Two buses, A and B, are going from city P to city Q and back without a

Hi,
You will have to look into the answer again.
It will not be 2800 every time for two reasons.
They start from opposite ends, so, they will meet for the first time after covering 1400m between themselves.
Also there will be times when both are moving in same direction, and at that time they would not have completed a lap between themselves.

Sir, they both started from P and i'd read an article during my CAT prep that if the \(v_1 < v_2 < 2v_1\), then they both have to travel 2D combined for their each subsequent meeting. I never tried to get into proof tho lol. Now i'll definitely give a try.

Everytime A and B meet, they would have traveled a total of 2800 kms combined.
Since the Ratio of Speed of A and B = Ratio of Distance traveled = 5:2, A would travel 2000 kms and B would travel 800 kms everytime they meet.
Now, at the 22nd time, A would have traveled 2000 x 22 = 44,000 kms
At every 2800 kms, A would be at Point P. Therefore, at 42,000 kms, A would be at Point P. For remaining 2000 kms, he would be 2000 - 1400 (distance P to Q) = 600 kms from Point Q.

Sorry, there was another question which had both starting from opposite ends. But the point remains the same..
You can check on the time they meet for 3rd and 4th time, and you will get the answer.

Yes, when they meet at the same point again, the average distance per meet for all the times they meet during this period would be 2800.

chetan2u

I agree to you Sir. Actually, i took the value of the speed of slower bus wrong (30). In that case, faster bus never catches the slower one two times in a lap. Hence, for each subsequent meetings, both have to travel 2D combined.

You are absolutely correct about this question. Since velocity of the faster bus is more than two times of the slower one, it will catch the slower one two times in few laps. Hence, for each subsequent meeting, the relative distance travelled is not always equals to 2D.

Thanks

After they set off from P, bus A (BA) doesn’t meet bus B (BB) on its (BA’s) 1st lap since it is faster than BB and always stays ahead until it reaches Q and turns back for its 2nd lap. On each of its subsequent laps BA meets BB once because: (a) it is impossible for BA to avoid meeting BB as BB will always be somewhere on the road when BA begins a fresh lap so BA will either overtake BB or cross it and (b) it is also not possible for BA to have more than one meeting with BB in a single lap because, after one meeting, BA will be moving away from BB until it completes the lap. This sequence (BA meeting BB once during each of its laps except the first) continues for (BA’s) 10 laps on completion of which both buses are back at P (BB having completed 4 laps in the same time) and the cycle begins again. So, during each cycle (BA’s 10 laps and BB’s 4), there are 9 (10 – 1) meetings. First two cycles and 18 meetings later both buses are back at P and the third cycle begins. For reasons explained earlier, there is no meeting on BA’s 1st lap of the fresh cycle and one meeting during each of its subsequent laps. So the 4th meeting of the third cycle (or 22nd meeting from the beginning) will take place on BA’s 5th lap of the third cycle. We have to determine exactly where on the PQ road this meeting takes place:
After the end of its 4th lap BA is at city P having completed 4 laps (4L). During the same time, BB will have completed 4L*(2/5) = (8/5)L = 1.6L. That means it has completed one lap (from P to Q), turned around and completed (3/5)th of the distance towards P. So, at this point, BA is at P and BB is (2/5)L from P. Since the two buses are moving towards each other and BA:BB=5:2, BA will cover (5/7)th of the separating distance or (5/7)*(2/5)L=(2/7)L=(2/7)*1400km=400km. Since BA starts out from P, it meets BB 1000kms (1400 – 400) from city Q.
ANS: C

P.S. Since this answer is different from the OA and other answers I was afraid that my reasoning was flawed or I might have overlooked something so I worked out, step by step, the location of each meeting for an entire cycle but the answer was the same. I would request Bunuel, chetan2u and nick1816 to kindly look into my solution and point out if and where I might have gone wrong.

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