## 5 Quick Shortcuts to Solve Time and Distance Problems

Time, Speed and Distance (TSD) is one of the most frequently occurring topics in quantitative aptitude section of many competitive exams. Here I’m putting together 5 short tricks which might come handy while answering various type of TSD questions. Let us start with some absolutely basic concepts –

### Shortcut #1: Basic Concepts of Time, Speed and Distance

• Speed =Distance/Time
• Time =Distance/Speed
• Distance = Speed x Time
• To convert km/hr to m/sec
• x km/hr =>x  * (5/18) m/sec
• To convert m/sec to km/hr
• y m/sec = y * (18/5) km/hr
• If the ratios of the speeds of A and B is a :b, then the ratio of the times taken to cover the same distance is (1/a) :(1/b) or b:a

### Shortcut #2: Finding out the Average Speed when Equal Distances are covered at Different Speeds

Lot of us make mistakes in calculation of average speed when the same distance is covered at different speeds.  We simply take the average of the given speeds. However, that gives absolutely wrong answer. So now get ready to find out what will give you the correct solution.

Theorem: If a certain distance is covered at x km/hr and the same distance is covered at y km/hr, then

Average speed =[2xy/(x+y)] km/hr

This is basically harmonic mean of the two speeds, i.e. 2/(1/x+1/y)

Example – If a car travels at 40 km/hr on a trip and at 60 km/hr on return trip. What is its average speed for the entire trip?

First thing we should be careful is we shouldn’t just average the 2 speeds. Overall average speed is not (S1+S2)/2.  From the above direct formula the answer will be

Solution: (2x40x60)/(40+60) = 48 km/hr

If we’ve to find the average of more than 2 speeds, average speed will be the harmonic mean of all such speeds

N / (1/a + 1/b + 1/c + 1/d)

Here N = 4, i.e. the number of variables (speeds in this case)

### Shortcut #3: Finding out the Distance when Equal Distances Covered at Different Speeds and Total Journey Time is given

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T hours, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= T * {S1*S2/(S1+S2)}

Example: A boy goes to school at a speed of 3 km/hr and returns to village at 2 km/hr. If he takes 5 hours, what is the distance between the school and the village?

Solution: Distance between school and village is 5 * (3*2) / (3+2)=6 km

### Shortcut #4.1: Finding out the Distance when Equal Distances Covered at Different Speeds

Theorem: A person goes to a destination at a speed of S1 km/hr and returns to his place at a speed of S2 km/hr, if he takes T1 and T2  hours respectively, then

One way distance = Total time taken x (Product of two speeds) / (Addition of two speeds)

= (T1 – T2) * {S1*S2/(S1-S2)}

### Shortcut #4.2: Shortcut for “Early and Late to Office” Type Problems

The same shortcut used above can be used in these type of problems. Here you go –

TheoremA person covers a certain distance having an average speed of x km/hr, he is late by x1 hours but with a speed of y km/hr, he reaches his destination yhours earlier, hence

Required distance = Product of two speeds x Difference between arrival times/Difference of two speeds

Example: A man covers a certain distance between his house and office on bike. Having an average speed of 30 km/hr, he is late by 10 minutes. However, with a speed of 40 km/hr, he reaches his office 5 minutes earlier. Find the distance between his house and office?

In the above case, the required distance = (30×40)x0.25/(40-30) =  30 km

Please note: 10+5 = 15 minutes = 15/60 hours = 0.25 hours

Now since you know few time, speed and distance shortcuts, let’s try this time and speed puzzle

### Shortcut #5: Finding Speed or Time Required after Crossing Each Other

Theorem: If two persons or trains A and B start their journey at the same time from two points P and Q towards each other and after crossing each other they take a and b hours in reaching Q and P respectively, then

Using this relationship you can find out the missing variables which can be either speed or time. Once these are known you can easily find the distance.

Example: Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

Solution: using the above relationship, the ratio of their speed is √16/√9 = 4/3 or 4:3

Do you know more shortcuts for time, speed and distance problems?

## Time and Speed Puzzle

Raja’s office is 9 km from his home. One day, he decided to jog to work in order to increase his fitness. He estimates the journey time to be a little over 1 hour 45 minutes.

At first he feels fine, but before long his energy beings to fade and he realizes he will not be able to run the whole distance. A passing taxi is the answer to his prayer. Raja hails the cab and completes his journey, arriving at work one hour 10 minutes earlier than he had anticipated.

Assuming Raja could at 1/6 the speed of the cab, how far did he run before he hailed the taxi?

## How to Find the Average Speed?

Lot of students gets confused while finding out the average speed, when various distances are travelled with different speed. Say for example, trip to Agra from Delhi is made at an average speed of 40 km/hr and the trip back at an average speed of 60 km/hr. Find their average speed for the entire trip. (Hint: It’s Not 50 Km/hr)

Rule: If a person travels a distance at a speed of x km/hr and the same distance at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

Or, If a person travels half the distance at a speed of x km/hr and the other half at a speed of y km/hr, then the average speed during the whole journey is given by 2xy/x + y km/hr.

So answer to the above question:

(2*60*40)/(60+40)  = 48 Km/hr

If a person travels three equal distances at a speed of x km/hr, y km/hr and z km/hr respectively, then the average speed during the whole journey is 3xyz/xy + yz + xz km/hr.

This problem can also be dealt with assuming some hypothetical distance.

You can find thousands of such amazing faster calculation tricks in the MAGICAL BOOK on QUICKER MATHS – by M. Tyra

If you’ve a question related to time and speed or any other topic in maths, you can post it on QuickerMaths.com’s question answer platform