Time and distance is one of the main sources for tricky quantitative questions in the well fame, well designed B-school entrance exams. An aspirant can expect 4 to 6 questions from this topic in CAT and CAT level exams. MAT and CMAT are playing a lot with this topic. For a better performance in this area, aspirants are required to acquire a certain level of awareness and skills which are mentioned below:
- Well awareness on basic concepts (ie. the relationship between the main parameters such as Distance, Speed and Time)
- Effective application of the concepts such as Ratio and Inverse & Direct variations
- Quick and intellectual percentage calculations
- Logical thinking and visualization/imagination skills
In this chapter we will look at problems in the following different dimensions:
- Basic time speed distance relation based on problems
- Basic Problems of trains
- Inverse direct variation problem
- Average Speed
Basic Concept (Relation between time, speed and distance)
If the distance between 2 cities X and Y is 100km and a car is travelling from X to Y at a constant speed of 50 kmph then it will take 2 hrs to reach Y. Here
- `"Distance = 100km"`
- `"Speed = 50 kmph"`
- `"Time = 2h"`
Let's see how these three parameters are connected with each other.
`"Time" = "Distance"/"Speed"`
`"Speed" = "Distance"/"Time"`
Standard units of parameters
Distance normally expressed in form of kilometer(km) or meter(m), while speed has the standard unit expressions such as kilometer per hour or kmph(km/h)or meter per second (m/s)and time is in the standard units of hours(hrs) and seconds(s).
Conversion of units of speed
`1 " kmph" = (1 " km")/(1 " hr") = (1000 " m ")/ (3000 " s ") = 5/18 "m/s"`
SImilarly, `1 "m/s" = 18/5 "kmph"`
`q " m/s" = q xx 18/5 " kmph"`
`"Speed " = 18 kmph = 5 m/s`
`"Time required to cover the distance" = "Speed"/"Distance" = 1000/5 = 200 sec = 3 min 20 sec`
Method I: Using Algebra
Let the distance he covered by bus is `x km`
`:.` Distance covered by train is `620 - x`
`-> x/60 + (620 - x)/80 = 9`
`-> 4x + 1860 - 3x = 2169`
`-> x = 300km`
Note: Work back method always start with a middle quantity as per the ascending order of option values. We can start the substitution with 300 or 320.Speed of bus is 60 kmph and the total time required in an integer value. So most probably the distance covered by bus is an integral multiple of 60(not always!). So here we can start our substitution from 300.
If 300 km is the distance covered by bus then the remaining 320 km covered by train.
`:.` Time taken by bus `= 300/6 = 5hr`
And time taken by train `= 320/80 =4hr`
Total time `= 5 + 4 = 9hr`, hence verified the answer our substitution is correct.
`"Distance covered by bus" = 300 km"`
Average speed of entire journey `= 620/9`
The resultant ratio is; `80/9 : 100/9 -> 4:5`, This is the ratio of time distribution for each mode of travel.
`:.` Time duration for the journey by bus `=5 hr`
Distance covered by bus `= 5 xx 60 = 300 km`
Basic problems on trains
This is the basic train related problems (or the first version of the train problems). In this area we are dealing with the kind of problems in which train crosses a stationary object rather than other moving objects. Most of these types of questions are intended to test the relationship between the basic parameters only and more often the distance covered is the sum of the lengths of the train and that of the covering object. Depends upon the situation, the length (width) of the covering objects may be negligible.
You will get a clear idea about the concept through the following set of concept review questions.
`"Time" = 200/10 = 20 " m/s"`
` = 20 xx 18/5 " kmph" = 72 " kmph"`
`= 2x + x = 3x`
Speed of the train is `135 " kmph" = 135 xx 5/18 "m/s"`
We know that, `"distance" = "speed" xx "time"`
`3x = 135 xx 5/18 xx 10 xx 1/3 = 375m`
`-> "Length of train, 2x" = 250m`
Data Sufficiency questions on time and distance
In these types of question, we can analyze if the given data is enough to arrive at solution for the problem in hand using basic concepts of time and distance. Let's look at few examples to understand how to tackle data sufficiency type of problems on time and distance.
