X and Y are two runners who run for the same duration of time on the same circular track. They started running at the same time in the same direction with uniform speeds. When X completed 7 rounds, Y did exactly 5. After completing 5 rounds, Y changed his direction and started running in the opposite direction with speed which is double of his earlier speed. On the other hand, X continued to run with the same speed. They stopped running when X completed exactly 21 rounds. How many times did X and Y meet after they had started and before they finally stopped?

EXPLANATION

The correct option is (a) - 35.

[as per provisional answerkey]

Solution

Step 1: Determine initial speeds
Let the length of the circular track be L.
X and Y run for the same duration. In the first phase, X completes 7 rounds while Y completes 5 rounds.
Ratio of speeds $V_{x} : V_{y} = 7 : 5$ .
Let $V_{x} = 7$ units/sec and $V_{y} = 5$ units/sec. Let the track length $L = 1$ unit.

Step 2: Phase 1 (Same direction)
This phase ends when Y completes 5 rounds. Since $V_{x} : V_{y} = 7 : 5$ , when Y has done 5 rounds, X has done 7 rounds.
Relative speed in same direction = $V_{x} - V_{y} = 7 - 5 = 2$ units/sec.
Number of meetings in same direction = (Relative distance covered) / (Track length).
Relative distance = $(V_{x} - V_{y}) \times Time$ . Since $V_{y} \times Time = 5$ rounds, $Time = 5 / 5 = 1$ unit.
Relative distance = $2 \times 1 = 2$ units. This means X overlapped Y exactly 2 times.
However, they started at the same point. In "same direction" races, the start is usually not counted as a "meeting" unless specified. They meet at the end of this phase (at the start line) because 7 and 5 are integers.
Meetings in Phase 1: At relative distances 1 and 2. (Total 2 times).

Step 3: Phase 2 (Opposite direction)
X has 14 rounds left to complete (from round 7 to 21).
Y doubles his speed and reverses direction. New $V_{y} = 5 \times 2 = 10$ units/sec.
Relative speed in opposite direction = $V_{x} + V_{y} = 7 + 10 = 17$ units/sec.
Time taken for X to finish remaining 14 rounds = Distance / $V_{x} = 14 / 7 = 2$ units of time.
Relative distance covered in Phase 2 = (Relative Speed) $\times$ Time = $17 \times 2 = 34$ units.
In opposite direction movement, the number of meetings = Relative distance / Track length = $34 / 1 = 34$ times.

Step 4: Total Meetings
Total meetings = Meetings in Phase 1 + Meetings in Phase 2.
Phase 1: 2 times (at the end of X's 3.5th and 7th round).
Phase 2: 34 times.
Wait, we must check the transition point. At the end of Phase 1, they are at the same spot (the start line). This is the 2nd meeting of Phase 1 and the 0th point of Phase 2. To avoid double counting:
Meetings = (Relative distance in Phase 1) + (Relative distance in Phase 2) - (overlap at transition if any).
Actually, the question asks for meetings "after they started" and "before they stopped".
Phase 1 relative distance = 2. Phase 2 relative distance = 34.
Total = $2 + 34 = 36$ .
Since the very last meeting occurs exactly when X completes the 21st round (the moment they stop), and the question says "before they finally stopped", we exclude the final meeting.
Total = $36 - 1 = 35$ .

Why the other options are incorrect

Option (b) - 34: This result is obtained if one fails to account for the meetings during the first phase or incorrectly calculates the relative speed in the second phase.
Option (c) - 31: This is a common error if the speed of Y is not doubled or if the relative speed calculation for opposite directions is handled as a subtraction instead of an addition.
Option (d) - 29: This value does not correlate with the relative distances covered in either phase and likely stems from a miscalculation of the time duration for the second phase.

Key Concept

The number of meetings on a circular track is determined by the total relative distance covered divided by the length of the track, using $(V_{1} - V_{2})$ for the same direction and $(V_{1} + V_{2})$ for opposite directions.