Finding Distance Covered by Train

Video Explanation

Question

A train covers a certain distance at a uniform speed. If its speed were 10 km/h faster, it would take 2 hours less. If its speed were 10 km/h slower, it would take 3 hours more. Find the distance covered by the train.

Solution

Step 1: Concept

Time = Distance / Speed

Step 2: Let Variables

Let original speed = \(x\) km/h

Let distance = \(d\) km

Step 3: Form Equations

Original time = \( \frac{d}{x} \) — Faster case:

\[ \frac{d}{x + 10} = \frac{d}{x} – 2 \quad (1) \]

Slower case:

\[ \frac{d}{x – 10} = \frac{d}{x} + 3 \quad (2) \]

Step 4: Convert into Linear Form

Let:

\[ a = \frac{d}{x} \]

Then:

\[ \frac{d}{x + 10} = \frac{ax}{x + 10}, \quad \frac{d}{x – 10} = \frac{ax}{x – 10} \]

— From (1):

\[ \frac{ax}{x + 10} = a – 2 \]

\[ ax = (a – 2)(x + 10) \]

\[ ax = ax + 10a – 2x – 20 \]

Cancel \(ax\):

\[ 0 = 10a – 2x – 20 \]

\[ 5a – x = 10 \quad (3) \]

— From (2):

\[ \frac{ax}{x – 10} = a + 3 \]

\[ ax = (a + 3)(x – 10) \]

\[ ax = ax – 10a + 3x – 30 \]

Cancel \(ax\):

\[ 0 = -10a + 3x – 30 \]

\[ 3x – 10a = 30 \quad (4) \]

Step 5: Solve Linear Equations

\[ 5a – x = 10 \]

\[ 3x – 10a = 30 \]

Multiply first by 3:

\[ 15a – 3x = 30 \quad (5) \]

Add (4) and (5):

\[ 5a = 60 \]

\[ a = 12 \]

Substitute into (3):

\[ 5(12) – x = 10 \]

\[ 60 – x = 10 \]

\[ x = 50 \]

Step 6: Find Distance

\[ a = \frac{d}{x} = 12 \Rightarrow d = 12 \times 50 = 600 \]

Conclusion

\[ \text{Distance} = 600 \text{ km} \]

Verification

Faster speed = 60 km/h → time = 10 hrs

Original = 12 hrs → 2 hours less ✔

Slower speed = 40 km/h → time = 15 hrs

Original = 12 hrs → 3 hours more ✔