Time, Speed & Distance Concepts & Solved Examples for CAT  Nitin Gupta, AlphaNumeric (Part 1/2)

In the entire TSD chapter, the only theory you need to know is Distance = Speed * Time. 70% of the TSD questions appearing in CAT can be solved using Proportionality b/w Time, Speed and Distance. This is probably the most important and the most thoughtintensive concept in this chapter. It provides the foundation to solve many tough problems without using complicated equations. So in this article, we will learn how to use this concept to solve TSD questions in a minimum time.
Usually, TSD Questions comes in two variants. In both variants, one among the three variables, S, D and T, is a constant. If one of these variables is a constant, we can use the proportionality relation among the other two to solve the question.
 When speed is constant, D is directly proportional to T.
 When Distance is Constant, S is inversely proportional to T.
Some examples given below (Complete questions are not given. This is just to get the idea)
Today I travelled from home to office 20% faster than my usual speed.... The two scenarios are ‘usual (everyday)’ and ‘today’. And while it is directly given that speeds are not constant, you need to realize the stated assumption that distance covered, home to office, will not change between ‘everyday’ and ‘today’ and thus, distance is constant
Two friends started simultaneously from their homes towards each other to meet ……
The two scenarios are that of the two individuals. Since they start simultaneously and they meet, both of them are travelling for the same time. Thus, the time is constant when we consider the case of the two individuals separately.A train takes 10 seconds to cross a pole and 15 seconds to cross a platform. While we will see this scenario in details later on, the two scenarios are obvious – one is ‘train crossing pole’ and other is ‘train crossing platform’. And in the two scenarios, the speed of the train is going to be constant.
Distance being constant
Time is inversely proportional to speed, when distance is constant. Over a same distance, if the ratio of speeds is a : b, the ratio of the time taken will be b : a. This should be obvious, right? because over the same distance, if I double my speed (ratio of speeds 1 : 2), the time taken will be half (ratio of time 2 : 1).
If I travel at 1/3rd the usual speed (ratio of speed 3 : 1), I would take thrice the time taken earlier (ratio of time 1 : 3) If I reduce my speed to 3/5th of the usual speed (ratio of speed 5 : 3), the time taken will be 5/3 times the usual time (ratio of time 3 : 5).
A boy walks at 1/3rd of his usual speed and reaches school 20 minutes late. Find the usual time taken by the boy and the time taken at the reduced speed. (Solve it in your mind  No pen/pencil) [ Actual CAT problem ]
In such problems, ‘late by 20 minutes’ implies that 20 more minutes will be taken to travel the same distance, or in other
words, the difference in the time taken at the usual speed and the reduced speed will be 20 minutes. Had my usual speed been S, the reduced speed would be S/3. We will find the ratio of usual speed to reduced speed (From next problem onwards, this step will be done directly).Since time is inversely proportional to speed, the ratio of the time is 3 : 1. We also know that the difference in the time taken will be 20 minutes. Thus we are looking for two numbers that are in the ratio 3 : 1 and the difference between them is 20 minutes.
3 : 1 (there is a gap of 2 but we need a gap of 20. So, multiply by 10) > 30 : 10.Travelling at 3/7th of his usual speed, a person is 24 minutes late in reaching his office. Find the usual time taken by him to cover this distance. (You don't need a pen for this, right?)
speed (3 <  > 7) ====> time (7 <  > 3) gap of 4 but we need a gap of 24 so multiply ratio by 6 ===> 42 <  > 18
If a man walks at the rate of 30 kmph, he misses a train by 10 minutes. However, if he walks at the rate of 40 kmph, he reaches the station 5 minutes before the departure of the train. Find the distance to the station.
The ratio of speeds is 3 : 4 and since distance is constant, the ratio of the time taken will be in the ratio 4 : 3 .
Missing the train by 10 minutes and reaching early by 5 minutes implies that the time taken at speed of 40 kmph is 15 minutes less than the time taken at speed of 30 kmph.
