Time, Speed & Distance Concepts & Solved Examples for CAT  Nitin Gupta, AlphaNumeric (Part 2/2)

In this article, we will cover concepts in Boats, Trains, Escalators and Circular motion in detail.
For Part 1, Please refer  Time, Speed & Distance Concepts & Solved Examples for CAT  Nitin Gupta, AlphaNumeric (Part 1/2)
Problems on Boats & Streams
When a boatman is rowing in still water, say a lake, he would be moving at a speed at which he can row. This speed is called the speed of boat in still water or simply speed of boat. But consider the same boatman in a stream. Because of the current he is either aided (if rowing in the direction of the stream, this is called Downstream) or will be opposed (if rowing against the stream, called Upstream).
If the speed of boatman in still water is B and the speed of the stream is S, we have
Downstream Speed (D) = B + S
Upstream Speed (U) = B – SDo not confuse with relative speed. In relative speed we saw that when two objects travel in same direction with speeds S1 and S2, the relative speed is S1 – S2. In the case of boats and streams, when boat is travelling in the direction of stream, the downstream speed of the boat is taken as B + S. Since they are in same direction, should we not consider relative speed to be B – S? No.
In case of relative speeds, the two objects are moving independently i.e. say police running to catch thief, the speed of police does not get ‘transferred’ or affects the speed of the thief. Whereas in the case of Boats and Streams, the speed of the Stream gets ‘transferred’ to the speed of the boat, the boat is travelling ‘on’ the stream. This is not a case of relative speed. Here the stream is ‘aiding’ the boat, unlike the case of police and thief where neither ‘aids/ hinders’ the other.
If the speed of boatman is lesser than the speed of the current or stream, the upstream speed will be negative i.e. he is trying to row upstream, but rather than move in that direction, he is taken in opposite direction by the stream. But such situations do not occur in math problems on this topic.
A boat covers a distance of 16 km in 2 hours when rowing downstream and in 4 hours if rowing upstream. What is the speed of the boat in still water?
Answer is 6 kmph.
The rowing speed of a man in still water is 7.5 kmph. In a river flowing at 1.5 kmph, it takes the same boatman 50 minutes to row a certain distance and come back. Find the distance.
Here Speed Downstream = 7.5 + 1.5 = 9 km/hr and Speed
Upstream = 7.5 – 1.5 = 6 km/hrApproach 1: Formulaic approach
Let the required distance be x. Equating the time taken, x/6 + x/9 = 5/6 hr
Solving we get, x = 3. Hence the place is 3 km away.Approach 2: Use of proportionality.
In going upstream and downstream, the distance covered is same. Hence time taken are in inverse proportion to their speeds. Ratio of upstream and downstream speeds is 6 : 9 i.e. 2 : 3.
Thus, ratio of time travelling upstream and downstream is 3 : 2.
And we know the total time is 50 minutes. Thus time travelled upstream is 30 minutes and time travelled downstream is 20 minutes.
Now the distance can be found using speed × time i.e. 6 kmph × 30 min = 6 kmph × 1/2 hr = 3 km
(it could also be found using downstream speed and time i.e. 9 kmph × 20 min = 9 kmph × 1/3 hr = 3 km)A boatman rows to a place 48 km distant and back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. Find the rate of the stream
Consider the statement, ‘he can row 4 km with the stream in the same time as 3 km against the stream’. Clearly the time is constant and thus, downstream speed and upstream speed are in the ratio of distances covered i.e. 4 : 3.
Next, consider travelling 48 km downstream and 48 km upstream. Again since distance is constant, time taken will be inversely proportional to speed i.e. time taken to travel downstream and that taken to travel upstream are in ratio 3 : 4. And we know that the total time taken is 14 hours.
Thus, 6 hours is taken to travel 48 km downstream (speed = 8 kmph) and 8 hours is taken to travel 48 km upstream (speed = 6 kmph).
Thus, speed of stream = 1 = (du)/2 = (86)/2A man travels downstream for 5 hours and again upstream for 5 hours. Yet it is at a distance of 2 kms from the place it started. What is the speed of stream?
