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# Time And Distance

Time and Distance quantitative aptitude problems are sure questions in any competitive exam or any other campus recruitment tests. Time and distance aptitude questions are easy to score if we know the very basic formula that we learnt in high school. This article provides Tips and Tricks to Solve “Time and Distance” Aptitude Problems. This article surely helps you crack any aptitude test on time and distance. After reading this article carefully and completely you can able to solve the time and distance problems easily.

## Important Formulas:

Most of the aptitude questions on time and distance can be solved if you know the basic formulas.

Speed: The distance covered per unit time is called Speed. Speed is directly proportional to distance and inversely to time.

• Speed = Distance ⁄ Time
• Time = Distance / Speed
• Distance = Speed×Time

Units :

• Time : Seconds,Minutes,Hours
• Distance : Meter,Kilometer
• Speed : Km/hr,m/sec

Conversion Of Units :

• 1 km/hr = 5/18 meter/second
• 1 meter/second = 18/5 km/hr
• 1 km/hr = 5/8 mile/hr
• 1 mile/hr = 22/15 foot/second

Average Speed :

Average speed is always equal to total distance travelled to total time taken to travel that distance.
Average speed = Total distance / Total time

Relative Speed :

• If two objects are moving in the same direction with speeds a and b then their relative speed is |a−b|
• If two objects are moving is opposite direction with speeds a and b then their relative speed is (a+b)

## Short Cuts :

• Finding out the Average Speed when Equal Distances are covered at Different Speeds.

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 a 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 the return trip. What is its average speed for the entire trip?

The 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, the 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).

•  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 the 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.

•  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 for “Early and Late to Office” Type Problems
The same shortcut used above can be used in this type of problems.

Theorem: A 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 y1 hours 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 a 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?

Solution : 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

•  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

A′ Speed / B ’ Speed = √b / √a

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