Thus, we need to find two numbers in ratio 4 : 3 with a difference of 15. Thus, a difference of 1 on the ratio scale is a difference of 15 in actual values. Hence the multiplying factor is 15.
Thus time taken at 30 kmph is 4 × 15 = 60 minutes and at 40 kmph is 3 × 15 = 45 minutes.
Now you can use D = S * T = 30 * 60/60 = 30 kmA train meets with an accident and travels at 4/7th of its regular speed hereafter and hence it reaches its destination 36 minutes late. Had the accident occurred 30 kms further, the train would have been late by only 21 minutes. Find the regular speed of the train.
method 1  Let’s say the accident occurred at point A when the train was late by 36 minutes and at point B when the train was late by 21 minutes. Let the destination be D.
Comparison over distance AD: The two scenarios are ‘at regular speed’ and ‘at reduced speed’. Ratio of speeds 7 : 4. Ratio of time 4 : 7. Difference in time is 36 minutes. Thus, time taken at regular speed for the distance AD is 4 × 12 = 48 minutes.
Comparison over distance BD: The two scenarios are ‘at regular speed’ and ‘at reduced speed’. Ratio of speeds 7 : 4. Ratio of time 4 : 7. Difference in time, over this distance is 21 minutes. Thus, time taken at regular speed for the distance BD is 4 × 7 = 28 minutes.
At it’s regular speed, the train takes 48 minutes to travel A to D and takes 28 minutes to travel B to D. Hence it must be taking 20 minutes to travel from A to B, a distance of 30 kms. Thus, it’s regular speed = 30/(20/60) = 90 km/hrmethod 2  Shortcut:
While in the above solution we had the comparisons (at regular speed & at reduced speed) made twice, once over distance AD and once over distance BD. We could make do with just one comparison as well.
Consider the stretch AB. When accident occurs at A, this stretch is travelled at reduced speed. And when accident occurs at B, this stretch is travelled at regular speed. And the difference in the time taken just over this stretch is 36 – 21 = 15 minutes. (At destination, why the train is late only by 21 mins, when accident occurs at B as compared to 36 mins late when accident occurs at A? Because AB is the causing the difference)
Comparing the two scenarios, ratio of speeds over AB is 4 : 7. Hence ratio of time will be 7 : 4 and difference in time taken over AB is 15 minutes. The multiplying factor to go from ratio scale to actual values will be 5. Thus, at regular speed it would take 4 × 5 = 20 minutes to cover AB (a distance of 30 km) ==> regular speed = 90 km/hrIn previous Question, find the total distance
To be late by 15 min distance travelled is 30: so to late by 1 min  distance travelled is 30/15 = 2 ; so to be late by 36 min ( since beginning)  distance travelled is 36 * 2 = 72 km
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon? ( CAT 2004)
ratio of speed = 10 : 15 = 2 : 3 ===> ratio of time = 3 : 2 ( there is a gap of 1 but we need a gap of 2 hrs as 11am & 1pm) so time is 6hrs & 4hrs. now let speed at 12noon is "s" > 10/s = 5/6 ===> s= 12 km/hr.
Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? (CAT 2006)
Let A, B, K represents speed & a,b,k Represents time of Arun, Barun & Kiranmal respectively.
Here distance is constant
(1) A/B = b/a ===> b/a = 3/4 (gap of 1 but we need gap of 2 as barun started 2 hrs late) so b= 6, a =8 ;
(2) A/K =k/a ===> 30/60 = k/8 ==> k = 4hr ===> kiran mala started 4hrs late.Time being constant
Distance is directly proportional to speed, when the time is same.
This is to say, if time is constant and the ratio of speeds is a : b, the ratio of the distance covered will also be a : b.The most common case of time remaining same would be when two persons, trains or objects start from two points simultaneously and meet each other. In this case the time that the two objects are travelling is the same for both the objects. So they will cover distances in proportion to their speeds.
