Time and Distance - Aptitude Test Tricks, Shortcuts & Formulas

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.

To enhance your knowledge and skills to solve Time and Distance aptitude test problems, go through the tutorial on Time and Distance.

Time and Distance Tutorial
Part I: Genral Concepts, Average Speed and Variation of Parameters
Part II: Relative Speed, Linear Races, Circular Races, Meeting Points

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${\displaystyle Speed=\frac{\text{Distance Covered}}{\text{Time taken}}}$
• Ratio of the varying components when other is constant: Consider 2 objects A and B having speed Sa${\displaystyle S_a}$, Sb${\displaystyle S_b}$. Let the distance travelled by them are Da${\displaystyle D_a}$ and Db${\displaystyle D_b}$ respectively and time taken to cover these distances be Ta${\displaystyle T_a}$ and Tb${\displaystyle T_b}$ 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 Sa=Sb${\displaystyle S_a=S_b}$, then DaDb=TaTb${\displaystyle \frac{D_a}{D_b}=\frac{T_a}{T_b}}$
  2. When time is constant speed is directly proportional to the distance travelled.
     ie; If Ta=Tb${\displaystyle T_a=T_b}$, then SaSb=DaDb${\displaystyle \frac{S_a}{S_b}=\frac{D_a}{D_b}}$
  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 Da=Db${\displaystyle D_a=D_b}$, then SaSb=TbTa${\displaystyle \frac{S_a}{S_b}=\frac{T_b}{T_a}}$
• 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${\displaystyle x}$ km/hr = x518${\displaystyle x\frac{5}{18}}$ m/s
  2. x${\displaystyle x}$ m/s = x185${\displaystyle x\frac{18}{5}}$ km/hr

Average Speed

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

Average speed=Total distanceTotal time${\displaystyle \text{Average speed}=\frac{\text{Total distance}}{\text{Total time}}}$

• Distance Constant 
  If distance travelled for each part of the journey, ie d1=d2=d3=...=dn=d${\displaystyle d_1=d_2=d_3=...=d_n=d}$, then average speed of the object is Harmonic Mean of speeds.
  Let each distance be covered with speeds s1,s2,...sn${\displaystyle s_1,s_2,...s_n}$ in t1,t2,...tn${\displaystyle t_1,t_2,...t_n}$ times respectively.
  Then t1=ds1${\displaystyle t_1=\frac{d}{s_1}}$, t2=ds2${\displaystyle t_2=\frac{d}{s_2}}$, … tn=dsn${\displaystyle t_n=\frac{d}{s_n}}$
  Average Speed=d+d+d+…ntimes(ds1)+(ds2)+(ds3)+...(dsn)${\displaystyle \text{Average Speed}=\frac{d+d+d+\dots \text{ntimes}}{\left(\frac{d}{s_1}\right)+\left(\frac{d}{s_2}\right)+\left(\frac{d}{s_3}\right)+...\left(\frac{d}{s_n}\right)}}$
  Average Speed=n(1s1)+(1s2)+…+(1sn)${\displaystyle \text{Average Speed}=\frac{n}{\left(\frac{1}{s_1}\right)+\left(\frac{1}{s_2}\right)+\dots +\left(\frac{1}{s_n}\right)}}$
• Time Constant 
  If time taken to travel each part of the journey, ie t1=t2=t3=…tn=t${\displaystyle t_1=t_2=t_3=\dots t_n=t}$, then average speed of the object is Arithmetic Mean of speeds
  Let distance of parts of the journey be d1,d2,d3,...dn${\displaystyle d_1,d_2,d_3,...d_n}$ and let them be covered with speed s1,s2,s3,...sn${\displaystyle s_1,s_2,s_3,...s_n}$ respectively.
  Then d1=s1t${\displaystyle d_1=s_{1}t}$, d2=s2t${\displaystyle d_2=s_{2}t}$, d3=s3t${\displaystyle d_3=s_{3}t}$, ... dn=snt${\displaystyle d_n=s_{n}t}$
  Average Speed=s1t+s2t+...+sntt+t+...+ntimes${\displaystyle \text{Average Speed}=\frac{s_{1}t+s_{2}t+...+s_{n}t}{t+t+...+\text{ntimes}}}$
  
  Average Speed=s1+s2+s3+...+snn${\displaystyle \text{Average Speed}=\frac{s_1+s_2+s_3+...+s_n}{n}}$

Relative Speed

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

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: 
   Average speed=Total distance travelledTotal time taken${\displaystyle \text{Average speed}=\frac{\text{Total distance travelled}}{\text{Total time taken}}}$ 
   Let the distance covered be d km. 
   Given d km be covered with speed a km/hr in time t1${\displaystyle t_1}$ hour => t1=da${\displaystyle t_1=\frac{d}{a}}$ 
   Given next d km be covered with speed b km/hr in time t2${\displaystyle t_2}$ hour => t2=db${\displaystyle t_2=\frac{d}{b}}$ 
   Average speed=2dda+db${\displaystyle \text{Average speed}=2\frac{d}{\frac{d}{a}+\frac{d}{b}}}$
   