Think of this problem as a person in a moving train. If he walks for five minutes in the direction the train is moving and then reverses direction and again walks for 5 minutes, he would come back to his original position in the train. However if he (and the train) was at New Delhi when he started to walk, and now he is at Faridabad (10 kms away from Delhi), is it not obvious that in the 10 minutes the train has taken him from New Delhi to Faridabad and speed of train is 1 km/min. Thus in the above problem also had the stream been stationary, after rowing 5 hours in either direction he would have come back to the original spot. But he is away from the original spot by 2 kms means the stream is moving and has taken him 2 kms downstream in 10 hours.
A swimmer jumps from a bridge over a canal and swims 1 kilometer stream up. After that first kilometer, he passes a floating cork. He continues swimming for half an hour and then turns around and swims back to the bridge. The swimmer and the cork arrive at the bridge at the same time. The swimmer has been swimming with constant speed. How fast does the water in the canal flow?
It is obvious that the cork does not move relatively to the water (i.e. has the same speed as the water). So if the swimmer is swimming away from the cork for half an hour (up stream), it will take him another half hour to swim back to the cork again. Because the swimmer is swimming with constant speed (constant relatively to the speed of the water!) you can look at it as if the water in the river doesn't move, the cork doesn't move, and the swimmer swims a certain time away from the cork and then back. So in that one hour time, the cork has floated from 1 km up stream to the bridge.
Conclusion: The water in the canal flows at a speed of 1 km/h.S1 and S2 started off from points A and B respectively along course of river at 7m/s and 3m/s resp. towards each other. Distance b/w A and B is 200m. After crossing each other S1 reaches B and S2 reaches A. After that they reverse their direction to comeback to their resp. starting points. When they meet for 1st time, ratio of distance covered by them is 4:1. After how much time they meet for 2nd time?
200 * 4/5 = 160 covered by S1 with speed of 7 + x
200 * 1/5 = 40 covered by S2 with speed of 3  x
relative speed = 10 m/s
200/10 = 20 sec
so , speed of stream will be 1 m/s
40/8 = 5 sec taken by S1 to reach B
In 5 sec S2 will swim 10 mtr more
now distance between them is 50 mtr
so , they will meet after 50/4 = 12.5
total time taken from starting = 20 + 5 + 12.5 = 37.5A cyclist drove one km, with the wind in his back, in three minutes and drove the same way back, against the wind in four minutes. If we assume that the cyclist always puts constant force on the pedals, how much time would it take him to drive one km without wind? [SNAP 2008]
1/(v + u) = 3 min and 1/(v  u) = 4 min => v = 7u => 1/u => 24 min => 1/7u = 24/7 minutes.
Problems on Trains
Train crossing a pole/man/platform/bridge/train
Whenever one object of finite length crosses another object of finite length, in crossing the object completely it covers a distance equal to the sum of the lengths. The distance to be covered is independent of the direction in which it is crossing i.e. whether the two objects are crossing each other in opposite directions or if one is overtaking the other i.e. in same direction, the distance to be covered for one to completely cross the other is the sum of the lengths.
In the problems asked, usually a train is crossing either a pole or a man (stationary or walking) or a platform or bridge or another train (obviously moving). For all such cases you could use the following with the modifications as noted below:
Time taken to cross = (L1 + L2)/(S1 + S2)If one of the object is a pole or a man, its length will be 0 (zero)
If one of the object is stationary (e.g. pole, platform, bridge), its speed will be 0 (zero) and the relative speed will just be the speed of the moving objectA train running at 72 kmph crosses a telephone pole in 7 sec. What is the length of the train?
Since a telephone pole is a stationary object of negligible length, the distance the train covers is just its own length.
Converting 72 kmph into m/s, since 72 is 4th multiple of 18,
speed in m/s is 4th multiple of 5 i.e. 20 m/s
Distance = 20 × 7 = 140 mA train crosses 2 platforms of length 400 m and 600 m in 6 seconds and 8 seconds respectively. What is the length of the train?
