Time, Speed & Distance Concepts for CAT - Proportionality
QA/DILR Mentor | Be Legend
Authored by Nitin Gupta, Founder, Director at AlphaNumeric.
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 thought-intensive 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 km
A 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/hr
method 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/hr
In 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 Ghaziabad-Hapur-Meerut EMU and the Meerut-Hapur-Ghaziabad 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 = 444
Two 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 police-man 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 (one-third 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?
let speed of train be x. when train runs x then cat runs 3 and when cat runs 5 the train runs x + 8
ratio of speeds =12 : 3 i.e. 4 : 1
Only 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 : 6
Let 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 = 1760
Alternative 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/(w-720) = (w+400)/(2w-400) on solving this W = 1760
I hope that funda of proportionality of T-S-D is clear to everyone. from now onwards, while solving TSD question, look for the part which is constant & then apply proportionality.