A car did a journey in t hours. Had the average speed been x kmph greater, the journey would have taken y hours less. How long was the journey?
x(t-y)ty -1
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:
- Defined variables for original speed and time.
- Expressed original distance using speed and time.
- Defined variables for new speed and time based on the given changes (\(x\) kmph greater speed, \(y\) hours less time).
- Expressed new distance using the new speed and time.
- Equated the original distance and new distance expressions because the journey is the same.
- Solved the resulting equation to find an expression for the original speed (\(v\)) in terms of \(x\), \(t\), and \(y\).
- Substituted the expression for \(v\) back into the original distance formula (\(D = vt\)).
- Simplified the resulting expression to find the total distance of the journey.
- Compared the derived distance formula with the given options to find the matching form.
| Variable/Formula | Description |
|---|---|
| Original Speed | \(v\) kmph |
| Original Time | \(t\) hours |
| Original Distance | \(D = vt\) km |
| New Speed | \(v+x\) kmph |
| New Time | \(t-y\) hours |
| New Distance | \(D = (v+x)(t-y)\) km |
| Distance Relationship | \(vt = (v+x)(t-y)\) |
| Original Speed \(v\) (derived) | \(v = \frac{x(t-y)}{y}\) kmph |
| Total Journey Distance \(D\) (derived) | \(D = \frac{x(t-y)t}{y} = x(t-y)ty^{-1}\) km |
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.
Similar Questions
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Important Questions from Relative Speed
In a circular race of 2500 m, a man and a woman start from a point towards opposite directions with speeds of 37 km/h and 35 km/h, respectively. After how much time from the start of the race will they meet for the first time?
- X and Yrun a 3 km race along a circular course of length 300 m. Their speeds are in the ratio 3 : 2. If they start together in the same direction, how many times would the first one pass the other (the start-off is not counted as passing)?
A train of length 600 meters passes another train of length 1000 meters moving in the opposite direction in 96 seconds. If the speed of both the trains is the same, then what is the speed of each train?
The distance between Delhi and Patna is $480$ km. A train starts from Delhi and travels towards Patna at a speed of $60$ kmph. Another train starts from Patna towards Delhi at a speed of $80$ kmph, but starts $1$ hour after the first train. In how much time, after the first train started, will they meet each other?
Ram traveling at the speed of 3 km/h reaches 15 minutes late. Had he walked at 4 km/h, he would have reached 15 minutes earlier. How much distance does Ram have to cover?