Approach 1:
(400 + L)/6 = (600 + L)/8 => L = 200Approach 2: Using proportionality
In the two scenarios ‘crossing platform of 400 m in 6 sec’ and ‘crossing platform of 600 m in 8 sec’, the speed is the same. Hence distance will be proportional to time taken. The distances covered are 400 + x and 600 + x i.e. a difference of 200. And the time taken are 6 sec & 8 sec. Thus ratio of time is 3 : 4. This will also be the ratio of distances and we know that the difference in distances is 200. Thus, the distances covered in the two
scenarios is 600 m and 800 m. Thus, length of train is 200 mA train crosses two persons who are walking at 2 kmph and 4 kmph, in the same direction in which the train is going, in 9 and 10 seconds respectively. Find the length of the train.
In the two scenarios, ‘train crossing man walking at 2 kmph’ and ‘train crossing man walking at 4 kmph’, the distance covered will be the length of the train itself i.e. will be same in both the cases. Thus, ratio of speed will be inverse of ratio of time taken. Since ratio of time taken is 9 : 10, ratio of speed will be 10 : 9. Further we also know that the difference in the speeds in the two scenario will be 2 kmph (speeds will be S – 2 and S – 4). Thus, the speeds in the two scenario are 20 kmph and 18 kmph. At 18 kmph i.e. 5 m/s, distance covered in 10 sec will be 50 m. This has to be the length of the train.
A tunnel measuring 4 km and 636 meters is designed specifically for two trains to pass simultaneously in the same or opposite directions. Therefore two express trains of length 400 m each, travel through the
tunnel at the rate 56 kmph and 80 kmph.
Q1) Assuming that both the trains enter the tunnel at the same point of time (t = 0) from the two different ends, then the minimum value of ‘t’ such that both the trains have cleared the tunnel will be.
a. 2 min 45 sec
b. 2 min 24 sec
c. 2 min 36 sec
d. None of these
Q2) Assuming that the guard starts walking immediately as he enters the tunnel and the train is traveling at the speed of 80 kmph, what is the time by taken a guard walking at the rate of 5 kmph along the
corridor towards the engine to clear the tunnel?
a. 3 min 46 sec
b. 3 min 53 sec
c. 4 min
d. None of theseQ1) Distance to be covered 4636 + 400 = 5036 m
Speed of the slower train = 56 kmph = 56 × (5 / 18) m
∴ sec
∴ time taken = 5.39 min.
Q2) Time taken = 4636/(80+5)*5/18 = 196 sec. = 3 min, 16 secProblems on Escalators
Escalator is similar to Boats & Streams
Distance = no. of steps in escalator (when escalator is not moving)
Here speed is usually given as no. of steps per second
Let speed of escalator is e steps/sec
Let speed of man/woman = m steps per second
Case 1: when escalator & man are moving in same direction, effective speed = (m + e) steps/sec
Case 2: when escalator & man are moving in opposite direction, effective speed = (m  e) steps/secAnother important concept:
Case 1: when escalator & man are moving in same direction,  No. of steps covered by man is always less than actual no. of steps in escalator.
Case 2: when escalator & man are moving in opposite direction,  No. of steps covered by man is always more than actual no. of steps in escalator.You walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end. You turn around and run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator. How many steps do you need if the escalator stands still?
Approach 1:
Let N = no. of steps in escalator (when escalator is not moving)
Case 1: walk upwards on an escalator, with a speed of 1 step per second. After 50 steps you are at the end
N /(e+1) = 50/1(1)
Case 2: run downwards with a speed of 5 steps per second. After 125 steps you are back at the beginning of the escalator
N/(5e) = 125/5 (2)
Dividing eq 1 by eq 2 , you will get e= 1
Putting e = 1 you will get N= 100 steps.Approach 2:
Remember
Case 1: when escalator & man are moving in same direction,  No. of steps covered by man is always less than actual no. of steps in escalator.