The GhaziabadHapurMeerut EMU and the MeerutHapurGhaziabad EMU start at the same time from Ghaziabad and Meerut and proceed towards each other at 16 km/hr and 21 km/hr, respectively. When they meet, it is found that one train has travelled 60 km more than the other. The distance between two stations is:
a) 445 km
b) 444 km
c) 440 km
d) 450 km [IIFT 2007]Difference between the speeds is 5 and difference between the distances travelled is 60
=> Total distance = (60/5) * (21 + 16) = 12 * 37 = 444Two trains start simultaneously, one from Bombay to Kolkata and other from Kolkata to Bombay. They meet each other at Nagpur which is at a distance of 700 kms from Bombay. If the distance between Bombay and Kolkata is 1600 km, find the ratio of their speeds.
Since the trains started simultaneously, the time they have been traveling till they meet is equal. And hence the distance they cover will be in ratio of their speed. Since the train from Bombay has covered 700 km and the train from Kolkata has covered 1600 – 700 = 900 kms, the ratio of their speeds will be 700 : 900 i.e. 7 : 9
A policeman starts chasing a thief. The ratio of the speeds of the thief and the policeman is 9 : 11 and when the policeman catches the thief it is found that the policeman has covered 60 meters more than the thief. How much distance did the police have to run to nab the thief?
Since the chase starts with both of them running simultaneously, from this point onwards to the time the police
has caught the thief, they are running for same duration. Thus the distance covered will be proportional to their speeds. So we are searching for two distances in the ratio 9 : 11 and the difference being 60 meters i.e. 2 of the ratio scale corresponds to 60 mts, implying that the multiplying factor is 30. Hence distance run by police will be 11 × 30 = 330.In the movie Ghulam, Aamir is able to spot the approaching train when it is 2 km away. He has to run towards the train and reach the red kerchief hung on a pole 400 meters away from him before the train reaches the pole. How fast must Aamir run if the speed of the train is 36 kmph so that he just manages to reach the kerchief at the same time as the train reaches it?
From the point when the distance between them is 2000 m, Aamir has to cover a distance of 400 m and the train will cover the rest of the 1600 m i.e. the distance will be in the ratio 1 : 4. Since they are running for the same duration, the speed will be proportional to the distance covered. Since the speed of the train is 36 kmph, the speed of Aamir should be 9 kmph.
Speed being constant
Distance is directly proportional to time when speed is constant At same speed, if the ratio of speed is a : b, the ratio of time will also be a : b. By now, you would have got the hang of solving questions based on proportionality and thus we will solve just one example for this proportionality.
A car overtakes an auto at point A at 9 am. The car reaches point B at 11 am and immediately turns back. It again meets the auto at point C at 11:30 am. At what time will the auto reach B?
Since the car takes 2 hours to travel AB and 0.5 hours to travel BC, The ratio of the distances AB : BC will be 2 : 0.5 = 4 : 1. (Distance is proportional to time)
Since C lies in between A and B, the ratio of the distance AC to CB will be 3 : 1.
The auto has traveled AC in 2.5 hours (from 9 am to 11:30 am).
To travel CB (onethird distance of AC) he will take 2.5 hrs /3 = 150 min/ 3 = 50 more minutes.
Thus the auto will reach B at 12:20 pm.There is a tunnel connecting city A and B. There is a CAT which is standing at 3/8 the length of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the CAT meet. In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?
