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How did you arrive at the conclusion that after 1st meet, they both reach at opposite ends at the same time. Have you considered speed same for both the bodies.
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Can you post an answer for this?
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Distance from P

P---0.2---0.4---0.6---0.8---Q

Distance from Q

P---0.8---0.6---0.4---0.2---Q


When P is at 0.6, Q is at 0.4 => they meet 1st time

P---0.2---0.4---0.6---0.8---Q
P---0.8---0.6---0.4---0.2---Q


When P travels more 0.6 and comes at 0.8 (0.6 to Q = 0.4 and + 0.2), Q is at 0.8

P---0.2---0.4---0.6---0.8---Q
P---0.8---0.6---0.4---0.2---Q


When P travels more 0.6 from 0.8 and comes at 0.2, Q will travel 0.4 from its 0.8 and come back at 0.8 (0.8 to P = 0.2 and + 0.2) => they meet the 2nd time

P---0.2---0.4---0.6---0.8---Q
P---0.8---0.6---0.4---0.2---Q

Now there is a symmetry in this, every odd time they will meet at 0.6 from P, and every even time they will meet at 0.2 from P

Since 4th time is an even time, they will meet at 0.2 from P i.e. 0.2D

Answer A.
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sset92
If two bodies start to move towards each other from opposite points P and Q, they meet at ­0.6D from P. Find the point of their 4th meeting. ­

A) 0.2D
B) 0.4D
C) 0.6D
D) 0.8D
E) 0.3D

If this question is a sub-part of a set, the data of the set needs to be given along with it. Otherwise, it makes no sense.
Assuming A and B are moving around a circle and start from diametrically opposite points P and Q and circumference is 2D and that options give distance from P.
If their first meeting while moving toward each other is at a distance of 0.6D from P, it means A covered 0.6D and B covered 0.4D. So ratio of their speeds is 3:2. So every time A covers 3/5th of the circle i.e. 1.2D and B covers 2/5th of the circle i.e. 0.8D.
Their fourth meeting takes place when A has covered 3.6D distance from the present point i.e. 1 full circle of 2D (and it comes back to 0.6D from P) and another 1.2D which means it reaches 0.2D ahead of P.

Answer (A)

Check here: https://youtu.be/Jz-mSB1xNFY
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