Case 2: when escalator & man are moving in opposite direction,  No. of steps covered by man is always more than actual no. of steps in escalator.
say escalator speed x steps/sec.
so total steps = 50 + 50x (from upward condition, in 50 sec escalator will cover 50x).
Total time to reach up is 50 sec.
total time to reach down = 25 sec.(125 steps, 5 steps/sec)
total steps = 125  25x (in 25 sec escalator will cover 25 x)
50 + 50x = 125  25x
75 x = 75 => x =1,
so total steps = 50 + 50 * 1= 100You all must know both approach but use approach 2 alwyas as it is easy!
A walks down an upescalator and counts 150 steps. B walks up the same escalator and counts 75 steps. A takes three times as many steps in a given time as B. How many steps are visible on the escalator?
Approach 1:
Let N = no. of steps in escalator (when escalator is not moving)
Speed of a / speed of b = 3:2 , let speed of a = 3x & speed of b = 2x
Case 1: “A “ walk down on up escalator,
N /(3xe) =1 50/3x(1)
Case 2: “B“ walk up on up escalator,
N/(2x+e) = 75/2x (2)
Dividing eq 1 by eq 2 , you will get e= x
Putting e = x you will get N= 120 steps.Approach 2:
Remember
Case 1: when escalator & man are moving in same direction,  No. of steps covered by man is always less than actual no. of steps in escalator.
Case 2: when escalator & man are moving in opposite direction,  No. of steps covered by man is always more than actual no. of steps in escalator.
Let T be time B takes to make 25 steps. Then B takes 3T to make 75, and A takes 2T to make 150. Suppose the escalator has N steps visible and moves n steps in time T. Then A covers N + 2n = 150, N  3n = 75.
Hence N = 120, n = 15.Colin takes the underground train to work and uses an escalator at the railway station. If Colin runs up 8 steps of the escalator, then it takes him 37.5 seconds to reach the top of the escalator. If he runs up 14 steps of the escalator, then it takes him only 28.5 seconds to reach the top. How many seconds would it take Colin to reach the top if he did not run up any steps of the escalator at all?
Approach 1:
If he runs up 8 steps, then he needs 37.5 seconds to reach the top.
If he runs up 14 steps, then he needs 28.5 seconds to reach the top.
The 6 additional steps take 9.0 seconds.
Therefore, each step takes 1.5 seconds.
Total steps in escalator = 8 + 37.5 / 1.5 = 33 or Total steps in escalator = 14 + 28.5 / 1.5 = 33.
If Colin did not run up any steps at all,
he would reach the top of the escalator in 49.5 seconds (i.e., 33 steps × 1.5 seconds/step).Approach 2 : Alternative Solution through Equations:
Let the total number of steps in the escalator be x.
The escalator moves at a constant speed given by
Speed of escalator = (x − 8)/37.5 = (x − 14)/28.5
The above equation may be solved as follows.
28.5 (x − 8) = 37.5 (x − 14); or
x = (14 × 37.5 − 8 × 28.5) / (37.5 − 28.5) = 33.
Now, Speed of escalator = (33 − 8)/37.5 = (33 − 14)/28.5 = 1/1.5 steps/second.
Time to reach top = Total Steps / Speed = 49.5 seconds.Approach 3 :
Let escalator speed = e steps/sec
No. of steps = 8+37.5x = 14+28.5x
X = 2/3
No. of steps = 8 + 37.5*2/3 = 33 steps
Now, Speed of escalator = (33 − 8)/37.5 = (33 − 14)/28.5 = 1/1.5 steps/second.
Time to reach top = Total Steps / Speed = 49.5 seconds.Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom’s steps. Shyama gets to the top of the escalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up? (cat 2001)
a. 40
b. 50
c. 60
d. 8025 + 25E/3S = 20 + 20E/2S
take E/S = k
25 + 25k/3 = 20 + 10k
k = 3
Thus steps = 25 + 25 * 3/3 = 502 kids, John and Jim, are running on an escalator (a moving stairway). John is running three times as fast as Jim, and by the time they are off the escalator, John has stepped on 75 stairs while Jim has stepped on 50 stairs. How many stairs does the escalator have? How is its speed related to the speed of the boys? Were they running with or against the escalator?
