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.
• Ratio of the varying components when other is constant: Consider 2 objects A and Bhaving speed ${\displaystyle S_a}$, ${\displaystyle S_b}$. Let the distance travelled by them are ${\displaystyle D_a}$ and ${\displaystyle D_b}$ respectively and time taken to cover these distances be ${\displaystyle T_a}$ and ${\displaystyle T_b}$ 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 ${\displaystyle S_a=S_b}$, then ${\displaystyle \frac{D_a}{D_b}=\frac{T_a}{T_b}}$
  2. When time is constant speed is directly proportional to the distance travelled.
     ie; If ${\displaystyle T_a=T_b}$, then ${\displaystyle \frac{S_a}{S_b}=\frac{D_a}{D_b}}$
  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 ${\displaystyle D_a=D_b}$, then ${\displaystyle \frac{S_a}{S_b}=\frac{T_b}{T_a}}$
• 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 = ${\displaystyle x\frac{5}{18}}$ m/s
  2. $x$ m/s = ${\displaystyle x\frac{18}{5}}$ km/hr

Average Speed

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

• Distance Constant 
  If distance travelled for each part of the journey, ie ${\displaystyle d_1=d_2=d_3=...=d_n=d}$, then average speed of the object is Harmonic Mean of speeds.
  Let each distance be covered with speeds ${\displaystyle s_1,s_2,...s_n}$ in ${\displaystyle t_1,t_2,...t_n}$ times respectively.
  Then ${\displaystyle t_1=\frac{d}{s_1}}$, ${\displaystyle t_2=\frac{d}{s_2}}$, … ${\displaystyle t_n=\frac{d}{s_n}}$
  
• Time Constant 
  If time taken to travel each part of the journey, ie ${\displaystyle t_1=t_2=t_3=\dots t_n=t}$, then average speed of the object is Arithmetic Mean of speeds
  Let distance of parts of the journey be ${\displaystyle d_1,d_2,d_3,...d_n}$ and let them be covered with speed${\displaystyle s_1,s_2,s_3,...s_n}$ respectively.
  Then ${\displaystyle d_1=s_{1}t}$, ${\displaystyle d_2=s_{2}t}$, ${\displaystyle d_3=s_{3}t}$, ... ${\displaystyle d_n=s_{n}t}$
  
  

Relative Speed

• If two objects are moving in same direction with speeds a and b then their relative speed is${\displaystyle |a-b|}$
• If two objects are moving is opposite direction with speeds a and b then their relative speed is ${\displaystyle (a+b)}$

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: 
    
   Let the distance covered be d km. 
   Given d km be covered with speed a km/hr in time ${\displaystyle t_1}$ hour => ${\displaystyle t_1=\frac{d}{a}}$ 
   Given next d km be covered with speed b km/hr in time ${\displaystyle t_2}$ hour => ${\displaystyle t_2=\frac{d}{b}}$ 
   
   
    
   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 
     Also 
     => ${\displaystyle \frac{2ab}{a+b}=2\frac{d}{T}}$
     
     ${\displaystyle d=T\left(a\frac{b}{a+b}\right)}$ km
   • If the difference between the individual time taken are given that is, if distance d is covered in ${\displaystyle t_1}$ hours with speed a km/hr and same distance is covered with speed b km/hr in ${\displaystyle t_2}$ hours, then to find d:
     
     Also ${\displaystyle t_1=\frac{d}{a}}$ and ${\displaystyle t_2=\frac{d}{b}}$
     So ${\displaystyle t_1–t_2=\frac{d}{a}–\frac{d}{b}}$
     
     ${\displaystyle d=(t_1–t_2)\left(\frac{ab}{b–a}\right)}$ km
3. If a person covers ${\displaystyle p^{th}}$ part of a distance at x km/hr, ${\displaystyle q^{th}}$ part of the distance at y km/hr, ${\displaystyle r^{th}}$part of the distance at zkm/hr, then average speed is
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 ${\displaystyle t_a}$ time to reach Q and B takes ${\displaystyle t_b}$time to reach B. If speed of A and B are ${\displaystyle S_a}$ and ${\displaystyle S_b}$ respectively.
   
   • Then ${\displaystyle \frac{S_a}{S_b}}$ is: Let ${\displaystyle PQ=d}$ and also let ${\displaystyle PR=l}$ => ${\displaystyle RQ=d–l}$ 
     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, ${\displaystyle \frac{S_a}{S_b}=\frac{PR}{QR}=\frac{l}{d–l}}$ 
     Also, B takes ${\displaystyle t_b}$ time to cover PR => ${\displaystyle PR=S_b{t}_b}$ => ${\displaystyle l=S_b{t}_b}$
     A takes ta time to cover RQ => ${\displaystyle RQ=S_a{t}_a}$ => ${\displaystyle d–l=S_a{t}_a}$ 
     Substituting these values in above equation, we get ${\displaystyle \frac{S_a}{S_b}=\frac{S_b{t}_b}{S_a{t}_a}}$
     => ${\displaystyle \frac{S_a}{S_b}=\sqrt{\frac{t_b}{t_a}}}$
   • Then time taken by A and B to meet at point R is:
     We know ${\displaystyle t=\frac{PR}{S_a}}$
     From previous analysis we also know ${\displaystyle PR=S_b{t}_b}$ and ${\displaystyle \frac{S_a}{S_b}=\sqrt{\frac{t_b}{t_a}}}$
     So ${\displaystyle t=\frac{S_b{t}_b}{S_a}=t_b\sqrt{\frac{t_b}{t_a}}}$
     
     Thus ${\displaystyle t=\sqrt{t_a{t}_b}}$
   • Both equations are valid even if A and B start at 2 different times from P and Q towards each other where A takes ${\displaystyle t_a}$ time to reach R and B takes ${\displaystyle t_b}$ 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 ${\displaystyle S_1}$ and ${\displaystyle S_2}$ respectively. They reach their respective destinations and reverse their directions. They continue this to and fro motion. If ${\displaystyle S_1}$ > ${\displaystyle S_2}$and ${\displaystyle S_1}$ < ${\displaystyle 2S_2}$ and D is the initial distance separating them, then,
   • 
     
     ${\displaystyle =\frac{D}{S_1+S_2}}$
     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, .
     
   • 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 ${\displaystyle =S1\frac{(2n-1)D}{S1+S2}}$
     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.