A train was scheduled to cover the distance between two cities in 10 hours. After the first 9 hours at the planned speed it reduced its speed by 7 km/h and arrived 6 minutes late. Find the train's original speed.

Step-by-step solution

1. 1

   Let the unknown be the speed, and express the distance through it. Set $v$ = original speed in km/h. Since the plan was $10$ hours at speed $v$, the total distance is

   $S=10v$

   Choosing $v$ rather than $S$ as the unknown is what keeps the algebra linear — the distance then cancels out on its own.

2. 2

   Split the journey into the two phases. In the first $9$ hours the train covers $9v$ km, so the remaining distance is

   $10v-9v=v\ \text{km}$

   This is the neat consequence of the setup: the leftover leg is numerically equal to the speed.

3. 3

   Write the delay as an equation. The remaining leg is driven at $v-7$ km/h, taking $\dfrac{v}{v-7}$ hours. Six minutes is $\tfrac{6}{60}=\tfrac{1}{10}$ hour, so the actual trip took $10+\tfrac1{10}=\tfrac{101}{10}$ hours:

   $9+\frac{v}{v-7}=\frac{101}{10}$

4. 4

   Solve the resulting rational equation. Subtract $9$:

   $\frac{v}{v-7}=\frac{101}{10}-\frac{90}{10}=\frac{11}{10}$

   Cross-multiplying, $10v=11(v-7)=11v-77$, so $v=77$ km/h. The extraneous case $v=7$ is excluded because it makes the reduced speed zero.

5. 5

   Check against the original story. Distance $=10\cdot77=770$ km. First $9$ hours cover $693$ km, leaving $77$ km at $77-7=70$ km/h, which takes $77/70=1.1$ h $=1$ h $6$ min. Total $9+1.1=10.1$ h, i.e. exactly $6$ minutes past the planned $10$ hours ✓.