The answers are: the length is 100 stairs, the boys were running along the escalator which was moving with the same speed as the slow boy. Solution: in the time the fast boy stepped on 75 stairs, the slow one could step on only 25, so, since he stepped on 50, he spent twice as much time on the escalator as the fast one. Therefore his speed relative to the ground was half that of the fast boy, therefore the escalator's speed was the same as the speed of the slow boy, and he counted exactly half the stairs. Another way is to use algebra (omitted).
Approach 2:
Assume A takes 1 step per unit time. Then B will take 3 steps per same unit time. Also, assume the the escalator is moving at E steps per unit time.
Let T be the total number of steps.
Let ta be time taken by A on the escalator, tb = time taken by B on the escalator.
Since A takes 50 steps  therefore we have:
50 = T/(1+E) units of time.
similarly,
75 /3= T/(3+E) units of time.
Solving for T, E we get E = 1 step per unit time; T = 100 stepsAn escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down faster and takes 90 steps to reach the bottom.If B takes 90 steps in the same time as A takes 10 steps then how many steps are visible when the escalator is not operating?
Method 1:
B = 9
A = 1
50 + 50x = 90 + 10x
=> x = 1
total steps = 100Method 2:
There are 100 steps in the escalator.
Case 1 : When A & B walk on unmoving surface
10 Steps of A = 90 Steps of B .................... [Eq.1]
=> Speed of A : Speed of B :: 1 : 9
=> Time of A : Time of B :: 9 : 1
Case 2 : When A & B walk on escalator
Steps taken by B : 90
Time taken by B : T
Steps taken by A : 50
Time taken by A : 5T [From [Eq.1]]
Time of A : Time of B :: 5 : 1
let e be the number of steps moved by the escalator when A takes 1 step
So
(1+ e) / (9 + e) = 1/5
=> e = 1 step for every step of A
A takes 50 steps
=> Escalator has moved by 50 steps
Total number of steps = 50 + 50 = 100On an upward moving escalator Amit, Sanjeev and Vicky take 10 steps, 8 steps and 5 steps respectively to reach the top. On the same upward moving escalator Amit takes 30 steps to come down from the top.Find the ratio of the time taken by Sanjeev and Vicky to reach the top
no. of steps = 10+x = 30  3x ===> 4x = 20 ===> x = 5; so total no. of steps in escalator is 15; now sanjeev takes 8 steps (remaining 7 are taken by escalator on his behalf); amit5 steps (remaining 10 are taken by escalator on his behalf); so required ratio is 7:10
pavan walked up a descending escalator and took 154 steps in 100 sec to reach the top. Rishabh started simultaneously from the top taking 3 steps for every 4 steps taken by pavan, reach the bottom in 40sec. How many steps from bottom were they when they crossed each other?
pavan's speed = 4x m/s
escalator speed = y m/s
400x  100y = D
rishabh's speed = 3x m/s
120x + 40y = D
400x  100y = 120x + 40y
280x = 140y
ratio = 2:1
it means P has to cover twice of step
P's speed = 4x
E's speed = 2x
R's speed = 3x
so , distance = 154/2 = 77
77/(2x+5x) = 11/x sec
so they will meet after 11/x secs that means (11/x) * 2x = 22 steps from bottomProblems on Circular Motion
These types of problems deal with athletes running on a circular track and questions being asked about when, where and how often would two or more athletes cross each other.
Meeting for the First Time :
When running in opposite directions:
Consider two friends, A and B, who are separated by 1500 meters and they move towards each other at speeds of 40 m/s and 10 m/s. After how much time would they meet?
