Amethi and Bellari are 645 km apart. Sonia and Priyanka start simultaneously from Amethi and Bellari toward each other and meet after 15 hours. After meeting, Sonia increases her speed by 3 km/h and Priyanka reduces hers by 3 km/h; both reach their destinations at the same time. What were their initial speeds?

Introduction / Context:
Two-way travel with a meeting point gives the sum of initial speeds. The symmetric post-meeting arrival-on-same-time condition, after adjusting speeds by ±3 km/h, fixes the individual initial speeds uniquely.

Given Data / Assumptions:

• Total distance D = 645 km; meeting time = 15 h.
• Initial speeds s (Sonia), p (Priyanka) so s + p = D/15 = 43 km/h.
• After meeting: Sonia's speed = s + 3; Priyanka's = p − 3.
• They arrive at endpoints at the same time after meeting.

Concept / Approach:
Remaining distances after meeting are D − 15s (for Sonia to Bellari) and D − 15p (for Priyanka to Amethi). Equal arrival times imply (D − 15s)/(s + 3) = (D − 15p)/(p − 3), together with s + p = 43.

Step-by-Step Solution:
s + p = 43.(645 − 15s)/(s + 3) = (645 − 15p)/(p − 3).Solving yields s = 20 km/h, p = 23 km/h.

Verification / Alternative check:
Check remaining distances: Sonia has 645 − 300 = 345 km; Priyanka has 645 − 345 = 300 km. Times: 345/(20 + 3) = 15 h; 300/(23 − 3) = 15 h; equal as required.

Why Other Options Are Wrong:
Other pairs do not satisfy both s + p = 43 and the equal post-meeting arrival condition.

Common Pitfalls:
Using s − p or averaging speeds instead of leveraging the equal-time condition with adjusted speeds.

Final Answer:
20 km/h and 23 km/h