5 Types Of Time & Distance Problems With Detailed Solutions

Dear Reader, below you will find 5 important types of time and distance problems which you can expect in IBPS, SBI and RRB exams.

You can use the comments section at the end of the article for doubts. Let us now start the tutorial…

Type I: Unit Conversion Time And Distance Problems (They Are The Most Basic Type)

This type is very easy to solve. You will get speed in one unit (e.g kilometer per hour or km/h) and you have to convert into another unit (e.g metres per second or m/s).

To solve such problems, you have to remember 2 formulas.

Example Question 1: A bus moves at a speed of 81 km/h. What is the speed of the bus in metres per second?

Solution:
In question, you can see that the speed is given in km/h.
You can write, X = 81 km/h
Therefore, speed in m/s (according to our formula) = X x 5/18
= (81 x 5/18)
= 22.5 m/s

Type II: Average Speed When Travelling To A Place And Returning

In this type of questions, a person will move from one place to another at certain speed. Then he will return to the starting place at a different speed. You may be asked to find average speed, distance, etc.

To solve such problems, you have to remember the formula given below.

Example Question 2: Ram walked from his home to the bank at the rate of 20 km/h and returned back at the rate of 5 km/h. If he took 4 hours and 30 minutes for the whole journey, find the distance of the bank from his home.

Solution:
Part 1: Finding Average Speed
From the question, you know the following:
Speed of Ram from home to bank = X = 20 km/h
Speed of Ram from bank to home = Y = 5 km/h
Also, you know that the average speed for the whole journey = 2XY/X+Y
= (2 x 20 x 5) / (20+5)
= 200 / 25 = 8 kmph.

Part 2: Finding Distance
From the question, you know that the time taken for the whole journey = 4 hrs and 30 min
4 hrs and 30 min can be written as 4 ½ hours or 9/2 hours
In part 1 of the solution, you have found that the average speed = 8 kmph
You know the familiar formula that Speed = Distance / Time
Therefore, total distance travelled by Ram = Average Speed X Time Taken
= 8 x 9/2 = 36 Km

Note: But, this distance is the total distance travelled by Ram from his home to bank plus distance he travelled from bank to home. Therefore, to calculate the distance from home to bank, you have to divide the above value by 2.
Therefore, distance from home to bank = 36/2 = 18 Km

Type III: Problems Based On Changing Time And Changing Speed

This type is based on a simple fact that if a person increases his speed he will reach his destination faster and if he decreases his speed he will reach his destination slower.

Though this type looks tough at first, you can solve this type easily if you carefully read and understand the below example.

Example Question 3: If a cyclist rides at a speed of 4 km/h, he reaches the office by 5 minutes late. However, if he rides at a speed of 5 km/h, he reaches 4 minutes earlier. Find the distance covered by him to reach office?

Solution:
Assume that the distance travelled by cyclist (from home to office) to be D km.

Case I: Cyclist’s speed is 4 km/h
Time taken to cover D km at 4 km/h = Distance covered by cyclist / Speed of cyclist= D/4 hours
Note: Above equation is based on the simple formula: Speed = Distance / Time

Case II: Cyclist’s speed is 5 km/h
Time taken to cover D km at 5 km/h = Distance covered by cyclist / Speed of cyclist = D/5 hours.
Time difference between case I and II
You know that when cyclist travels at 4 km/h, he reaches 5 minutes LATE but when he travels at 5 kmph, he reaches 4 minutes EARLIER.
Therefore, difference in time taken between case I and II = (5 minutes + 4 minutes)
= 9 minutes or 3/20 hours

But we know that the time taken in case I is D/4 and that in case II is D/5. Therefore, above equation becomes,
D/4 – D/5 = 3/20
(5D-4D) / 20 = 3/20
D = 3 km

Type IV: Time And Speed Problems On Trains

This type is very popular in bank and other government exams. In this type, starting time, speed, etc., of trains will be given. You will asked to find the time of their crossing.