   Average speed=2aba+b${\displaystyle \text{Average speed}=\frac{2ab}{a+b}}$ 
   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${\displaystyle \text{average speed}=\frac{\text{total distance travelled}}{\text{total time taken}}}$
     Also average speed=2aba+b${\displaystyle \text{average speed}=\frac{2ab}{a+b}}$
     => 2aba+b=2dT${\displaystyle \frac{2ab}{a+b}=2\frac{d}{T}}$
     
     d=T(aba+b)${\displaystyle d=T\left(a\frac{b}{a+b}\right)}$ km
   • If the difference between the individual time taken are given that is, if distance d is covered in t1${\displaystyle t_1}$ hours with speed a km/hr and same distance is covered with speed b km/hr in t2${\displaystyle t_2}$ hours, then to find d:
     Difference between individual time=t1–t2(ift1>t2)${\displaystyle \text{Difference between individual time}=t_1–t_2(if{t}_1>t_2)}$
     Also t1=da${\displaystyle t_1=\frac{d}{a}}$ and t2=db${\displaystyle t_2=\frac{d}{b}}$
     So t1–t2=da–db${\displaystyle t_1–t_2=\frac{d}{a}–\frac{d}{b}}$
     
     d=(t1–t2)(abb–a)${\displaystyle d=(t_1–t_2)\left(\frac{ab}{b–a}\right)}$ km
3. If a person covers pth${\displaystyle p^{th}}$ part of a distance at x km/hr, qth${\displaystyle q^{th}}$ part of the distance at y km/hr, rth${\displaystyle r^{th}}$part of the distance at zkm/hr, then average speed is
   Average speed=1(px)+(qy)+(rz)${\displaystyle \text{Average speed}=\frac{1}{\left(\frac{p}{x}\right)+\left(\frac{q}{y}\right)+\left(\frac{r}{z}\right)}}$
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${\displaystyle t_a}$ time to reach Q and B takes tb${\displaystyle t_b}$ time to reach B. If speed of A and B are Sa${\displaystyle S_a}$ and Sb${\displaystyle S_b}$ respectively.
    
   • Then SaSb${\displaystyle \frac{S_a}{S_b}}$ is: Let PQ=d${\displaystyle PQ=d}$ and also let PR=l${\displaystyle PR=l}$ => RQ=d–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, SaSb=PRQR=ld–l${\displaystyle \frac{S_a}{S_b}=\frac{PR}{QR}=\frac{l}{d–l}}$ 
     Also, B takes tb${\displaystyle t_b}$ time to cover PR => PR=Sbtb${\displaystyle PR=S_b{t}_b}$ => l=Sbtb${\displaystyle l=S_b{t}_b}$ 
     A takes ta${\displaystyle t_a}$ time to cover RQ => RQ=Sata${\displaystyle RQ=S_a{t}_a}$ => d–l=Sata${\displaystyle d–l=S_a{t}_a}$ 
     Substituting these values in above equation, we get SaSb=SbtbSata${\displaystyle \frac{S_a}{S_b}=\frac{S_b{t}_b}{S_a{t}_a}}$
     => SaSb=tbta−−−√${\displaystyle \frac{S_a}{S_b}=\sqrt{\frac{t_b}{t_a}}}$
   • Then time taken by A and B to meet at point R is:
     We know t=PRSa${\displaystyle t=\frac{PR}{S_a}}$
     From previous analysis we also know PR=Sbtb${\displaystyle PR=S_b{t}_b}$ and SaSb=tbta−−−√${\displaystyle \frac{S_a}{S_b}=\sqrt{\frac{t_b}{t_a}}}$
     So t=SbtbSa=tbtbta−−−√${\displaystyle t=\frac{S_b{t}_b}{S_a}=t_b\sqrt{\frac{t_b}{t_a}}}$
     
     Thus t=tatb−−−√${\displaystyle t=\sqrt{t_a{t}_b}}$
   • Both equations are valid even if A and B start at 2 different times from P and Q towards each other where A takes ta${\displaystyle t_a}$ time to reach R and B takes tb${\displaystyle t_b}$ time to reach R. After meeting at R they take the same time t${\displaystyle t}$ 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${\displaystyle S_1}$ and S2${\displaystyle S_2}$ respectively. They reach their respective destinations and reverse their directions. They continue this to and fro motion. If S1${\displaystyle S_1}$ > S2${\displaystyle S_2}$ and S1${\displaystyle S_1}$ < 2S2${\displaystyle 2S_2}$ and D is the initial distance separating them, then,
   • Total distance covered till nth meeting=(2n−1)D${\displaystyle \text{Total distance covered till nth meeting}=(2n-1)D}$
   • Time taken by them to meet for the nth time=(2n−1)DS1+S2${\displaystyle \text{Time taken by them to meet for the nth time}=\frac{(2n-1)D}{S_1+S_2}}$
     Time taken by the persons to meet first time=Distance travelled by A + Distance travelled by BSpeed of A + Speed of B${\displaystyle \text{Time taken by the persons to meet first time}=\frac{\text{Distance travelled by A }+\text{ Distance travelled by B}}{\text{Speed of A }+\text{ Speed of B}}}$
     =DS1+S2${\displaystyle =\frac{D}{S_1+S_2}}$
     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, by nth meeting, they will have covered a total distance=(2n−1)D${\displaystyle \text{by nth meeting, they will have covered a total distance}=(2n-1)D}$.
     
     Time taken by them to meet for the nth time=(2n−1)DS1+S2${\displaystyle \text{Time taken by them to meet for the nth time}=\frac{(2n-1)D}{S_1+S_2}}$
   • 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${\displaystyle =S1\frac{(2n-1)D}{S1+S2}}$
     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