This is a straight question on relative speed and the time taken to meet = 1500/(40+10) = 30 secWhen running in the same direction:
Consider a police traveling at 40 m/s chasing a thief traveling at 10 m/s. If the distance between the police and thief when the chase starts is 1500 meters, find the time after which the police catches up with the thief.
Again, the solution should be known by now as 1500/(4010) = 50sec
Time taken to meet = Track Length /Relative Speed
Frequency of meeting :
Say two athletes start running from a same point on a circular track and meet after t units of time. When they meet they are again together at a point (not necessarily the same point where they started but nevertheless they are together). And if they continue running in the direction they were running and at their respective speeds, then this instance can be thought of as a new beginning, two athlete starting to run from a same point. And thus, from this instant onwards, they would again meet after t units of time. And the reasoning
could be again argued similarly at their 2nd meeting. So considering the 2nd meeting as a fresh start, they will again meet after t duration of time. Thus, the time after which they meet for the first time is also the frequency with which they keep meet if they continue running at their respective speeds and respective directions. If the first meeting takes place after t duration after start, the nth meeting will take place after n × t duration after start.Where would the first meeting take place?
Once the time is known after which two athletes, running on a circular track, meet, we can find the distance run individually by them (rather by any one of them) and find the exact position on the track where the meeting takes place. Rather than measuring this distance in absolute terms, it is more beneficial to measure this distance in terms of the track length.
Meeting for the First Time at Start :
The question of identifying after how much time would two or more athletes meet at the starting point is an application of LCM rather than a problem of Time Speed Distance. The key word here is meeting at the starting point. Please realize that this may not be the first time that they meet. The could have crossed each other (met) at some other point on the track but then that would not be counted as a meeting point for this question as it has not occurred at the starting point.
Consider two athletes, A and B running on a circular track and taking x and y units of time to complete one full circle. A would reach the starting point for the first time after x units and thereafter would be at the starting point after every x units. Similarly B would reach the starting point for the first time after y units and thereafter would be at the starting point after every y units. Thus, A and B would both be at the starting time after common multiple of x and y and the first time that this would occur would be the LCM of x and y.
For these types of problems, it does not matter in which direction the two athletes are running. Even if both are running clockwise or if one is running clockwise and other anticlockwise, the time when they would be at the starting point would remain the same.
Relative Distances or Rounds run :
When running in opposite directions : they meet whenever together they have covered one full round and hence at the nth meeting, together they would have covered n rounds. This is easy to understand. Since both start from common point and run in opposite directions, at 1st meeting, if one runs f fraction of the track, the other has to run (1 – f ) fraction of the track so that he is at same point. Thus, sum of distances run = f + (1 – f ) i.e. 1 full round. This is new beginning, and from now onwards, till 2nd meeting they would together again cover 1 full round i.e. they would together cover 2 rounds since start. And so on.
When running in same directions : They meet whenever the faster one has covered 1 round more than the slower one and hence at the nth meeting, the faster one would have covered n rounds more than the slower one. Consider both start from same point and run in same direction. Focus on the gap between them. The faster one will race ahead of the slower one and a gap will start emerging between them. As time passes the gap will start increasing
Three or more people meeting :
To find time when three or more people meet, find the time after which pairs of athletes meet, such that at least one athlete is common to all the pairs (say, A & B, A & C, A & D, ……). The respective pairs will keep meeting after any multiple of the time found. At the LCM of these durations of time, we are sure that A & B have met, that A and C have met, that A & D have met and so on. But the only way that A could have met all these people at the same instant is when all of them are together.
Position of the meeting points :
These types of questions do not deal with when or how often do the athletes meet when they are running on a circular track. These questions pertain to the number of points and their placement on the circular track where the athletes can possibly meet.
When running in the opposite directions : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a + b
When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a – b
Number of the meeting points ( for 3 or more people) :
Find the hcf of meeting points of all possible pair.
Two men , Jain & Sood, walk round a circle 1200 metres in circumference. Jain walks at the rate 150 meters/minute , and sood walks @80 minutes per minute. if both start at the same time from the same point & walk in the same direction.
