AHSME 1955)

Question

I was trying to figure out this question:

A train travelling from town $A$ to town $B$ meets with an accident after 1 hour. The train is stopped for 30 minutes, after which it proceeds at four-fifths of its usual rate, arriving at town $B$ 2 hours late. If the train had covered $80$ miles more before the accident, it would have just been one hour late. What is the usual rate of the train?

$T=$ time(usual time)

$D=$ distance

$S=$ speed

So first:

$(T+2)-t=2$(2 hours late)

$T+2=$distance/time so:

In the first hour it travels $s*1=d$, or $s$ distance in 1hr

The rest, it gets slowed down by $4/5$ of usual speed, so it already travelled $S$ distance, so the rest of the distance = $d-s$

So $t+2 = d-s/4/5*s+1$

$t = d/s$

so like that I did the second equation as well but added $80$ with $s$, it travels $80$ miles more in $1$ hour for being an hour earlier than the previous one. Thus more speed.

So my $2$ equations are shown below in the image: Image of my 2 equations

It is showing me no solution when I put it to Photomath. What have I done wrong?

Answer 1

Actually the problem can be very simply solved, (actually mentally), and if you are ever planning to appear for tests for Management, etc, you must be alert for methods to solve such problems as simply as possible.

Let V mph be the original speed of the train

In the $80$ extra miles it travels at full speed, the $\frac15V$ extra speed representing $16$ miles helps to cut delay by $1$ hr. at $\frac45V$ speed .

Thus $16 = \frac45V\times 1 \Longrightarrow V = 20\, mph$

Answer 2

Let $D$ be the distance between towns $A$ and $B$. Let $s$ be the usual speed of the train, and let $T$ be the usual time it takes the train to get from $A$ to $B$.

Firstly, we have

$ D = s T \tag{1}$

Secondly, we're told that the train is delayed $2$ hours, after a $30min$ pause because of the accident, which happens $1$ hour after leaving town $A$, and because its speed is reduced to $\dfrac{4}{5} s$ after the accident.

Hence,

$ D = s \times 1 + \dfrac{4}{5} s (T + 2 - 1 - 0.5) = s + \dfrac{4}{5} s (T + 0.5) \tag{2}$

Thirdly, we're told that if the train had travelled $80 \ miles$ more before the accident, AND paused for $30 min$, then continued at $\dfrac{4}{5} s$ of its usual speed, that it would be delayed by $1$ hour.

Hence,

$$ D = s (\dfrac{80}{s}+ 1) + \dfrac{4}{5} s ( T + 1 - \left(\dfrac{80}{s}+1\right) - 0.5 ) $$

This is equal to

$ D = 16 + \dfrac{4}{5} s T + \dfrac{3}{5} s \tag{3}$

Substituting $(1)$ in $(2)$ we get

$$ s T = s + \dfrac{4}{5} s (T + 0.5) $$

Dividing by $s$, we get

$$ T = 1+\dfrac{4}{5} (T + 0.5) $$

So

$$ \dfrac{T}{5} = 1.4 $$

$$ T = 7 \hspace{3pt} hours $$

Substituting $(1)$ and $T$ into $(3)$, we get

$$ 7 s = 16 + \dfrac{28}{5} s + \dfrac{3}{5} s $$

From which

$$ s = 20 \ \ mph $$

Verification of solution:

The total distance between town $A$ and town $B$ is

$ D = s T = 20 (7) = 140 \ miles$

In the first scenario, the train travels for $1 hour$ covering a distance of

$ d_1 = s (1) = 20 \ miles $

It is delayed by $0.5 \ hour$ and then continues at $\frac{4}{5} (20) = 16 \ mph $ for the remaining time which is $ T + 2 - 1.5 = 7.5 \ hours $. The second distance travelled is

$ d_2 = 7.5 (16) = 120 \ miles $

Thus the total distance travelled is $20 + 120 = 140 = D$

In the second scenario, the train first travels $d_1 = 20 \ miles$ in an hour, and then travels an additional $80 \ miles$ and this takes an additional $ \dfrac{80}{20} = 4 \ hours $. So total time for first part of the trip is $ 5 \ hours$. It stops for $0.5 hour$, making the total time elapsed before it continues $5.5 \ hours$. We're given that the total trip is delayed by $1 \ hour$, so the time for the second part of the trip (after the $30\text{-minute wait}$) is $7 + 1 - 5.5 = 2.5 \ hours $. Therefore the distance travelled in the $2.5 \ hours$ is

$d_2 = 2.5 \left( \dfrac{4}{5} \right) \left(20 \right) = 40 \text{ miles} $

Thus the total distance travelled is $20 + 80 + 40 = 140 = D$