Solving the Distance Problem: Speed, Time, and Difference

This problem involves the concepts of speed, time, and distance. We are given the speeds of two individuals, P and Q, and the difference in their arrival times. We need to find the distance they covered.

Understanding the Given Information

• Speed of P = 3 km/h
• Speed of Q = 3.75 km/h
• Q reaches 0.5 hours (30 minutes) before P.
• Both start from the same place A and go to the same place B.

Let the distance between place A and place B be \(d\) km. Since they both travel from A to B, the distance covered by both P and Q is the same, which is \(d\).

Relating Speed, Time, and Distance

The fundamental relationship between speed, time, and distance is:

\(\text{Distance} = \text{Speed} \times \text{Time}\)

This can be rearranged to find time or speed:

• \(\text{Time} = \frac{\text{Distance}}{\text{Speed}}\)
• \(\text{Speed} = \frac{\text{Distance}}{\text{Time}}\)

Setting up the Equations

We can calculate the time taken by P and Q to cover the distance \(d\) using the formula \(Time = \frac{Distance}{Speed}\).

• Time taken by P (\(t_P\)) = \(\frac{d}{3}\) hours
• Time taken by Q (\(t_Q\)) = \(\frac{d}{3.75}\) hours

We are told that Q reaches half an hour before P. This means the time taken by P is 0.5 hours more than the time taken by Q.

So, the difference in their times is:

\(t_P - t_Q = 0.5\) hours

Solving for the Distance (d)

Now, substitute the expressions for \(t_P\) and \(t_Q\) into the time difference equation:

\(\frac{d}{3} - \frac{d}{3.75} = 0.5\)

To solve this equation for \(d\), we can first work with the speeds. It's often easier to work with fractions or decimals consistently. Let's use decimals for now.

The equation is:

\(\frac{d}{3} - \frac{d}{3.75} = 0.5\)

To combine the terms on the left side, find a common denominator or use cross-multiplication idea:

\(\frac{d \times 3.75 - d \times 3}{3 \times 3.75} = 0.5\)

\(\frac{3.75d - 3d}{11.25} = 0.5\)

\(\frac{0.75d}{11.25} = 0.5\)

Now, multiply both sides by 11.25:

\(0.75d = 0.5 \times 11.25\)

\(0.75d = 5.625\)

Finally, divide by 0.75 to find \(d\):

\(d = \frac{5.625}{0.75}\)

To simplify the division, multiply numerator and denominator by 1000:

\(d = \frac{5625}{750}\)

This can be simplified. Both are divisible by 25, then 3, etc. Or notice that \(750 \times 7 = 5250\) and \(750 \times 0.5 = 375\), so \(750 \times 7.5 = 5250 + 375 = 5625\).

So, \(d = 7.5\)

Alternatively, using fractions:

\(3 = \frac{3}{1}\), \(3.75 = 3 \frac{3}{4} = \frac{15}{4}\), \(0.5 = \frac{1}{2}\)

\(\frac{d}{3} - \frac{d}{15/4} = \frac{1}{2}\)

\(\frac{d}{3} - \frac{4d}{15} = \frac{1}{2}\)

Find a common denominator for 3 and 15, which is 15:

\(\frac{5d}{15} - \frac{4d}{15} = \frac{1}{2}\)

\(\frac{5d - 4d}{15} = \frac{1}{2}\)

\(\frac{d}{15} = \frac{1}{2}\)

Multiply both sides by 15:

\(d = \frac{1}{2} \times 15\)

\(d = \frac{15}{2}\)

\(d = 7.5\)

The distance covered by each person (P and Q) is 7.5 km.

Verification

Let's check if this distance results in a 0.5-hour time difference.

• Time taken by P = \(\frac{\text{Distance}}{\text{Speed of P}} = \frac{7.5 \text{ km}}{3 \text{ km/h}} = 2.5\) hours
• Time taken by Q = \(\frac{\text{Distance}}{\text{Speed of Q}} = \frac{7.5 \text{ km}}{3.75 \text{ km/h}} = 2\) hours

Time difference = Time taken by P - Time taken by Q = \(2.5 - 2 = 0.5\) hours.

This matches the condition given in the problem.

Therefore, the distance covered is 7.5 km.

Revision Table: Speed, Time, and Distance Concepts

Additional Information: Solving Time and Distance Problems

Problems involving speed, time, and distance are common in quantitative aptitude. They often require setting up equations based on the given information, similar to the problem we just solved. Here are some related concepts and tips:

• Unit Consistency: Always ensure that units are consistent. If speed is in km/h, distance should be in km and time in hours. If speed is in m/s, distance should be in meters and time in seconds. Convert units if necessary (e.g., minutes to hours).
• Relative Speed: When objects move towards each other, their relative speed is the sum of their speeds. When they move in the same direction, their relative speed is the difference between their speeds. This concept is useful for problems involving objects meeting or overtaking.
• Average Speed: Average speed is calculated as total distance divided by total time. It is NOT simply the average of different speeds if the time taken for each segment is different.
• Solving with Time Difference: Problems like this one often involve setting up an equation based on the difference in time taken to cover the same distance, or the difference in distance covered in the same time.
• Algebraic Approach: Using variables (like \(d\) for distance or \(t\) for time) and setting up equations is a standard method to solve these problems systematically.

Practice with different variations of speed, time, and distance problems helps build confidence in applying the formulas and setting up the correct equations.