Two horses start trotting towards each other from points A and B, which are 50 km apart. They meet after 1 hour. After meeting, the first horse reaches B exactly 5/6 hour before the second horse reaches A. What is the speed of the slower horse?

Difficulty: Medium

Correct Answer: 20 km/h

Explanation:


Introduction / Context:
This is a relative speed problem involving two objects starting from opposite ends and then continuing to their destinations after they meet. The key idea is that the distances from the meeting point to each end are related to the distances travelled before the meeting, and the difference in the times taken after the meeting gives a relation between their speeds.

Given Data / Assumptions:

  • Distance between A and B = 50 km.
  • The two horses meet after 1 hour of trotting towards each other.
  • Horse 1 reaches B exactly 5/6 hour before Horse 2 reaches A.
  • We must find the speed of the slower horse.

Concept / Approach:
Let speeds of horses be v1 and v2 km/h, with v1 > v2. They meet after 1 hour, so:
v1 + v2 = 50.
After meeting, the remaining distance for Horse 1 to B equals the distance Horse 2 travelled before meeting, which is v2. So time taken by Horse 1 after meeting is t1 = v2 / v1. The remaining distance for Horse 2 to A equals v1, so its time after meeting is t2 = v1 / v2. The given condition is:
t2 - t1 = 5/6 hour.
This gives a relation in terms of the ratio r = v1 / v2.

Step-by-Step Solution:
Step 1: v1 + v2 = 50 because they meet after 1 hour. Step 2: Let r = v1 / v2, so v1 = r * v2. Step 3: Time after meeting for Horse 1, t1 = v2 / v1 = 1 / r. Step 4: Time after meeting for Horse 2, t2 = v1 / v2 = r. Step 5: Given t2 - t1 = 5/6, so r - 1 / r = 5 / 6. Step 6: Multiply both sides by r: r^2 - 1 = (5 / 6) r, so 6r^2 - 6 = 5r. Step 7: Rearranging: 6r^2 - 5r - 6 = 0. Solving gives r = 3/2 (taking the positive root). Step 8: Thus v1 = (3/2) v2. Using v1 + v2 = 50: (3/2)v2 + v2 = 50, or (5/2)v2 = 50, so v2 = 20 km/h.
Verification / Alternative check:
With v2 = 20 km/h and v1 = 30 km/h, the total distance in 1 hour is 30 + 20 = 50 km, matching the separation. After the meeting, Horse 1 has remaining distance 20 km and takes 20 / 30 = 2/3 hour. Horse 2 has remaining distance 30 km and takes 30 / 20 = 3/2 hour. The difference is 3/2 - 2/3 = (9 - 4) / 6 = 5/6 hour, exactly as given.

Why Other Options Are Wrong:
  • 30 km/h: This is the speed of the faster horse, not the slower horse.
  • 15 km/h: Too low; it does not satisfy v1 + v2 = 50 with a reasonable positive partner speed.
  • 25 km/h: Leads to inconsistent meeting and time difference conditions.

Common Pitfalls:
  • Assuming the times after meeting are equal instead of using the given 5/6 hour difference.
  • Mixing up which horse is faster and which is slower when interpreting the conditions.
  • Not converting the ratio equation correctly when solving r - 1 / r = 5 / 6.

Final Answer:
The speed of the slower horse is 20 km/h.

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