No no….don’t get it wrong.
Here, we are talking about one of the interesting topics in CAT-preparation (and all others MBA entrance tests also) i.e. Circular motion – a small but very important part of a broader topic: Time, Speed and Distance.
We are specifically focusing here on the meetings of the runners on a circular track (or any closed track for that matter). As we go on, I will post some questions and in the explanation I’ll discuss few important concepts that’ll remain with you at the end of this session.
To get the best benefit, first solve the questions on your own before seeing the solutions.. 🙂
When two runners run in SAME direction:
Problem 1. Shyam and Shreya start running simultaneously from a point O on a circular track in same direction. If the ratio of their speeds is 7 : 4 respectively, then how many times do they meet each other till they reach O again?
(1) 2
(2) 3
(3) 4
(4) 7
Answer: (2) 3
Solution: No need to apply traditional long methods. Just understand a very important and basic concept here.
Whenever fast runner travels one extra round than the slower one, they meet exactly once. Similarly after that, this meeting point act as new starting point for next meet. And again when difference in their distances travelled becomes one more round, they meet for next time and so on.
Now here, when Shyam travels 7 rounds and reach O, by same time Shreya travels 4 rounds and reach O. {Mind that their speed ratio is 7 : 4, so in equal time they travel distances also in the ratio 7 : 4}.
So Shyam travels 7 – 4 = 3 extra rounds than Shreya. Hence, by the time both reach O again simultaneously, they meet each other 3 times.
Problem 2. Two persons, Ali and Baba, start running simultaneously from a point O on a circular track in the same direction. If the ratio of their speeds is 16 : 11 respectively, how many times are they diametrically opposite to each other before they meet at O for the next time?
(1) 3
(2) 4
(3) 5
(4) 6
(5) 7
Answer: (3) 5
Solution: Whenever fast runner travels one extra round than the slower one, they meet exactly once. Similarly after that, this meeting point act as new starting point for next meet. And again when difference in their distances travelled becomes one more round, they meet for next time and so on. And in the duration of one meeting point to next meeting point the two runners are diametrically opposite to each other exactly once.
Now, it is given that their speed ratio is 16 : 11, i.e. when the faster one travels 16 rounds, slower one is able to travel 11 rounds only and they reach starting point O simultaneously after start. By this time the difference in their distances travelled is = 16 – 11 = 5 rounds. So they were diametrically opposite of each other exactly 5 times.
Problem 3. Ali and Baba start simultaneously at the same point on a circular track and run along the track in the same direction. The point on the track at which they meet for the 29th time is same as that at which they meet for the 41st time. If the ratio of the speed of the faster boy to that of the slower one is n : 1 where n is a natural number. Which of the following is not a possible value of n?
(1) 2(2) 4
(3) 5
(4) 6
Answer: (4) 6
Solution: Continuing from the previous question and solution, the two runners are going to meet exactly (n – 1) times till they reach starting point again after start simultaneously. And it’s interesting that this reaching simultaneously at starting point is the first repetition of a meeting point. Hence, it can be safely concluded that before this meeting at starting point, all meeting points were DISTINCT…. And after reaching the starting point simultaneously, all further meeting points are going to replicate the previous ones only.
Thus in the present question, we just need to see that for which value of ‘n’, 29 and 41 will yield different remainders when divided by (n – 1). It can be easily seen now, that for n = 6, the two numbers yield different remainders, hence cannot be a possible value of n.
Problem 4. Two motorcyclists, Angad and Bali, start simultaneously from a point S on a circular track and drive around the track in same direction with speeds of 9 km/hr and 5 km/hr respectively. Every time Angad overtakes Bali (anywhere on the track), both of them decrease their respective speeds by 1 km/hr. If the length of the track is 1 km, how many times do they meet at the starting point before Bali comes to rest?
(1) 0(2) 1
(3) 2
(4) 4
Answer: (2) 1
Solution: Following the concepts of previous two questions only, we get that number of distinct meeting points depends on the speed ratio of two runners and also the position of the meeting point on the track.
For example, if two runners start running in same direction simultaneously at speed ratio 2 : 5, then they will meet at exactly 5 – 2 = 3 distinct meeting points and there locations on the track will be: one-third, two-third and starting point.
