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Time and Distance Aptitude Problems, Important Formulas, Shortcuts and Tricks.

Quantitative aptitude problems on time and distance are sure questions in any competitive exam like GRE, GMAT, CAT, XAT, MAT, Bank exams or any other campus recruitment tests for that matter. These problems are often lengthy and test takers often skip them, thinking these questions will consume most amount of time. This is a mistake. Time and distance aptitude questions are easy to score area if you know the very basic formula that you learnt in high school. All shortcuts to solve time and distance problems can be easily derived and learnt. This article provides Tips and Tricks to Solve "Time and Distance" Aptitude Problems with Important Formulas, Shortcuts, Core Concepts to crack Placement Test and Competitive exams.
Do not simply byheart the shortcuts or the tips and tricks. You should actually take time and learn how one arrives at these shortcuts, try it out yourself, then solve as many problems you can. By this you will automatically use the shortcut when you solve questions later on. We have compiled time and distance concepts and formulas for you in this section. We have also given how we derive each of these formulas. They should surely help you crack any aptitude test on time and distance.

Basic Concepts of Time and Distance

Most of the aptitude questions on time and distance can be solved if you know the basic correlation between speed, time and distance which you have learnt in your high school class.
  • Relation between time, distance and speed: Speed is distance covered by a moving object in unit time.
    Speed=Distance CoveredTime taken
  • Ratio of the varying components when other is constant: Consider 2 objects A and Bhaving speed SaSb. Let the distance travelled by them are Da and Db respectively and time taken to cover these distances be Ta and Tb respectively. 
    Let's see the relation between time, distance and speed when one of them is kept constant
    1. When speed is constant distance covered by the object is directly proportional to the time taken.
      ie; If Sa=Sb, then DaDb=TaTb
    2. When time is constant speed is directly proportional to the distance travelled.
      ie; If Ta=Tb, then SaSb=DaDb
    3. When distance is constant speed is inversely proportional to the time taken ie if speed increases then time taken to cover the distance decreases. 
      ie; If Da=Db, then SaSb=TbTa
  • We know that when distance travelled is constant, speed of the object is inversely proportional to time taken.
    1. If the speeds given are in Harmonic progression or HP then the corresponding time taken will be in Arithmetic progression or AP
    2. If the speeds given are in AP then the corresponding time taken is in HP
  • Unit conversion:While answering multiple choice time and dinstance problems in quantitative aptitude test, double check the units of values given. It could be in m/s or km/h. You can use the following formula to convert from one unit to other
    1. x km/hr = x518 m/s
    2. x m/s = x185 km/hr

Average Speed

Average speed is always equal to total distance travelled to total time taken to travel that distance.
Average speed=Total distanceTotal time
  • Distance Constant 
    If distance travelled for each part of the journey, ie d1=d2=d3=...=dn=d, then average speed of the object is Harmonic Mean of speeds.
    Let each distance be covered with speeds s1,s2,...sn in t1,t2,...tn times respectively.
    Then t1=ds1t2=ds2, … tn=dsn
    Average Speed=d+d+d+ntimes(ds1)+(ds2)+(ds3)+...(dsn)
    Average Speed=n(1s1)+(1s2)++(1sn)
  • Time Constant 
    If time taken to travel each part of the journey, ie t1=t2=t3=tn=t, then average speed of the object is Arithmetic Mean of speeds
    Let distance of parts of the journey be d1,d2,d3,...dn and let them be covered with speeds1,s2,s3,...sn respectively.
    Then d1=s1td2=s2td3=s3t, ... dn=snt
    Average Speed=s1t+s2t+...+sntt+t+...+ntimes
    Average Speed=s1+s2+s3+...+snn

Relative Speed

  • If two objects are moving in 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)

