Question

After covering a distance of 30 km with a uniform speed there is some defect in a train engine and therefore, its speed is reduced to $\frac{4}{5}$ of its original speed. Consequently, the train reaches its destination late by 45 minutes. Had it happened after covering 18 kilometers more, the train would have reached 9 minutes earlier. Find the speed of the train and the distance of the journey.

• A. Speed=40 km/hr, Distance=160 km
• B. Speed=20km/hr, Distance=100km.
• C. Speed=30km/hr., Distance=120 km.
• D. Speed=35km/hr., Distance=140 km.

Answer 1

The correct option is C

Speed=30km/hr., Distance=120 km.

Step 1: Use the given information to obtain the equations.

Let the speed of a train be $p$ km/hr, and the length of the journey be $q$ km.

Time taken by the train is $\frac{q}{p}$.

Here, the speed of the train for the first 30 km is $p$ km/hr.

After covering a distance of 30 km. with uniform speed, there is some defect in the train engine. Therefore, its speed is reduced to $\frac{4}{5}$ of its original speed.

The speed of the train for the remaining journey is $\left(q-30\right)=\frac{4p}{5}\text{km/hr}$.

Time taken by the train to cover 30kms equal to $\frac{30}{p}$ and that of remaining $\left(q-30\right)\text{km}$ equal to $\frac{5\left(q-30\right)}{4p}$.

Now, the train reaches its destination late by 45 minutes i.e., $\left(\frac{45}{60}hours\right)$.

$\begin{array}{rcl}& \Rightarrow & \frac{30}{p}+\frac{5}{4p}\left(q-30\right)=\frac{q}{p}+\frac{45}{60}\\ & \Rightarrow & \frac{30}{p}+\frac{5q-150}{4p}=\frac{q}{p}+\frac{3}{4}\\ & \Rightarrow & 120+5q-150=4q+3p\\ & \Rightarrow & 3p-q=-30...\left(1\right)\end{array}$

After covering 18 kilometers more, the train would have reached 9 minutes earlier.

Here, the speed of the train for the first 48 km is $p$ km/hr.

After covering a distance of 30 km. with uniform speed, there is some defect in the train engine. Therefore, its speed is reduced to $\frac{4}{5}$ of its original speed.

The speed of the train for the remaining journey: $\left(q-48\right)\text{km}=\frac{4}{5}\text{km/hr}$

Time taken by the train to cover 48kms is equal to $\frac{48}{p}$ and that of remaining $\left(q-48\right)\text{km}$ equal to $\frac{5\left(q-48\right)}{4p}$.

Now, the train reaches its destination earlier by 9 minutes i.e., $\left(\frac{36}{60}\text{hours}\right)$.

$\begin{array}{rcl}& \Rightarrow & \frac{48}{p}+\frac{5}{4p}\left(q-48\right)=\frac{q}{p}+\frac{36}{60}\\ & \Rightarrow & \frac{48}{p}+\frac{5q-240}{4p}=\frac{q}{p}+\frac{3}{5}\\ & \Rightarrow & \frac{192+5q-240}{4p}=\frac{5q+3p}{5p}\\ & \Rightarrow & \frac{5q-48}{4p}=\frac{5q+3p}{5p}\\ & \Rightarrow & 25q-240=20q+12p\\ & \Rightarrow & 12p-5q=-240...\left(2\right)\end{array}$

Step 2: Solve the system of equations by the elimination method.

Multiply the equation $\left(1\right)$ by $5$ to get:

$15p-5q=-150...\left(3\right)$

Now, subtract the equation $\left(2\right)\text{from}\left(3\right)$ to get:

$\begin{array}{rcl}& \Rightarrow & 3p=90\\ & \Rightarrow & p=30\end{array}$

By equation (2),

$\begin{array}{rcl}& \Rightarrow & q=\frac{12\left(30\right)+240}{5}\\ & \Rightarrow & q=\frac{600}{5}\\ & \Rightarrow & q=120\end{array}$

Thus, the speed of the .train is $30\text{km/hr}$ and the length of the journey is $120\text{km}$.

Hence, option (c) is correct.