[CAT 2002]let speed of train be x. when train runs x then cat runs 3 and when cat runs 5 the train runs x + 8
equating, x=12
ratio of speeds =12 : 3 i.e. 4 : 1Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N andS increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day? [CAT 2002]
a) 1 : 3
b) 1 : 4
c) 1 : 5
d) 1 : 6Let the speed of superfast & passenger train is N & S respectively. Since there is one single track the combined time taken by superfast & passenger train is 1 hr. Ratio of speed N/S = 4:1 ===> ratio of time = 1:4===> time taken by superfast train = 1/5 * 60 = 12 min & time taken by passenger train = 48 min. Now one day train started 20 min late since speed of superfast is doubled time taken is  6 min . so time taken by passenger train to reach on time = 60  (20 + 6) = 34 min : so new ratio is 6 : 34 = 1:6 (approx)
Two boats start from opposite banks of river perpendicular to the shore. One is faster then the other. They meet at 720 yards from one of the ends. After reaching opposite ends they rest for 10mins each. After that they start back. This time on the return journey they meet at 400yards from the other end of the river. Calculate the width of the river. Also find ratio of speed of boats.
when they meet for the first time , together they will cove 1 round in which 1st boat will contribute 720; when they meet for the 2nd time together they will contribute 3 round  so contribution by 1st boat is 720 * 3 = 2160.
So length is 2160  400 = 1760Alternative method: The distances travelled by the boats are proportional to their speeds. Since the speeds are constant, the ratio of the distances is constant.
let w="width of river" ___ river is > 720 yard wide
let x and y be the boat distances
at first meeting ___ x = 720, y = w  720
at second meeting ___ x = w + 400, y = 2w  400
ratio of distances is constant, so 720/(w720) = (w+400)/(2w400) on solving this W = 1760I hope that funda of proportionality of TSD is clear to everyone. from now onwards, while solving TSD question, look for the part which is constant & then apply proportionality.
Application of AM & HM in TSD Problems
Quite often in questions we find that the given speeds (or time taken) are in an Arithmetic Progression. And if distance covered at the speeds is constant, then time taken (or speeds) will be inversely proportional i.e. they will be in Harmonic Progression.
For those who have forgotten, the Arithmetic Mean of a and b is (a + b)/2 and the Harmonic Mean of a and b is 2ab/(a + b)
Though it requires a little trained eyes to identify the above, it will be useful if you keep a watch for it. See the following data to realise that either time taken or speeds are in an Arithmetic Progression.
A, B, C leave point P, one after the other in the given order, with equal time intervals between their departure. If all three simultaneously meet at Q, given that speed of A and C is 40 kmph and 60 kmph, find speed of B.
The time taken by A, B, C over constant distance PQ will be of the type t, t – x and t – 2x i.e. in an AP. Thus, their speeds will be in a Harmonic Progression.
The required speed will be the HM of 30 and 60 i.e. 2 * 30 * 60/90 = 40 kmphIf I travel at 10 kmph, I reach office at 10:30 am, if I travel at 15 kmph, I reach office at 10:00 am. At what speed should I travel so that I reach office at 10:15. Assume I leave home at same time and take the same route.
Leaving at same time and reaching at 10 am, 10:15 am and 10:30 am suggests that the time travelled are in AP. Thus, speeds are in HP and required speed is the HM of 10 & 15
i.e.2 * 10 * 15/25 = 12 kmphArun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start? [CAT 2006]
a) 3
b) 3.5
c) 4
d) 4.5
e) 5speed 304060 (AP)
wheras time t(t2)  ???
since speed is in HP, so time Must be in AP, so ans is t4 ===> 4hrsA man can row UP stream in 84 minute. He can row the same distance in 9 min less than he could row it in still water. How long will he take to row down with the stream?
upstream = b  r, still water = b, downstream = b + r ;
so time must be in hp ;
84, t, t9 are in hp. solving this u will get t = 72; so ans is 72  9 = 63A boy is walking along the direction of 2 parallel railway tracks. on one of these tracks, trains are going on 1 direction at equal intervals. on the other track, trains are going in the opp direction at the same equal intervals. the speed of every train is same . In one direction, a train crosses the boy every 20 mins. and in the opp direction the train passes the boy every 30 mins. if the boy stands still beside the tracks, at intervals of how many will two consecutive trains going in the same direction cross him?
when boy & train are travelling in same direction  relative speed = tb;
for opposite direction =t+b;
when boy is stand stil, train will move at = t ;
so here tb, t, t+b are in AP ===> time must be in HP ==> t = 2 * 20 * 30/50 = 24 minEveryday I cover the distance from my home and office at a usual speed and take a certain time at the usual speed. When I increase my usual speed by 5 kmph, I take 10 minutes less than usual. If I reduce my usual speed by 5 kmph, I take 15 minutes more than usual. Find the distance from home to office.