Important shortcuts to solve time and distance problems quickly

Using the shortcuts provided below, you can solve the aptitude problems on time and distance quickly
  1. Given a person covers a distance with speed a km/hr and further covers same distance with speed b km/hr, then the average speed of the person is: 
    Average speed=Total distance travelledTotal time taken 
    Let the distance covered be d km
    Given d km be covered with speed a km/hr in time t1 hour => t1=da 
    Given next d km be covered with speed b km/hr in time t2 hour => t2=db 
    Average speed=2dda+db
    Average speed=2aba+b 
    Shortcut: As discussed in "Basic Concepts" section, average speed is the HM (Harmonic Mean) of speeds a & b
  2. Given a person covers a certain distance d km with speed a km/hr and returns back to the starting point with speed b km/hr.
    • If the total time taken for the whole journey is given as T hours, then to find d:
      We know average speed=total distance travelledtotal time taken
      Also average speed=2aba+b
      => 2aba+b=2dT
      d=T(aba+b) km
    • If the difference between the individual time taken are given that is, if distance d is covered in t1 hours with speed a km/hr and same distance is covered with speed b km/hr in t2 hours, then to find d:
      Difference between individual time=t1t2(ift1>t2)
      Also t1=da and t2=db
      So t1t2=dadb
      d=(t1t2)(abba) km
  3. If a person covers pth part of a distance at x km/hr, qth part of the distance at y km/hr, rthpart of the distance at zkm/hr, then average speed is
    Average speed=1(px)+(qy)+(rz)
  4. Two persons A and B start at the same time from two points P and Q at the same time towards each other. They meet at a point R and A takes ta time to reach Q and B takes tbtime to reach B. If speed of A and B are Sa and Sb respectively.
    • Then SaSb is: Let PQ=d and also let PR=l => RQ=dl 
      Time taken by A to cover PR is same as time taken by B to cover QR. 
      We know that when time is constant, speed is directly proportional to distance covered.
      So, SaSb=PRQR=ldl 
      Also, B takes tb time to cover PR => PR=Sbtb => l=Sbtb
      A takes ta time to cover RQ => RQ=Sata => dl=Sata 
      Substituting these values in above equation, we get SaSb=SbtbSata
      => SaSb=tbta
    • Then time taken by A and B to meet at point R is:
      We know t=PRSa
      From previous analysis we also know PR=Sbtb and SaSb=tbta
      So t=SbtbSa=tbtbta
      Thus t=tatb
    • Both equations are valid even if A and B start at 2 different times from P and Q towards each other where A takes ta time to reach R and B takes tb time to reach R. After meeting at R they take the same time t to reach Q and P respectively
  5. Two persons A and B start at the same time from two points P and Q at the same time towards each other with speeds S1 and S2 respectively. They reach their respective destinations and reverse their directions. They continue this to and fro motion. If S1 > S2and S1 < 2S2 and D is the initial distance separating them, then,
    • Total distance covered till nth meeting=(2n-1)D
    • Time taken by them to meet for the nth time=(2n-1)DS1+S2
      Time taken by the persons to meet first time=Distance travelled by A + Distance travelled by BSpeed of A + Speed of B
      =DS1+S2
      After meeting, they continue to Q and P respectively. When they reach their destinations, they have together covered 2D distance.
      Then they reverse directions. By the time they meet for second time, they will have covered 3D distance. Total distance covered by A and B for their 3rd meeting is 5D.
      With this logic, by nth meeting, they will have covered a total distance=(2n-1)D.
      Time taken by them to meet for the nth time=(2n-1)DS1+S2
    • Point of meeting when they meet for the nth time
      1. Distance covered by A till nth meeting = Speed of A * Time taken by A till nth meeting =S1(2n-1)DS1+S2
      2. Divide distance obtained in step 1 by 2D, if value > 2D.
      3. Remainder obtained in step 2 will give you the distance of meeting point from P
When you learn time and distance, you should keep in mind to cover following sections:
  • Time and distance
  • Relative speed and average speed
  • Trains
  • Boats and streams
  • Races
  • Clocks

All these are just different variations of that one simple formula speed = distance / time.

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