Understanding the Car Journey Problem

This problem asks us to find the total distance of a car journey. We are given the original time taken (\(t\)) and told how the journey time would change (\(y\) hours less) if the average speed were increased by a certain amount (\(x\) kmph).

We need to use the fundamental relationship between distance, speed, and time:

Distance = Speed \( \times \) Time

Let's define the variables based on the problem description:

• Original time = \(t\) hours
• Let the original speed = \(v\) kmph
• Original distance = \(v \times t\) km

Now consider the hypothetical situation:

• New speed = Original speed + \(x\) kmph = \(v + x\) kmph
• New time = Original time - \(y\) hours = \(t - y\) hours
• New distance = \((v + x) \times (t - y)\) km

Since the journey is the same in both cases, the distance must be equal.

Setting Up the Equation

We can set the expressions for the original distance and the new distance equal to each other:

\(vt = (v + x)(t - y)\)

Solving for the Original Speed

We need to find the distance, which is \(vt\). However, we don't know \(v\). We can use the equation above to solve for \(v\) in terms of \(x\), \(t\), and \(y\).

Expand the right side of the equation:

\(vt = vt - vy + xt - xy\)

Subtract \(vt\) from both sides of the equation:

\(0 = -vy + xt - xy\)

Move the term with \(v\) to the left side:

\(vy = xt - xy\)

Factor out \(x\) on the right side:

\(vy = x(t - y)\)

Now, solve for \(v\) by dividing both sides by \(y\):

\(v = \frac{x(t - y)}{y}\)

Calculating the Journey Distance

The total distance of the journey is the original distance, which is \(vt\). Now that we have an expression for \(v\), we can substitute it into the distance formula:

Distance \(D = vt\)

\(D = \left(\frac{x(t - y)}{y}\right) \times t\)

Multiply the terms:

\(D = \frac{x(t - y)t}{y}\)

This expression gives the total distance of the car journey in terms of \(x\), \(t\), and \(y\).

This can also be written using negative exponents for \(y\):

\(D = x(t - y)t y^{-1}\)

Verifying the Solution Form

Let's compare our derived distance formula to the options provided.

• Option 1: \(x(t-y)ty\)
• Option 2: \(x(t-y)ty^{-1}\)
• Option 3: \(x(t-y)ty^{-2}\)
• Option 4: \(x(t+y)ty\)

Our derived formula \(D = x(t - y)t y^{-1}\) matches the form of Option 2.

Thus, the length of the journey was \(x(t - y)t y^{-1}\) km.

Summary of Steps

Here's a quick recap of how we solved the problem:

1. Defined variables for original speed and time.
2. Expressed original distance using speed and time.
3. Defined variables for new speed and time based on the given changes (\(x\) kmph greater speed, \(y\) hours less time).
4. Expressed new distance using the new speed and time.
5. Equated the original distance and new distance expressions because the journey is the same.
6. Solved the resulting equation to find an expression for the original speed (\(v\)) in terms of \(x\), \(t\), and \(y\).
7. Substituted the expression for \(v\) back into the original distance formula (\(D = vt\)).
8. Simplified the resulting expression to find the total distance of the journey.
9. Compared the derived distance formula with the given options to find the matching form.

Revision Table: Car Journey Calculations

Reviewing the core concepts used in this car journey calculation problem:

• The direct relationship: Distance = Speed \( \times \) Time.
• If speed increases, time decreases for a fixed distance.
• Setting up equations based on given conditions is crucial.
• Solving algebraic equations to find unknown variables is a key skill.

Additional Information: Speed, Time, and Distance

Problems involving speed, time, and distance often require setting up equations to represent different scenarios of the same journey. The core formula \(D = S \times T\) (Distance = Speed \( \times \) Time) can be rearranged as \(S = D/T\) or \(T = D/S\). Understanding how changes in one variable affect others while distance remains constant is important for solving these types of quantitative aptitude questions.

For example, if a journey takes time \(T_1\) at speed \(S_1\) and the same journey takes time \(T_2\) at speed \(S_2\), then \(S_1 T_1 = S_2 T_2\).

In our problem, the original speed is \(v\) and time is \(t\), and the new speed is \((v+x)\) and time is \((t-y)\). So, \(vt = (v+x)(t-y)\), which is exactly the equation we started with.