Can you realise that if usual speed is s, then the three speeds are (s – 5), s and (s + 5) i.e. in AP?
Thus, the time taken are in HP.
The time taken as per the data in question is (t + 15), t and (t – 10) and hence we have t = 60 min;
Thus time taken are 75 mins, 60 mins and 50 mins.
Speeds will be in ratio of 1/75 : 1/60 : 1/50 i.e. 4 : 5 : 6.
And we know the difference in speeds are 5 kmph. Thus speeds are 20 kmph, 25 kmph and 30 kmph.
Now distance can be found using any combination of speed and time.Problems on Relative Speed
When two objects are moving simultaneously with speeds S1 and S2, the speed of any object when observed from the other object’s perspective is called the relative speed. And it is distinct from S1 or S2.
E.g.: Consider yourself to be sitting in a moving train and another train passes your train. If the other train is in opposite direction to your train, the speed of the other train appears to be far far more than it actually is. And if the other train is in the same direction as yours, it just appears to be inching ahead of your train very very slowly.This observed/perceived speed is called as relative speed and is calculated as:
If two objects are moving with speeds S1 and S2, their relative speed is
S1 + S2, if they are moving in opposite direction and
S1 – S2, if they are moving in same direction.Relative speed is usually considered when one has to find the time taken to meet or catch and it can be found as follows:
Time taken to meet/catch = Initial distance separating them/Relative Speed i.e. S1 ± S2A thief escapes from a prison at 2 pm and travels away at a speed of 30 kmph. The police realize the escape at 3:30 pm and start the chase then at a speed of 40 kmph. At what time will the police catch the thief? At what distance from the prison is the thief caught?
Approach 1 : Using Relative Speed
When the chase starts i.e. at 3: 30 pm, the thief has already run 30 × 1.5 = 45 kms.
Thus, this is the distance that separated the police and the thief and so the time taken to catch is 45/ (40  30) = 4.5 hours.
This 4.5 hours is measured from the time chase starts i.e. from 3:30 and thus the thief is caught at 8 pm.
The distance run by the police is 40 kmph × 4.5 hours = 180 kms
So the thief is caught at a distance of 180 kms from the prison.Approach 2: Using Proportionality
Method 1: Considering distance constant
Thief escapes from police station and say is caught at point X.
The police also run the same distance, from police station to X.
Thus, the distance run by police and thief is the same. Hence the time for which they are running will be inversely proportional to their speeds. Since ratio of speeds of police and thief is 4 : 3, time for which they are running will be 3 : 4. Also the difference in the time for which they are running is 1½ hour (thief starts running
1½ hours earlier). Thus, police and thief have been running for 3 × 1.5 = 4.5 hrs and 4 × 1.5 = 6 hrs respectively. And so the thief is caught at 3:30 pm + 4.5 hrs i.e. 8 pm (or 2 pm + 6 hrs i.e. 8 pm)
Method 2: Considering time as constant
If we consider the time interval for which the chase is on i.e. from the time police start chasing to the time thief is caught, then the time run will be the same. However in this case the distance run is different (because we are considering the event from 3:30 pm when thief has a headstart of 30 × 1.5 = 45 km).
