To solve the problem, let’s assume the following:

• The uniform speed of the train is $v$ km/h.
• The length of the journey is $d$ km.
• The scheduled time to cover the distance at the uniform speed is $t$ hours.

From basic relations in kinematics:

$\text{Distance} = \text{Speed} \times \text{Time}$ $d = v \cdot t \quad \text{or} \quad t = \frac{d}{v}.$

Now, the problem provides two conditions:

1. If the train were 6 km/h faster, it would take 4 hours less to cover the journey: $t - 4 = \frac{d}{v + 6}.$

2. If the train were 6 km/h slower, it would take 6 hours more to cover the journey: $t + 6 = \frac{d}{v - 6}.$

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Step 1: Express $t$ from the first condition

We know $t = \frac{d}{v}$. Substituting into the first condition:

$\frac{d}{v} - 4 = \frac{d}{v + 6}.$

Multiply through by $v(v + 6)$ to clear the denominators (assuming $v > 0$):

$d(v + 6) - 4v(v + 6) = dv.$

Simplify and distribute terms:

$dv + 6d - 4v^2 - 24v = dv.$

Cancel out $dv$ from both sides:

$6d = 4v^2 + 24v.$

Rearranging gives:

4v^2 + 24v - 6d = 0. \tag{1}

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Step 2: Express $t$ from the second condition

From the second condition:

$\frac{d}{v} + 6 = \frac{d}{v - 6}.$

Substitute $t = \frac{d}{v}$, and multiply through by $v(v - 6)$:

$d(v - 6) + 6v(v - 6) = dv.$

Simplify:

$dv - 6d + 6v^2 - 36v = dv.$

Cancel $dv$ on both sides:

$-6d = -6v^2 + 36v.$

Divide through by $-6$ to simplify:

d = v^2 - 6v. \tag{2}

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Step 3: Solve the system of equations

We now have two equations:

1. $4v^2 + 24v - 6d = 0$
2. $d = v^2 - 6v$.

Substitute $d = v^2 - 6v$ into Equation (1):

$4v^2 + 24v - 6(v^2 - 6v) = 0.$

Simplify:

$4v^2 + 24v - 6v^2 + 36v = 0.$

Combine like terms:

$-2v^2 + 60v = 0.$

Factorize:

$-2v(v - 30) = 0.$

Thus:

$v = 0 \quad \text{or} \quad v = 30.$

Since $v = 0$ is not physically meaningful, we have:

$v = 30 \, \text{km/h}.$

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Step 4: Find $d$ (the length of the journey)

Substitute $v = 30$ into $d = v^2 - 6v$:

$d = 30^2 - 6(30).$

Simplify:

$d = 900 - 180 = 720 \, \text{km}.$

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Final Answer:

The length of the journey is 720 km.

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Verification:

1. Scheduled time at 30 km/h: $t = \frac{720}{30} = 24 \, \text{hours}$.
2. Faster speed (36 km/h): Time = $\frac{720}{36} = 20 \, \text{hours}$ (4 hours less).
3. Slower speed (24 km/h): Time = $\frac{720}{24} = 30 \, \text{hours}$ (6 hours more).

Everything checks out!

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Related Questions:

1. What happens to the time if the speed increases by 10 km/h instead of 6 km/h?
2. Derive a general formula for the distance based on the given conditions.
3. How would the problem change if the time difference were 5 hours instead of 4 and 6?
4. If the train’s speed doubles, how much time will it take for the journey?
5. What is the effect of reducing the speed by 20% on the time taken for the journey?

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Tip:

Always use algebraic equations to model word problems and ensure careful substitution and simplification for accuracy.