How to Solve Time, Distance & Speed Problems | 5 Easy Short tricks to Solve Time & Distance Questions

How to Solve Time, Distance & Speed Problems, 5 Easy Short tricks to Solve Time & Speed Questions, Time and Distance – Aptitude Test Tricks, Shortcuts and Formulas, Time and Distance Questions and Answers

How to Solve Time, Distance & Speed Problems : – Are you preparing for Banking, Insurance, Competitive Recruitment or Entrance exams? Aspirants will likely need to solve a section on Quantitative Aptitude for sure. Time Speed and Distance Quiz IBPS, RRB, & SBI will help you learn concepts on an important topic in Numerical Ability Speed, Time & Distance.

This Time Speed and Distance Questions is important for exams such as IBPS PO, IBPS Clerk, IBPS RRB Officer, IBPS RRB Office Assistant, IBPS Specialist Officer, SBI Probationary Officer, SBI Clerk, SBI Specialist Officer, IPPB Bank Scale I Officer, LIC Assistant Administrative officer, GIC AO, United India insurance AO, NIACL AO, NICL AO and others exam.

5 Easy Short tricks to Solve Time, Distance & Speed Problems

Time and Distance is an important chapter of quantitative aptitude. In every numerical ability exam, at least one question can be seen related to this chapter. Before solving questions below read Time and Distance concepts.

How to Solve Time, Distance & Speed Problems

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

(1).Relation between distance, time and speed:

Distance = Speed x Time

(2).To convert speed of any object from KMPH to MPS multiply the speed by = 1000 / 3600 = 5 / 18

(3).To convert speed of any object from MPS {Meter er second} to KMPH {Kilometer per hour} multiply the speed by = 3600 / 1000 = 18/ 5

(4).If the speed ratio of A and B is A : B then ratio of time to cover certain distance is = 1/A : 1/B = B : A

(5).If a person covers certain distance with speed x Kilometer Per Hours and return back with speed y  Kilometer Per Hours then his average speed throughout the journey is

Average speed = 2xy/(x + y) Kilometer per Hours

(6).If a certain distance is covered with 3 diffrent speed x KMPH, y KMPH and z KMPH then average speed throughout the journey is

Average speed = 3xyz/ (xy + yz + zx) Kilometer Per Hours

(7).If 2 different distances covered with speed x KMPH and y KMPH respectively but required same time the then average speed throughout the journey is

Average speed = (x + y)/2 Kilometer Per Hours

(8).If 2 trains start at the same time from different points suppose A and B respectively toward each other and after crossing if they take a and b seconds time resp to reach at B and A point then

(A’s speed) : (B’s speed) =    Squire-root B –  Squire-root A

 Formulae based on Train Problems

Relative Speed (Train Problems):

(9)If two trains are moving in the same direction with speed x KMPH and y KMPH where x>y in that case their relative speed is given as :(x-y) KMPH

(10)If two trains are moving in the opposite direction with speed x KMPH and y KMPH in that case their relative speed is given as: (x+y) KMPH

5 Easy Short tricks to Solve Time & Speed Questions

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)

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Let us take a look at some simple examples of distance, time and speed problems.

Example 1. A boy walks at a speed of 4 kmph. How much time does he take to walk a distance of 20 km?

Solution

Time = Distance / speed = 20/4 = 5 hours.

Example 2. A cyclist covers a distance of 15 miles in 2 hours. Calculate his speed.

Speed = Distance/time = 15/2 = 7.5 miles per hour

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)}

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 these type of problems. Here you go –

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

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

महत्वपूर्ण बिंदु

जब एक ही दिशा में दो ट्रेनें जा रही हैं, तो उनकी सापेक्ष गति दो गति दोनों ट्रेनों की गति का अंतर होता है

जब दो ट्रेनें विपरीत दिशा में आगे बढ़ रही हैं, तो उनकी सापेक्ष गति दो गतियों का योग है

स्थिर आदमी / खंबा/ लैंप पोस्ट / साइन पोस्ट को ट्रैन  क्रॉस करने की स्तिथि में- इन सभी मामलों में, ट्रेन जिस वस्तु को पार करती है, उस दौरान तय की गई यात्रा की दूरी ट्रेन की लंबाई होती है

जब दो ट्रेनें एक ही दिशा में आगे बढ़ रही हैं, तो उनकी गति घटा दी जाएगी।

जब दो ट्रेनें विपरीत दिशाओं में आगे बढ़ रही हैं, तो उनकी गति जोड़ दी जाएगी।

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

Upstream speed, SU=B−S and,

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

Downstream speed, SD=B+S.

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Solution

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

So, our equation would be d/2 + d/4 = 3 → solving which, we get d = 4 miles

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Solution

            Distance                   Speed                   Time

Train      d                             30                           t

Bus         100-d                     40                           3-t

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solving which gives d = 60, which is the distance travelled by train. 100-60 = 40 miles is the distance travelled by bus

Quantitative Aptitude Exam Syllabus pdf

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