A runs in the clockwise direction on a circular track whereas B runs

VeritasKarishma Bunuel chetan2u can someone explain this in more detail? Thanks.

since A and B both travel in opposite directions
time t meet when they both meet in 1st round
= \(\frac{L}{ (25+40)}\) where L is length of track
hence time to meet for 10th time would be\(\frac{ 10 L}{ (65)}\)
since we are asked in clock wise direction
hence how far will A be when he meet 10th ime
in \(\frac{10 L}{ (65)}\) time he would have travelled
\(\frac{10 L}{ (65) *25}\) metres
=\( \frac{50L}{13}\)
or \(\frac{39 +11 }{13} *L\)
or \(3L + \frac{11}{13} L\)
Hence they would have met in 11/13 L direction from clock wise position
D

Hi exc4libur

One should know that the distance covered will be in opposite ratio of their speed.
So A:B in speed is 25:40=5:8, so distances covered by them will be in ratio 8:5.

Now whenever both meet, they complete one circle, so they will cover 10 circles when they meet 10th time.
So A will cover 10*5/(5+8) =50/13=(39+11)/13=3 + 11/13

So A would have covered 3 complete circles and would be at 11/13 of 4th circle.

Thanks chetan2u!! I found what I was doing wrong, I was calculating the final distance for anti-clockwise, instead of clockwise.

Two person with speed x and y runs in a round track D.

Since Speed (A): Speed (B) = 40/25 = 8/5. It's easy to use the variable D as 13, so when A&B meet, A has run 8/13 round and B has run 5/13 round to meet each other.

So, after 10 times A&B meet, B has run 5/13*10D = 50/13 = 3 + 11/13D.

IMO D

Let the circumference of the track be 13 meters (sum of the ratios of the speeds of A and B). If they start from point S, when they meet the 1st time, A will have run 5m and B 8m (since the ratio of their speeds is 5:8). Let the point of the 1st meeting be denoted by S1 and subsequent meetings by S2, S3... etc. By the time they meet the second time at S2, A will have run a further 5m from S1 (i.e. 5*2m from S). So by the time they meet at S10, A will have run 5*10=50m which is 50/13 (3 & 11/13) revolutions of the circular track. So the 10th meeting takes place at a point which is 11/13th of the track length measured clock-wise from the first starting point S. ANS: D

Given: A runs in the clockwise direction on a circular track whereas B runs on the same track but in the anti-clockwise direction, but they start from the same point.

Asked: If their speeds are 25 m/s and 40 m/s respectively, at what fraction of the track-length, measured in clockwise direction from the starting point, does the 10th meeting take place?

Their relative speed = 25 + 40 = 65 m/s

Since A runs in clock-wise direction, portion of track covered by him when they first met = 25/65 = 5/13

Portion of track covered by him when they met for 10th time = 50/13 = 3 11/13 = 3 full rounds and 11/13th of the track in clock-wise direction

IMO D

If their first meeting occurs t seconds after they start running, we have:

distance = rate × time

distance_A = 25t

distance_B = 40t

The sum of their distances is equal to the length of the circular track:

track-length = 25t + 40t = 65t

When they first meet, person A, who runs in the clockwise direction, covers 25t/65t = 5/13 of a track-length.

Therefore, when their 10th meeting occurs, A covers a total of:

10 × 5/13 = 50/13 = 3 + 11/13 track-lengths.

In other words, A covers 3 full tracks and an additional 11/13 track length.

Answer: D