Time & Distance Tutorial I: Basic Concepts, Average Speed & Variation of Parameters

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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.

Ajay must get stationary products from either shop A or shop B .Which of the two shops is closer to Ajay's home?
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

Find out which of the below holds true
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 the speeds of three cars P, Q and R is `2: 4: 5` respectively and the distance covered by car P in 2 hrs is 20 km. Find the difference in the distances covered by Q and R in 3 hrs?
Speed is directly proportional to distance when time is constant.
`"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)`

A person covers a certain distance at `S_1 kmph` taking `t_1` minutes and covering the same distance at `S_2 kmph` and taking `t_2` minutes; Where `S_1 < S_2` and `T = (t_1 - t_2)/60` Then,
Then distance,`D = (S_1 xx S_2 xx T)/(S_2 - S_1)`
OR
D = `"Product of the speeds and time difference"/"Difference between speeds"`

Examples

Starting from home and travelling at a constant speed of 5 kmph, girl reaches school 10 minutes late. If she travels at a speed of 8 kmph, she will reach the school 5 minutes early. What is the distance from her home to school?
`S_1 = 5 kmph` and `S_2 = 8 kmph`
`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`
`= 3333.3m`
If Sam walks at a speed of 3 kmph, he is 20 minutes early at the office. If he walks at a speed of 2 kmph, he is late by 30 minutes at the office. What should be his speed, so as to reach the office on time?
`S_1 = 3 kmph` and `S_2 = 2 kmph`
`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`
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