With time being constant, the distances run will be proportional to speed. Thus, ratio of speed and distance will be 4 : 3. And we know that since 3:30, the police have to run a distance of 45 km more than the thief. Thus, 1 of ratio scale corresponds to 45 km and police will cover a total of 4 × 45 = 180 km in catching the thief. Time taken for this will be 180/40 = 4.5 hrs.Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 am and travels at 50 km per hour towards Baroda situated 100 kms away. At 7:00 am Howrah  Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kms per hour. At 7:30 Mr. Shah, the traffic controller at Baroda realises that both the trains are running on the same tack. How much time does he have to avert a headon collision between the two trains? [cat 1999]
a) 15 minutes
b) 20 minutes
c) 25 minutes
d) 30 minutesConsidering the reference time to be 7:30 am, we would need to find the distance between the two trains at 7:30 am. From 6:30 to 7:30, the Navjivan Express would have travelled 50 kms and from 7 to 7:30, the HowrahAhmedabad would have travelled 20 kms.
Thus the distance separating the two trains at 7:30 would be 100 – 50 – 20 = 30 kms.Approach 1: Using Relative Speed
Time to meet (collide) from 7:30 am is 30 /(50 +40) = 1/3 hours = 20 minutes.
Approach 2: Using proportionality
Since 7:30 to the time they meet, the two trains are travelling for same time. Thus, the distances covered will be in the same ratio as their speeds i.e. 5 : 4. And we know that the total distance covered by the two trains in this time is 30 km.
Thus the trains individually cover, 5/9 * 30 = 50/3 km & 4/9 * 30 = 40/3km.
The time taken for this is (50/3)/50 = 1/ 3 hr = 20 minutesA train X departs from station A at 11:00 am for station B, which is 180 km away. Another train Y departs from station B at 11:00 am for station A. Train X travels of an average speed of 70 kmph and does not stop anywhere until it reaches station B. Train Y travels at an average speed of 50 kmph, but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the point where the trains cross each other? (CAT 2001)
a. 112
b. 118
c. 120
d. None of theseMethod1: train Y stops for 15 min : So in 15 min 'Y' can travel 50 * 15/60 = 12.5 km.
So now we can say that "y" start 12.5 km behind station B. so now the initial distance between them 192.5km. now both trains are starting at the same time ===> whenever they meet time is constant ===> so ratio of distance is 7:5.
distance travelled by "X" = 7/12 * 192.5 = 112 kmMethod2: 60/50 = 1.2 or 1 hr 12 minutes + 15 minutes = 1 hr 27 minutes
in this mean time train X will travel 70+18.9 = 98.9
now distance between them = 21.xx km
21/(70 + 50) = 0.175
0.175 * 70 = 12.25
total distance from A = 98.9 + 12.25 = 111.xx or 112 km approxA car is climbing up a hill at a speed of 90kmph. A rabbit, sitting at the top of the hill spots the car when it is 1170 km away, and start running down the hill towards the car at a speed of 140kmph. As soon as it meets the car, the rabbit turns back towards the top of the hill at a speed of 120kmph. The rabbit continues this to & from motion from the top of the hill towards the car and again back at the top of the hill till the car raches the top of the hill. Find the total distance covered by the rabbit.
The car will reach to the top of the hill after 1170/90 = 13 hrs.
Let total time during which the rabbit climbed downhill & uphill are d & u hrs respectively. Obviously , during every to & fro motion, the distance covered downhill is equal to distance covered uphill.
Therefore, 140d = 120u
==> d/u = 6/7 , also d + u = 13 ==> d= 6 & u= 7
Total distance covered by rabbit = 14d + 120u = 840 + 840 = 1680km.Two trains 150 miles apart are travelling toward each other along the same track. The first train goes 60 miles per hour; the second train rushes along at 90 miles per hour. A fly is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until they collide. If the fly's speed is 120 miles per hour, how far will it travel? find the distance travelled by fly in the direction of 1st train & in the direction of 2nd train.
First part : We want to know the total distance that the fly covers, so let's use Distance = Rate * Time to solve the problem. We already know the fly's rate of flight. If we can find the time that the fly spends in the air, we can figure out how far it travels.
