# Time and Distance - Aptitude Test Tricks, Shortcuts & Formulas

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.

To enhance your knowledge and skills to solve Time and Distance aptitude test problems, go through the tutorial 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 Covered" / "Time taken"
• Ratio of the varying components when other is constant: Consider 2 objects A and B having speed S_a, S_b. Let the distance travelled by them are D_a and D_b respectively and time taken to cover these distances be T_a and 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 S_a = S_b, then D_a/D_b = T_a/T_b
2. When time is constant speed is directly proportional to the distance travelled.
ie; If T_a = T_b, then S_a/S_b = 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 D_a = D_b, then S_a/S_b = 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 = x5/18 m/s
2. x m/s = x18/5 km/hr
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## 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"
• Distance Constant
If distance travelled for each part of the journey, ie 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 s_1, s_2, ... s_n in t_1, t_2, ... t_n times respectively.
Then t_1 = d/s_1, t_2 = d/s_2, … t_n=d/s_n
"Average Speed" = (d + d + d +… "ntimes") / ((d/s_1) + (d/s_2) + (d/s_3) + ... (d/s_n))
"Average Speed" = n/((1/s_1) + (1/s_2) + … + (1/s_n))
• Time Constant
If time taken to travel each part of the journey, ie t_1 = t_2 = t_3 = … t_n = t, then average speed of the object is Arithmetic Mean of speeds
Let distance of parts of the journey be d_1, d_2, d_3, ... d_n and let them be covered with speed s_1, s_2, s_3, ... s_n respectively.
Then d_1 = s_1t, d_2 = s_2t, d_3 = s_3t, ... d_n = s_nt
"Average Speed" = (s_1t + s_2t + ... + s_nt) / (t + t + ... + "ntimes")
"Average Speed" = (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|
• 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 travelled" / "Total time taken"
Let the distance covered be d km.
Given d km be covered with speed a km/hr in time t_1 hour => t_1 = d/a
Given next d km be covered with speed b km/hr in time t_2 hour => t_2 = d/b
"Average speed" = 2d/(d/a + d/b)
"Average speed" = (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 travelled" / "total time taken"
Also "average speed" = (2ab) / (a+b)
=> (2ab)/(a+b) = 2d/T
d = T (ab/(a+b)) km
• If the difference between the individual time taken are given that is, if distance d is covered in t_1 hours with speed a km/hr and same distance is covered with speed b km/hr in t_2 hours, then to find d:
"Difference between individual time" = t_1 – t_2 (if t_1 > t_2)
Also t_1 = d/a and t_2 = d/b
So t_1 – t_2 = d/a – d/b
d = (t_1 – t_2) ((ab)/(b – a)) km
3. If a person covers p^(th) part of a distance at x km/hr, q^(th) part of the distance at y km/hr, r^(th) part of the distance at zkm/hr, then average speed is
"Average speed" = 1 / ((p/x) + (q/y) + (r/z))
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 t_a time to reach Q and B takes t_b time to reach B. If speed of A and B are S_a and S_b respectively.

• Then S_a/S_b is: Let PQ = d and also let PR = l => 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, S_a/S_b = (PR)/(QR) = l / (d – l)
Also, B takes t_b time to cover PR => PR = S_bt_b => l = S_bt_b
A takes t_a time to cover RQ => RQ = S_at_a => d – l = S_at_a
Substituting these values in above equation, we get S_a/S_b = (S_bt_b) / (S_at_a)
=> S_a/S_b = sqrt(t_b / t_a)
• Then time taken by A and B to meet at point R is:
We know t = (PR) / S_a
From previous analysis we also know PR = S_bt_b and S_a/S_b = sqrt(t_b / t_a)
So t = (S_bt_b )/S_a = t_b sqrt(t_b / t_a)
Thus t = sqrt(t_at_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 t_a time to reach R and B takes 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 S_1 and S_2 respectively. They reach their respective destinations and reverse their directions. They continue this to and fro motion. If S_1 > S_2 and S_1 < 2S_2 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)D) / (S_1 + S_2)
"Time taken by the persons to meet first time" = ("Distance travelled by A " + " Distance travelled by B") / ("Speed of A " + " Speed of B")
= 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.
"Time taken by them to meet for the nth time" = ((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)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.

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