(a) when will they 1st be together again at starting point.
(b) when will they be together again .
(c) no. of distinct meeting points.
(d) find the distance travelled & time required by Sood & Jain when they meet for 100th time.
(e) when they meet for 100th time what is the distance from the starting point to the point at which they will meet (in anticlock wise direction)a) lcm (time taken by jain to comlete 1 round, time taken by sood to complete 1 round)
= lcm (1200/150,1200/80) = 120 min.
c) since ratio of speed is 15 : 8 ===> no. of distinct point = 15  8 = 7
b) from a & c we can say that they are meeting 7 times in 120 min ==> 1st meeting will take after 120/7 min
d) ratio of speed = ratio of no. of rounds ===> no. of rounds covered by jain to sood = 15:8 ==> here there is a gap of 7 but for first meeting there must be a gap of 1 round so distance covered by jain & sood when they meet for the first time = 15/7 & 8/7( in terms of rounds) so at 100th meeting distance covered is 15/7 * 100 & 8/7 * 100 ( in terms of round 1 round =1200m) & time required = 120/7 * 100.( avoid using relative speed work on ratio)
(e) since there are 7 distinct points : 100 = 7k + 2; so 100th meeting point will coincide with 2nd meeting point & when they meet for 2nd time no. of rounds covered covered by sood = 8/7 * 2 = 16/7 = 2 + 2/7 ==> 2/7 clockwise or anticlockwise depending upon whether they started clockwise or anticlockwise respectively. So this cannot be determined.Consider athletes A , B & C running at speeds of 150 m/s , 70 m/s &110 m/s respectively on a circular track of 1000 meters, A & B running clockwise and C anticlockwise. If they keep running indefinitely, at how many distinct point on the circle would they meet? what is the distance covered by each when they meet for 100th time.
a : b = 15 : 7 so A & B will meet @ 8 points (as direction is same).
b : c = 7 : 11 so B & C will meet at 18 points (as direction is opposite)
so no. of distinct meeting points for all 3 = hcf(8,18) = 2In a circular track, there are two points P and Q which are diametrically opposite.C starts running clockwise with a speed of 4m/s from P. At the same time, from Q, A starts running clockwise with a speed of 3 m/s and B starts running anti clockwise with a speed of 5 m/s. If the length of the track is 300m, then after how much time will C be equidistant from A and B for the first time.
B and C meets after 150/(4 + 5) = 50/3 seconds.
At this point distance between A and C is 150 + 50  200/3 = 400/3
So, suppose that from this moment on wards it take them t time to reach the situation when C is equidistant from A and B
After time t, distance between A and C will be 400/3 + 3t  4t = 400/3  t and distance between B and C will be 9t
9t = 400/3  t
=> t = 40/3
So, after 50/3 + 40/3 = 30 seconds C will be equidistant from A and B for the first timeSarah and Neha start running simultaneously from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduces to half of their current speeds. If the length of the circular track is 1500 m, how many times will Sarah and Neha meet on the track?
(1) 6
(2) 9
(3) 11
(4) 7
(5) 8Total distance Sarah can cover = 10/60 * (15 + 15/2 + 15/4 + ...) = 5 km
Total distance Neha can cover = 10/60 * (25 + 25/2 + 25/4 + ...) = 25/3 km
First time they cover 750 m and subsequently they cover a distance of 1500 m to meet.
The total distance they cover together is 40/3 km.
Number of meetings possible is 1 + (40000/3  750)/1500 = 9.4
=> choice (B) is the right answerThree boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speeds of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card.What is the difference in the number of cards received by B and C if A distributes 33 cards in all?
