Question

A pedestrian and a cyclist start simultaneously towards each other from Aurangabad and Paithan which are 40 km apart and meet 2 hours after the start. Then they resumed their trips and the cyclist arrives at Aurangabad 7 hours 30 minutes earlier than the pedestrian arrives at Paithan. Which of these could be the speed of the pedestrian?

• A. 3 km/hr
• B. 4 km/hr
• C. 5 km/hr
• D. 6 km/hr
• E. 8 km/hr

Hint:

For complex problems involving multiple variables, setting up the equations correctly is the most critical step. When a quadratic equation appears, checking the given options (back-solving) can be much faster than solving the equation from scratch.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
A pedestrian and a cyclist travel towards each other from two points 40 km apart. We are given the time they take to meet and the difference in time they take to complete their respective journeys after meeting. We need to find the pedestrian's speed.
Step 2: Key Formula or Approach:
Let the speed of the pedestrian be \(S_p\) and the speed of the cyclist be \(S_c\).
1. Use the concept of relative speed to form the first equation: Distance = (Sum of speeds) \(\times\) Time to meet.
2. Calculate the time taken by each to complete the journey after meeting.
3. Form a second equation based on the given time difference.
4. Solve the two equations simultaneously.
Step 3: Detailed Explanation:
Part 1: Before meeting
Total distance = 40 km.
Time to meet = 2 hours.
When moving towards each other, their relative speed is \(S_p + S_c\).
\[ \text{Distance} = \text{Relative Speed} \times \text{Time} \] \[ 40 = (S_p + S_c) \times 2 \] \[ S_p + S_c = 20 \quad \cdots(1) \] Part 2: After meeting
Distance covered by the pedestrian before meeting = \(S_p \times 2 = 2S_p\). This is the remaining distance for the cyclist.
Distance covered by the cyclist before meeting = \(S_c \times 2 = 2S_c\). This is the remaining distance for the pedestrian.
Time taken by pedestrian to finish the journey after meeting (\(T_p\)) = \(\frac{\text{Remaining distance}}{\text{Speed}} = \frac{2S_c}{S_p}\).
Time taken by cyclist to finish the journey after meeting (\(T_c\)) = \(\frac{\text{Remaining distance}}{\text{Speed}} = \frac{2S_p}{S_c}\).
Part 3: Time difference
The cyclist arrives 7 hours 30 minutes (7.5 hours) earlier than the pedestrian.
\[ T_p - T_c = 7.5 \] \[ \frac{2S_c}{S_p} - \frac{2S_p}{S_c} = 7.5 \] \[ 2 \left( \frac{S_c^2 - S_p^2}{S_p S_c} \right) = 7.5 \] \[ \frac{(S_c - S_p)(S_c + S_p)}{S_p S_c} = \frac{7.5}{2} = 3.75 \] Part 4: Solving the equations
From equation (1), we know \(S_c + S_p = 20\) and \(S_c = 20 - S_p\). Substitute these into the equation from Part 3:
\[ \frac{((20 - S_p) - S_p)(20)}{S_p(20 - S_p)} = 3.75 \] \[ \frac{(20 - 2S_p)(20)}{20S_p - S_p^2} = 3.75 \] \[ 400 - 40S_p = 3.75(20S_p - S_p^2) \] \[ 400 - 40S_p = 75S_p - 3.75S_p^2 \] Rearranging into a standard quadratic equation form (\(ax^2 + bx + c = 0\)): \[ 3.75S_p^2 - 115S_p + 400 = 0 \] To simplify, multiply the entire equation by 4/15 (or divide by 3.75): \[ S_p^2 - \frac{92}{3}S_p + \frac{320}{3} = 0 \] Multiplying by 3: \[ 3S_p^2 - 92S_p + 320 = 0 \] Solving the quadratic equation (e.g., by factoring or using the quadratic formula). Let's test the options. If \(S_p=4\): \[ 3(4)^2 - 92(4) + 320 = 3(16) - 368 + 320 = 48 - 368 + 320 = 368 - 368 = 0 \] The equation holds true for \(S_p = 4\).
If \(S_p = 4\), then from equation (1), \(S_c = 20 - 4 = 16\) km/hr. These are reasonable speeds.
Step 4: Final Answer:
The speed of the pedestrian is 4 km/hr.