I: Travelling from home at a constant speed, Ajay takes 15 minutes to reach shop A
II: Travelling from shop A at a constant rate, he takes 15 minutes to reach shop B
A. Statement I alone sufficient statement II alone is not sufficient
B. Statement II alone sufficient statement I alone is not sufficient
C. Both statements together are sufficient but neither statement alone is sufficient
D. Each statement alone is sufficient
E. Statement I and II together are not sufficient to answer the question, and additional data are required
Statement I tell us the time taken by Ajay to reach the Shop A from his home. It doesn't contain any information about shop B .Hence the statement I alone is not sufficient.
Statement II tells us the time duration for the trip from shop A to shop B. It doesn't convey the information about the placing of his home as per the positions of both the shops. Hence statement II alone is not sufficient.
Clubbing statements I and II together doesn't give any inference about the placing of shop A and shop B. If the speeds which are mentioned in the statements are same then only one inference we can form. The distance between home and shop A is equal to the distance e between shop A and shop B
Consider the following possibilities
Possibility I: Shop A is closed to home
Possibility II: Shop B is closed to home
Possibility III: Shop A and shop B are equidistant from home
`:. ` A unique conclusion is not possible while clubbing the statements together, hence answer is `E`.
Variation wise relations of basic parameters
Whenever we can think about a time - speed - distance situation, there are only three parameters are valid. On in a different wording we can tell time, speed and distance are the foundation blocks of this topic for solving the questions effectively
One of main fundamental requirements is to understand the variation wise (proportionate) relationship between time, speed and distance.
Consider the different scenarios
- Scenario I: When distance is constant
When distance is constant, increase in speed will make a corresponding decrease in time, required covering a fixed distance. Both the parameters (speed and time) vary in opposite direction means increase in one parameter will make a corresponding decrease in the other and it is vice versa. Then we can say that,Speed is inversely proportional to time, when distance is constant. This can be expressed as `"Speed" \propto 1/"Time"`Let the Distance = 120 km
At 40kmph speed, we can cover it in 3 hours
At 60kmph speed, we can cover it in 2 hours.
`"Speed I"/"Speed II" = "Time II"/"Time I" -> 40/60 = 2/3`
- Scenario II: When time is constant
When time is constant, distance covered will increase or decrease as per the increase or decrease in the speed. Then we can say that,Speed is directly proportional to distance, when time is constant. This can be expressed as `"Speed" \propto "Distance"`Let the time taken = 2hr
At 40kmph speed, we can cover 80km.
At 60kmph speed, we can cover 120km.
`"Speed I"/"Speed II" = "Distance I"/"Distance II" -> 40/60 = 80/120`
- Scenario III: When speed is constant
When speed is constant, distance covered will increase or decrease as per the increase or decrease in the time. Then we can say that,Time is directly proportional to distance when speed is constant. This can be expressed as `"Time" \propto "Distance"`Let the speed be = 40kmph
With 2hrs, we can cover 80km.
With 3hrs, we can cover 120km.
`"Time I"/"Time II" = "Distance I"/"Distance II" -> 2/3 = 80/120`
Let's look at few examples to see how to use the above proportions to solve aptitude problems on time and distance.
`"Ratio of speed" = "Ratio of distance" = 2: 4: 5`
Given that Car P covers 20 km in 2 hrs, i.e. 10 km in 1 hr. Now, if we map it to the proportions, we get
`2 -> 10km`
`4 -> 20km`
`5 -> 25km`
Distance covered by car Q and R in one hour are 20 km and 25 km respectively. In three hours car Q and car R can cover `20 xx 3 = 60 km` and `25 xx 3 = 75 km` respectively.
`:.`Difference between the distances is `75 - 60 = 15 km`
Generalization of the concept
If a person travels from A to B at a speed of `S_1 " kmph"` then he reaches early by 'x' minutes and when he travels at `S_2 " kmph"`, then he reaches early by 'y' minutes.
Let the distance between A and B is D.
Time taken for the journey from A to B at `S_1 "kmph", t_1 = D/S_1`
Time taken for the journey at `S_2 " kmph", t_2 = D/S_2`
Here `S_2 > S_1` then `t_1 > t_2` and T is the time difference. `T = t_1 - t_2` or `(x + y) min = (x + y)/60 hr`.
