Solving the Train Speed and Time Problem

This problem involves the relationship between distance, speed, and time for a train journey. We are given information about how a change in speed affects the time taken for a specific distance and asked to find the time taken for a different distance at the original speed.

Understanding the Problem

Let's denote the usual speed of the train as \(v\) km/h and the usual time taken for the 300 km journey as \(T\) hours. The relationship between distance, speed, and time is: Distance = Speed × Time.

So, for the usual journey of 300 km:

\(300 = v \times T\)

When the speed is increased by 20 km/h, the new speed is \(v + 20\) km/h. The train takes \(2\frac{1}{2}\) hours less time for the same 300 km journey. \(2\frac{1}{2}\) hours is equal to 2.5 hours. The new time taken is \(T - 2.5\) hours.

For the journey with increased speed:

\(300 = (v + 20) \times (T - 2.5)\)

Setting up the Equations

From the first equation, we can express \(T\) in terms of \(v\):

\(T = \frac{300}{v}\)

Substitute this expression for \(T\) into the second equation:

\(300 = (v + 20) \times \left(\frac{300}{v} - 2.5\right)\)

Solving for the Usual Speed

Now we need to solve this equation for \(v\):

• Expand the right side of the equation:
• \(300 = v \times \frac{300}{v} + v \times (-2.5) + 20 \times \frac{300}{v} + 20 \times (-2.5)\)
• \(300 = 300 - 2.5v + \frac{6000}{v} - 50\)
• Subtract 300 from both sides:
• \(0 = -2.5v + \frac{6000}{v} - 50\)
• Rearrange the terms to form a standard quadratic equation. Multiply the entire equation by \(v\) to eliminate the fraction (assuming \(v \neq 0\), which is true for speed):
• \(0 \times v = -2.5v \times v + \frac{6000}{v} \times v - 50 \times v\)
• \(0 = -2.5v^2 + 6000 - 50v\)
• Move all terms to one side to make the coefficient of \(v^2\) positive:
• \(2.5v^2 + 50v - 6000 = 0\)
• To simplify, multiply the entire equation by 2 (or divide by 2.5):
• \(5v^2 + 100v - 12000 = 0\)
• Now, divide by 5:
• \(v^2 + 20v - 2400 = 0\)

This is a quadratic equation in the form \(av^2 + bv + c = 0\). We can solve this by factoring or using the quadratic formula.

Using factoring, we look for two numbers that multiply to -2400 and add up to 20. These numbers are 60 and -40.

• So, the equation can be factored as:
• \((v + 60)(v - 40) = 0\)
• This gives two possible solutions for \(v\):
• \(v + 60 = 0 \implies v = -60\)
• \(v - 40 = 0 \implies v = 40\)

Since speed cannot be a negative value, the usual speed of the train is \(v = 40\) km/h.

Calculating Time for 192 km at Usual Speed

The question asks for the time the train will take to cover a distance of 192 km at its usual speed (40 km/h).

• Distance = 192 km
• Speed = 40 km/h
• Time = \(\frac{\text{Distance}}{\text{Speed}}\)
• Time = \(\frac{192}{40}\) hours

Let's perform the division:

\(\frac{192}{40} = \frac{19.2 \times 10}{4 \times 10} = \frac{19.2}{4}\)

\(\frac{19.2}{4} = \frac{16 + 3.2}{4} = \frac{16}{4} + \frac{3.2}{4} = 4 + 0.8 = 4.8\)

So, the time taken to cover 192 km at the usual speed is 4.8 hours.

The final answer is 4.8 hours.

Revision Table: Train Speed and Time

Additional Information: Time and Distance Problems

Time and distance problems are common in quantitative aptitude and physics. They often involve understanding how changes in speed or distance affect the travel time. Key concepts include:

• Constant Speed: If speed is constant, time is directly proportional to distance, and distance is directly proportional to speed (for a fixed time).
• Average Speed: When speed varies over a journey, average speed is the total distance divided by the total time. It is NOT necessarily the average of different speeds.
• Relative Speed: Used when dealing with two moving objects (e.g., trains moving towards or away from each other). Relative speed is the sum of speeds if moving towards each other or in opposite directions, and the difference of speeds if moving in the same direction.
• Solving with Equations: Many problems can be solved by setting up algebraic equations based on the given information, often involving variables for unknown speeds or times.

In this problem, we used the relationship between time difference and the algebraic expressions for time based on speed, leading to a quadratic equation.