Here is an example question.

Example Question 4: Two trains start from stations A and B which are 400 km apart. First train starts from A at 9.00 am and travels towards B at 50 km/h. Another train starts from B at 10.00 am and travels towards A at 40 km/h. At what time do they meet?

Solution:
Assume that the two trains will meet each other at X hours after 9.00 am.
Case I: Distance travelled by train 1 before crossing train 2
First train starts at 9.00 am and travels X hours before meeting. Also, in question you can see that the speed of the first train is 50 km/h.
Let D1 be the distance travelled by the first train in X hours.
You can write the below equation based on the above data:
D1 = Speed of the first train x Time taken by the first train till crossing
D1 = 50X km …equation 1

Case II: Distance travelled by train 2 before crossing train 1
Second train starts at 10.00 am and travels (X – 1) hours before meeting the first train
Do you get a doubt that why we used X – 1 instead of X? Here is your answer:
Our assumption is that the trains meet each other at X hours after 9.00 am. Second train starts at 10.00 am i.e., 1 hour after 9. Therefore, it will take X – 1 hours to meet the first train.

Let D2 be the distance travelled by second train in X-1 hours.
Therefore, D2 = Speed of the second train x Time taken by the second train till crossing
D2 = 40(X-1) km …equation 2

Forming equation for total distance
See the below diagram carefully. Here D1 is the distance travelled by first train and D2 is the distance travelled by the second train before meeting.

Therefore, based on the diagram, you can say that
D1 + D2 = Total distance between A and B
From question, you know that the total distance between A and B is 400 Km
Therefore, D1 + D2 = 400

If you substitute the values from equations 1 and 2 in above equation, you will get,
50X + 40(X-1) = 400
50X + 40X – 40 = 400
90X = 360
X = 4 hours
So the two trains will meet each other after 4 hours after 9.00 am. i.e., at 1.00 pm.

Type V: Problems On Bus With Stoppages

Though the bus you are travels appears to travel fast, it will reach its destination very slow if it stops at many places. Type V deals with such cases. Below example will help you to understand clearly.

Example Question 5: A bus without any stoppage can travel a certain distance at an average of 70km/h and with stoppages covers the same distance at an average speed of 50 km/h. How many minutes per hour does the bus stop?

Solution:
Assume that the total distance travelled by the bus to be D km.
Case I: Without stoppages
Without stoppages, the average speed of the bus is 70 km/h.
Time taken by bus without stoppages = Distance covered by the bus / Speed of the bus = D/70 hours …equation 1

Case II: With stoppages
With stoppages, the average speed of the bus is 50 km/h.
Time taken by the bus with stoppages = Distance covered by the bus / Speed of the bus = D/50 hours …equation 2

Calculation of Stopping Time
If we subtract the values of time taken by bus with and without stoppages, we can find the total stopping time. In other words, total stopping time equals the difference in values of equations 1 and 2
Therefore, we can write
Time taken by bus for stoppages (D/50 – D/70) hours = 7D – 5D / 350
= 2D/350
= D/175 hours. …. equation 3

Now, have a look at case II. In case II, bus with stoppages travels at 50 km/h to cover the distance D.
Therefore, journey time of bus with stoppages = Distance / Speed = D/50 hours
Also we found that the total stoppage time = D/175 hours (see equation 3)

In D/50 hours of journey, the bus stops for D/175 hours. The total stoppage time for 1 hour of journey can be calculated using the direct proportion table method as given below.

D/50 = D/175X
Or, X = 50/175 = 2/7 hours
= 2/7 x 60 minutes = 17.14 minutes

Alternate Shortcut Method:
You know that the speed of the bus is reduced by 20 km/h due to stoppages
In other words, the time that the bus takes without stopping to drive 20 km/h will be equal to stoppage time per hour.
Time taken to cover 20 km = (20 km / Actual speed of bus without stoppage )
= 20/70 hours = 2/7 hours
= 20/70 x 60 minutes
= 17.14 minutes