Let speed of A , B , C = 5 , 1 , 1 km/hr
A and B runs in clock wise direction so , their total meeting points will be 4 and C runs in anti clock wise direction so , their meeting points will be 6
let the distance be 5km
=> after 1st 5 rounds B = 4 cards and C = 6 cards
=> after 2nd 5 rounds B = 8 cards and C = 12 cards
=> after 3rd 5 rounds B = 12 cards and C = 18 cards
=> after 4th round B = 13 cards and C = 20 cards
so , difference = 20  13 = 7Practice Problems:
A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. The speed of the boat in still water is: [IIFT 2008]
a) 3 km/hour
b) 4 km/hour
c) 8 km/hour
d) None of the aboveAt his usual rowing speed, Rahul can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing speed for this 24 mile round trip, the downstream 12 miles will then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour? (CAT 2001)
a. 7/3
b. 4/3
c. 5/3
d. 8/3Two trains are travelling in opposite direction at uniform speed 60 and 50 km per hour respectively. They take 5 seconds to cross each other. If the two trains had travelled in the same direction, then a passenger sitting in the faster moving train would have overtaken the other train in 18 seconds. What are the lengths of trains (in metres)?
a) 112.78
b) 97.78, 55
c) 102.78, 50
d) 102.78, 55Laxman and Bharat decide to go from Agra to Delhi for watching a cricket match and board two different trains for that purpose. While Laxman takes the first train the leaves for Delhi, Bharat decides to wait for some time and take a faster train. On the way, Laxman sitting by the windowseat noticed that the train boarded by Bharat crossed him in 12 seconds. Now the faster train can travel 180 km in three hours, while the slower train takes twice as much time to do it. Given this, mark all the correct options.
a. If the faster train has taken 30 seconds to cross the entire length of the slower train, the difference between the lengths of the two trains is 50 m.
b. If the faster train had been running twice as much faster, it would have taken 10 seconds to overtake the slower train.
c. Had the faster train taken 24 seconds to cross the entire length of the slower train, the length of the slower train would have been 100 m.
d. If the slower train had been running at one and a half times of its current speed, the faster train would have taken 24 seconds to overtake Laxman. (IIFT 2006)Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them carry flags of different colours and whenever two or more children meet, they place their respective flag at that point. However nobody places more than 1 flag at a same point. They are running in anticlockwise direction. How many flags will be there in total, when there will be no scope of putting more flags?
Three athletes A, B and C are running on a circular track of length 1200 meters with speeds 30 m/s, 50 m/s and 80 m/s. A is running clockwise and B and C are running anticlockwise. Find the time after which A and B will meet for the first time and the frequency (in seconds) after which they will keep meeting. Also find the sum of the distance (in fraction of the track length) run by them till their first meeting.
Which of the following cannot be the ratio of speeds of two joggers running on a circular jogging track if while running they meet at a diametrically opposite point to the point from where both of them started?
a) 3 : 5
b) 1 : 3
c) 1 : 5
d) 2 : 5

Practice Problem  5 : Seven children A, B, C, D, E, F and G started walking from the same point at the same time, with speeds in the ratio of 1 : 2 : 3 : 4 : 5 : 6 : 7 respectively and they are running around a circular park. Each of them carry flags of different colours and whenever two or more children meet, they place their respective flag at that point. However nobody places more than 1 flag at a same point. They are running in anticlockwise direction. How many flags will be there in total, when there will be no scope of putting more flags?
Solution : When running in the same direction : If the ratio of speeds of two athletes (in the most reducible form) is a : b, the number of distinct meeting points on the track would be would be a – b
A and B will meet at 1  2 = 1 point.
A and C will meet at 1  3 = 2 points
A and D will meet at 1  4 = 3 points
A and E will meet at 1  5 = 4 points
A and F will meet at 1  6 = 5 points
A and G will meet at 1  7 = 6 points
So A will put 1 + 2 + 3 + 4 + 5 + 6 = 21 flags.similarly B and C will meet at 2  3 = 1 point
B and D will meet at 2  4 = 2 points
B and E will meet at 2  5 = 3 points
B and F will meet at 2  6 = 4 points
B and G will meet at 2  7 = 5 points
So B will put 1 + 2 + 3 + 4 + 5 = 15 flagsSimilarly find for C, D, E and F.
We will get 21 + 15 + 10 + 6 + 3 + 1 = 56 flags