Consider an equilateral triangle ABC as given in the following diagram :
Two people start at the same time from points A and B with speeds 30 km per hour and 20 km per hour respectively, and move on the sides of the triangle in the clockwise direction. They meet each other for the first time at
point C
a point between C and A
a point between A and B
point A
Answer is Right!
Answer is Wrong!
This question was previously asked in
UPSC CAPF – 2018
Let the side length of the equilateral triangle be L. The total perimeter is 3L. Person 1 starts at A (speed 30 km/h) and Person 2 starts at B (speed 20 km/h), both moving clockwise. Person 1 moves along A->B->C->A… and Person 2 moves along B->C->A->B… The initial distance between them along the clockwise path from A’s perspective to B’s position is L. Their relative speed when moving towards each other along the perimeter is the sum of their speeds: 30 + 20 = 50 km/h. The time taken for them to cover the initial distance L and meet for the first time is Time = Distance / Relative Speed = L / 50 hours. In this time, Person 1 travels a distance of 30 * (L/50) = 3L/5 km from A. Person 2 travels a distance of 20 * (L/50) = 2L/5 km from B. Let’s track their positions: Person 1 starts at A (position 0), moves 3L/5 along A->B. This point is along AB. Person 2 starts at B (position L from A along A->B->C), moves 2L/5 along B->C. This point is along BC. This initial distance calculation (L) only considers the segment AB. A better approach for motion on a closed loop is relative speed on the entire loop. The relative speed is 50 km/h. They will meet when the difference in the total distance covered by them is a multiple of the perimeter (3L), or when their positions modulo 3L are the same. Let P1’s position be d1 and P2’s position be d2, measured from A clockwise. d1 = 30t mod 3L. P2 starts at B (distance L from A), so d2 = (L + 20t) mod 3L. They meet when 30t = L + 20t (mod 3L), which means 10t = L (mod 3L). The smallest positive t occurs when 10t = L, so t = L/10. At this time, P1 has traveled 30 * (L/10) = 3L km. Starting from A, traveling 3L km clockwise along the perimeter A->B->C->A means P1 is back at A. At this time, P2 has traveled 20 * (L/10) = 2L km. Starting from B, traveling 2L km clockwise along B->C->A->B means P2 travels B->C (L) then C->A (L), ending up at A. Thus, they meet for the first time at point A.For objects moving on a closed track, they meet when the difference in the distances they have traveled is a multiple of the track length, or by considering their positions modulo the track length. Relative speed can be the sum or difference of speeds depending on direction.