Then `D/S_1 - D/S_2 = T`
`D(S_2 - S_1)/S_1S_2 = T`
`D = (S_1 xx S_2 xx T)/(S_2 - S_1)`
D = `"Product of the speeds and time difference"/"Difference between speeds"`
`T = 10 min + 5 min = 15 min = 1/4 hrs`
`D = (S_1 xx S_2 xx T)/(S_2 - S_1) = (5 xx 8 xx 1/4)/(8 - 5) = 10/3 km`
`T = (20 + 30)/60 = 5/6 hrs`
`D = (S_1 xx S_2 xx T)/(S_2 - S_1) = (3 xx 2 xx 5/6)/(3 - 2) = 5 km`
If he is covering 5km at 2kmph then the time required to cover the distance `= 5/2 = 2.5 hrs`; given that this is 30 minutes more than the usual time consumption .
`:.` The usual time (i.e. the expected time duration required to cover the distance) `= 2.5 hrs - 0.5 hrs = 2hrs`
`"Distance" = 5 km, "Usual time duration" = 2 hrs, "Usual speed" = 2.5 kmph`
Consider the following situation: A train is travelling at `50 kmph` to cover a distance in between stations A and B .In its further journey, train took a speed of `70 kmph` to reach station C from station B. Find the average speed of the train in the total journey from station A to station C? You may easily reach a (obviously wrong!!!) conclusion; that the average speed of the entire journey `= (50+70)/2 =`
60 kmph. What is wrong with this conclusion?
In this question , the distance between stations are not specified, hence it is not possible to find a unique solution for the question. The above answer,
60 kmph, is actually average of speeds not average speed.
Suppose in the above example if distance between Station A and station B is 150 km and distance between Station B and C is 140 km. then,
Time taken to cover AB `= 150/50= 3 hrs`
Time taken to cover BC `= 140 /70= 2 hrs`
Total Distance (AB + BC) `= 150 + 140 = 290 km`
Total time taken for journey from A to C `= 3 + 2 = 5 hrs`
Average speed `= 290/5= 58 kmph`
Average Speed of Round Trip: When distance covered in to and fro segments of journey are same.
We can consider the situation in two ways. In a to and fro (round trip) journey both the distances are equal, i.e. AB = BA. So we can consider a similar situation in the form of a series trip such as A to B then B to C. If AB = BC (distances are equal) then the situations are similar.
Let the distance between A and B = `d km`, `:.` Distance between B and C = `d km`
Time taken for trip A to B = `d/u hrs` (`u` is the speed with which A to B is covered)
Time taken for trip B to C = `d/v hrs` (`v` is the speed with which A to B is covered)
We know that, `"Average speed" = "Total distance" / "Total time"`
`-> (d + d)/ (d/u + d/v) = (2uv)/(u+v) kmph`
Note: Harmonic mean of any two quantities u and v `= (2uv)/(u+v)`, ie. in the above context, average speed of the entire round trip is harmonic mean of the speeds.
Average Speed: When distance covered in two segments of the journey are different.
Let's consider following situation: Cities A, B and C situated as the three vertices of an equilateral triangle, i.e. the cities are equidistant to each other and connected by straight line roads such as the sides of an equilateral triangle. If a person starts his journey from A to B at 'x' kmph, then B to C at 'y' kmph and further C to A at 'z' kmph then stopped. What is the average speed of this entire round trip?
A car covered AB at a constant speed of 'u' kmph and BC at a constant speed of 'v' kmph. Then
`"Average speed of the entire journey from A to C via B" = ((x + y)uv)/(xv + yu) kmph`
Average Speed and Average of Speeds
Let's look at example to understand this concept in detail.
Let the speed from B to C `= p kmph`
Average speed `= (2 xx 70 xx p)/(70 + p)`
Average of the speeds `= (70 + p)/2`
Now, requirement is `"Average speed" > "Average of the speeds"`
`-> (2 xx 70 xx p)/(70 + p) > (70 + p)/2`
`-> (2 xx 70 xx p) xx 2 > (70+p)^2`
`-> p^2 - 140p + 4900 < 0`
`-> (p - 70)^2 < 0`, which is not possible since square of a natural number is always greater than or equal to 0.
This means, average of speeds will be always greater than or equal to average speed.