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Quantitative Aptitude · Time Speed and Distance

Basics- Formulas, Key Points and Examples

Explanation

🟠 1. BASIC CONCEPT Time, Speed and Distance are related by the fundamental formula:
Speed=DistanceTime \text{Speed}=\frac{\text{Distance}}{\text{Time}}
From this formula:
Distance=Speed×Time \text{Distance}=\text{Speed}\times\text{Time}
Time=DistanceSpeed \text{Time}=\frac{\text{Distance}}{\text{Speed}}
🟠 2. IMPORTANT FORMULAS Speed:
S=DT S=\frac{D}{T}
Distance:
D=S×T D=S\times T
Time:
T=DS T=\frac{D}{S}
🟠 3. UNITS OF SPEED Common units:
km/h \text{km/h}
m/s \text{m/s}
Conversion from km/h to m/s:
1 km/h=518 m/s 1\text{ km/h}=\frac{5}{18}\text{ m/s}
Therefore:
Speed in m/s=Speed in km/h×518 \text{Speed in m/s} = \text{Speed in km/h}\times\frac{5}{18}
Conversion from m/s to km/h:
1 m/s=185 km/h 1\text{ m/s}=\frac{18}{5}\text{ km/h}
Therefore:
Speed in km/h=Speed in m/s×185 \text{Speed in km/h} = \text{Speed in m/s}\times\frac{18}{5}
🟠 4. UNIT CONVERSION Important conversions:
1 km=1000 m 1\text{ km}=1000\text{ m}
1 hour=60 minutes 1\text{ hour}=60\text{ minutes}
1 minute=60 seconds 1\text{ minute}=60\text{ seconds}
Therefore:
1 hour=3600 seconds 1\text{ hour}=3600\text{ seconds}
🟠 5. AVERAGE SPEED Average speed is:
Average Speed=Total DistanceTotal Time \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}
Important: Do not simply take the average of two speeds unless the time spent at each speed is equal. 🟠 6. AVERAGE SPEED FOR EQUAL DISTANCES If a person travels equal distances at speeds aa and bb, then:
Average Speed=2aba+b \text{Average Speed} = \frac{2ab}{a+b}
🟠 7. AVERAGE SPEED FOR EQUAL TIMES If a person travels for equal amounts of time at speeds aa and bb, then:
Average Speed=a+b2 \text{Average Speed} = \frac{a+b}{2}
🟠 8. RELATIVE SPEED Relative speed is used when two objects are moving with respect to each other. For objects moving in the same direction:
Relative Speed=∣S1−S2∣ \text{Relative Speed}=|S_1-S_2|
For objects moving in opposite directions:
Relative Speed=S1+S2 \text{Relative Speed}=S_1+S_2
🟠 9. TWO OBJECTS MOVING TOWARDS EACH OTHER If two objects are moving towards each other:
Relative Speed=S1+S2 \text{Relative Speed}=S_1+S_2
Therefore:
Time=Distance Between ThemS1+S2 \text{Time} = \frac{\text{Distance Between Them}} {S_1+S_2}
🟠 10. TWO OBJECTS MOVING IN THE SAME DIRECTION If two objects are moving in the same direction:
Relative Speed=S1−S2 \text{Relative Speed}=S_1-S_2
Therefore:
Time=Distance Between ThemS1−S2 \text{Time} = \frac{\text{Distance Between Them}} {S_1-S_2}
🟠 11. SPEED AND TIME ARE INVERSELY PROPORTIONAL For a fixed distance:
S∝1T S\propto\frac{1}{T}
Therefore:
S1T1=S2T2 S_1T_1=S_2T_2
If speed increases, time decreases. If speed decreases, time increases. 🟠 12. PERCENTAGE CHANGE IN SPEED AND TIME For the same distance, if speed increases by x%x\%, the decrease in time is:
x100+x×100% \frac{x}{100+x}\times100\%
If speed decreases by x%x\%, the increase in time is:
x100−x×100% \frac{x}{100-x}\times100\%
🟠 13. TRAIN PROBLEMS When a train crosses a pole, person or tree:
Distance covered=Length of train \text{Distance covered}=\text{Length of train}
Therefore:
T=Length of TrainSpeed T=\frac{\text{Length of Train}}{\text{Speed}}