Ignore the fly for a minute, and concentrate on the trains. The first train is traveling at 60 miles/ hour and the second train is going 90 miles/ hour, so they are approaching each other at 60 miles/ hour + 90 miles/ hour = 150 miles/ hour. Now we know the rate at which the trains are closing in on each other and their distance apart (150 miles), so we can find the time until they crash:
Distance = Rate * Time
Time = Distance / Rate
= (150 miles) / (150 miles/ hour)
= 1 hour.
The fly spends the same amount of time traveling as the trains. It goes 120 miles/ hour, so in the one hour the trains take to collide, the fly will go 120 miles.now try to find distance travelled in the direction of first train & distance travelled in the direction of 2nd train
Approach 1: they are meeting after 1hr ===> first train has travelled a distance of 60 miles. since fly has started from 1st train  it has also covered those 60 . but total distance travelled by fly is 120 ==> fly has travelled 60 extra which will be equally distributed in the direction of 1st & 2nd train (as to & fro) ===> in the direction of first train = 60 +30 = 90 & in the direction of 2nd train 30.
Approach 2: let forward direction is F & backward direction is B ==> always remember F:B = (f+t1)/(ft1) where f = speed of fly & t1 is the speed of train from where fly has started initially . here f:b = (120+60):(12060)= 3:1 . so divide 120 miles in the ratio 3:1==> 90 & 30.
Approach 3: 150/150 = 1 hr
in 1 hour fly will travel = 120 miles
forward distance = x
backward distance = y
x + y = 120
x  y = 60
2x = 180
x = 90 = forward distance
y = 30 = backward distanceSimilar question (Credits : Abhishek Jain)
There is a robbery in a shop and when the police reaches the destination, the thief is 5Km away. The speed of the thief is 10m/s and that of the police is 20m/s. Also, the police has a dog which continuously runs between the police and the thief to give the police information about the thief.
a) Find the total distance covered by the dog.
b) Find the distance covered by the dog in the forward direction.
(The speed of the dog is 30m/s)a) relative speed = 36
5/36 * 108 = 15b) forward distance = X
backward distance = Y
X + Y = 15
X  Y = 10
solving X = 12.5 km
proof of X  Y = 10
look try to visualize things
M(Q)(Police)(O)(Dog)+(thief)
let police and dog both are at M dog runs (Q+O) km and touch thief in this mean time police also ran Q km
now DOG will run backward
M(Q)1st point(S)(dog+police)(D)thief
say dog catch police in Z minutes in this mean time police covers S km
distance covered by dog in forward direction = Q+O
distance covered by dog in backward direction = D
now what i have done is
(Q+O)  D should be equal to (Q+S) where as S+D = O
Q+O  O + S = Q+S
RHS = LHSA military column of 1 km length marches with a constant speed of 6 kmph. A courier from the back of the column is sent to the head of the column by bike to deliver a message. Arrived at the front, he instantly returns. When he returns again at the back, the column has travelled a distance of 1 km since the start of his errand. The courier drove with constant speed. How fast did the courier bike?
The relative speed of the courier (compared to the column) is v6 on the journey to the front, and v+6 on the way back. The distance travelled (relatively to the column) by the courier is in both cases 1 km, therefore the total time he is biking is 1 /(v  6) + 1/(v + 6)
The column travels 1 km in 10 minutes (1/6 hour), therefore
1 /(v  6) + 1/(v + 6) = 1/6
This can be written as a quadratic equation (v+6) + (v6) = (v6) × (v+6) / 6 with as only allowed solution v = 6 + 6 × sqrt(2) km/h (approximately 14.49 km/h).(Warning  Tough question)
Moreshwar and Ganesh started travelling towards each other from their hometowns, Hyderabad and Bangalore respectively. They met at point P in between for the first time. As soon as they met, they exchanged their cars (which could travel with their predefined speeds only) and turned back to travel towards their respective hometown cities. As soon as they reached their hometowns, they again started travelling back towards the other city and met at point Q for the second time. Note that after meeting at point P they did not meet each other before they reached their respective hometown cities. What was the ratio of their speeds such that the distance PQ was the highest?