In the given question, difference between the two speeds is 4 km/hr, so there can be at most 4 distinct and equidistant points where the two runners would be meeting. Let’s call them I, II, III, IV where IV is the same as the starting point.
Now when Bali travels 5 unit distance i.e. reaches at I circling the track once, then in same time Angad travels 9 unit distance and reaches at I after circling the track twice.
Similarly, we can form this table here quickly to get the desired answer.
In all of their meetings, there is exactly 1 instance when the runners are meeting at IV i.e. starting point.
Problem 5. How many times the angle between the minute hand and hour hand of a wall clock becomes 13-degree in a day?
(1) 13(2) 22
(3) 26
(4) 44
Answer: (4) 44
Solution: Baffled??? that what this is to do with the ongoing line of thought.
Easy!!! Here minute hand and hour hand of the wall clock are just two runners. And in 12 hours, say after midnight, minute hand travels 12 complete rounds while hour hand travels only 1 complete round. We know that both the runners (the two hands) are moving in same direction.
So applying our previous concept, we know that they meet each other exactly 12 – 1 = 11 times. And in each of the duration between two consecutive meets, there are exactly 2 instances when the angle between the two hands is 13-degree. Can you observe that?
Okay..let me offer you a small help. First instance is when minute hand is getting ahead of hour hand after meeting, and second is when minute hand is just approaching the hour hand for next meeting.
I hope it is clear. 🙂
So clearly, answer will be 44 as in 12 hours the required instance will occur 22 times and same scenario for next 12 hours also.
When two runners run in OPPOSITE direction:
Problem 6. Anu and Vibhu start simultaneously from the same point on a circular track and run around the track in opposite directions. The radius of the track is 70m and the speeds of Anu and Vibhu are 20m/s and 10m/s respectively. Find the difference in distances travelled by Anu and Vibhu when both meet for the 9th time.
(1) 440m
(2) 880m
(3) 1320m
(4) 1760m
Answer: (c) 1320m
Solution: Again we can go by traditional methods, but try to learn a handy concept here:
When the two runners run in opposite direction on a circular track starting simultaneously from same starting point, then they meet each other for the first time when they travel entire circumference once together. For next meet, previous meeting point acts as new starting point. Hence when they travel one more round together, they meet for the second time and so on.
So when Anu and Vibhu meet for the 9th time, they have together travelled 9 rounds completely.
As the distance travelled by them will be in the ratio of their speeds i.e. 2 : 1, so Anu travelled exactly 6 rounds and Vibhu travelled 3 rounds.
Thus the required difference is = 3 × 2 × pi × r = 3×2×(22/7)×70 = 1320m.
Problem 7. Arun and Barun start running simultaneously from same point on a circular track in opposite directions and with speeds in the ratio of 3 : 5 respectively. Every time they meet, they interchange their speeds and also reverse their directions. At how many distinct points on the track do they meet each other, if they run continuously?
(1) 2(2) 3
(3) 5
(4) 8
Answer: (4) 8
Solution: Interchanging the speeds and reversing the direction is as good as the two runners are just crossing each other.
So total number of meeting point is just = 3 + 5 = 8.
Remember that two runners running in opposite direction meet every time when sum of their distance travelled becomes one complete round. And when both the runners reach starting position simultaneously, the meeting points start repeating. And till this point, Arun will travel 3 rounds and Barun will travel 5 rounds. So a total of 8 rounds will be travelled by that time and hence 8 distinct meeting points.
Problem 8. Aalu and Bhalu start simultaneously from same point on a circular track, of length 4 km, and run in opposite directions. Their speeds are doubled every time they cross each other. Find the number of times they will meet within the first hour given that they started the race with respective speeds of 3 km/h and 5 km/h.
(1) 3(2) 5
(3) 8
(4) Infinitely many
Answer: (4) Infinitely many
Solution: They meet for the first time in 4/(3 + 5) = ½ hour.
Next time they’ll take half time i.e. ¼ hour as they have doubled their speeds.
Every next time their time taken will be halved.
So if they continue like this then for infinite meets their total time taken will be = ½ + ¼ + … = (1/2)/(1-1/2) = 1 hour.
Let’s summarise the learnings from this small mathematical excursion:
Huff!!! That’s all for the day. Enjoy!! 🙂
No no….don’t get it wrong.
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