a) 2 : 5
b) 2 : 1
c) 2 : 3
d) 5 : 6
e) 1 : 3Let the speed of Moreshwar and Ganesh be u and v respectively. Their relative speed was u + v and the time they took for the first meeting was d/(u + v) where d is the total distance between the two cities. Therefore, the distance of the point P from Hyderabad (H), where they met for the first time is
L(HP) = u * d/(u+v)  eq 1
Similarly, after meeting they exchange their cars and start travelling back to
their own cities. Now the condition given is that they would not meet each
other before reaching their respective cities. Let us say that u > v. (We will
get similar results and the same ratio even if we consider v > u) This means
that Ganesh would not overtake Moreshwar before Moreshwar reaches
Hyderabad.
Now time taken by moreshwar to reach Hyderabad = (u/v) * {ud/u+v}v
In this time the distance travelled by ganesh = [{ud/u+v}/v] * u = (u/v) * {ud/u+v}This distance is not more than the distance between the point P to Bangalore
and back to Hyderabad.
== (u/v) * {ud /u+v} < {vd/u+v} + dSolving this we get,
(u + v)(u − 2v) < 0
Therefore u has to lie between −v and 2v. But since we have assumed u > v,
the allowed interval for u is v to 2v, both inclusive.
Now we need to find the distance of the point Q from Hyderabad so that we
can find the distance PQ.
After leaving point P and meeting again at point Q they would have travelled a
distance of 2d and their relative speed would be same as (u + v).
Therefore time taken by them to meet again = 2d /(u+v)
In this time distance travelled by moreshwar = 2vd/(u+v)
This is same as the distance between point P and Hyderabad and the
distance between Hyderabad and point Q.
L(HP) +L(HQ) = 2vd/(u+v)eq2Now, l(PQ) = l(HP) − l(HQ)
Solving for l(PQ) from equations (i) and (ii) we get,
L(PQ)= 2d(uv) /(u+v)Looking at the expression, we can see that l(PQ) is an increasing function
with the value of u. Therefore given the range of u, we would see that the
value of l(PQ) is maximum when u = 2v.
Hence, option 2.Yash Chopra’s office is 10 km from his home, which he covers in 2 hours,. One day, as he was on his way to office, he passed his friend’s office who immediately pointed out that he had forgotten to wear his shoes. Yash Chopra therefore turned back, put on his shoes and set off the office. As the passed his friend’s office, she again pointed out that now he had forgotten his glasses. Yash chopra again went back, collected his glasses and reached the office 3 hours late. What is the distance between his home and friend’s office?
a. 3.75
b. 7.35
c. 8.5
d. Insufficient dataLet the distance between home and friend’s office is x. His speed = 10/2 =5 kmph. Extra time taken is 3 hrs. So the total extra distance covered in this time duration is 15 km. This distance is traveled because he goes from home to friends office and back twice.
Thus 4x = 15 or x = 3.75.Some problems for your practice. Use only AM/HM concept while solving them.
Anuva takes 20min less than the usual time to reach her office if her speed increases by 5km/hr , and takes 30 min more than the usual time if her speed decreases by 5 km/hr. What is her usual speed.
A man can walk up a moving “up” escalator in 30 second. The same can walk down this moving “up” escalator in 90 seconds. Assume his walking speed is same Upwards & downwards. How much time he will take to walk up the escalator when escalator is not moving ? (cat 1994)
If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon? [CAT 2004]
a) 11 km/hr
b) 12 km/hr
c) 13 km/hr
d) 14 km/hr
Part 2 : Time, Speed & Distance Concepts & Solved Examples for CAT  Nitin Gupta, AlphaNumeric (Part 2/2)