🟠 14. TRAIN CROSSING A PLATFORM When a train completely crosses a platform:
Distance covered=Length of Train+Length of Platform \text{Distance covered} = \text{Length of Train} + \text{Length of Platform}
Therefore:
T=Train Length+Platform LengthSpeed T= \frac{\text{Train Length}+\text{Platform Length}} {\text{Speed}}
🟠 15. TWO TRAINS CROSSING EACH OTHER For trains moving in opposite directions:
Relative Speed=S1+S2 \text{Relative Speed}=S_1+S_2
Distance covered:
D=L1+L2 D=L_1+L_2
Therefore:
T=L1+L2S1+S2 T= \frac{L_1+L_2}{S_1+S_2}
For trains moving in the same direction:
Relative Speed=∣S1−S2∣ \text{Relative Speed}=|S_1-S_2|
Therefore:
T=L1+L2∣S1−S2∣ T= \frac{L_1+L_2}{|S_1-S_2|}
🟠 16. BOAT AND STREAM Let:
B=Speed of boat in still water B=\text{Speed of boat in still water}
S=Speed of stream S=\text{Speed of stream}
Downstream speed:
D=B+S D=B+S
Upstream speed:
U=B−S U=B-S
🟠 17. SPEED OF BOAT AND STREAM If downstream and upstream speeds are known:
B=D+U2 B=\frac{D+U}{2}
S=D−U2 S=\frac{D-U}{2}
🟠 18. DOWNSTREAM AND UPSTREAM TIME Downstream:
T=DB+S T=\frac{D}{B+S}
Upstream:
T=DB−S T=\frac{D}{B-S}
🟠 19. RACES If A beats B by xx metres in a race of DD metres, then when A completes DD metres, B completes:
D−x D-x
Therefore:
SASB=DD−x \frac{S_A}{S_B} = \frac{D}{D-x}
🟠 20. RACE TIME RATIO For the same distance:
T1T2=S2S1 \frac{T_1}{T_2} = \frac{S_2}{S_1}
Higher speed means lower time. 🟠 21. CIRCULAR TRACK If two people run in the same direction on a circular track:
Relative Speed=S1−S2 \text{Relative Speed}=S_1-S_2
If they run in opposite directions:
Relative Speed=S1+S2 \text{Relative Speed}=S_1+S_2
Time to meet:
T=Track LengthRelative Speed T= \frac{\text{Track Length}} {\text{Relative Speed}}
🟠 22. MEETING PROBLEMS If two people start from two different places and move towards each other:
Time to Meet=Distance Between ThemSum of Speeds \text{Time to Meet} = \frac{\text{Distance Between Them}} {\text{Sum of Speeds}}
🟠 23. IMPORTANT KEY POINTS • Always keep distance, speed and time in compatible units. • Convert km/h to m/s when distance is given in metres and time in seconds. • Convert m/s to km/h when the answer is required in km/h. • For the same direction, subtract speeds. • For opposite directions, add speeds. • For a train crossing a pole, use only the train length. • For a train crossing a platform, use train length plus platform length. • Average speed is always:
Total DistanceTotal Time \frac{\text{Total Distance}}{\text{Total Time}}
• For a fixed distance, speed and time are inversely proportional. • In boat and stream problems, add the stream speed downstream and subtract it upstream. • In race problems, compare speeds using the distances covered in the same time.

Example

🟢 EXAMPLE 1: FINDING SPEED QUESTION: A car travels 240 km in 4 hours. Find its speed. SOLUTION:
S=DT S=\frac{D}{T}
S=2404 S=\frac{240}{4}
S=60 km/h S=60\text{ km/h}
ANSWER:
60 km/h 60\text{ km/h}
🟢 EXAMPLE 2: FINDING DISTANCE QUESTION: A car travels at 72 km/h for 5 hours. Find the distance travelled. SOLUTION:
D=S×T D=S\times T
D=72×5 D=72\times5
D=360 km D=360\text{ km}
ANSWER:
360 km 360\text{ km}
🟢 EXAMPLE 3: FINDING TIME QUESTION: A train travels 450 km at a speed of 75 km/h. Find the time taken. SOLUTION:
T=DS T=\frac{D}{S}
T=45075 T=\frac{450}{75}
T=6 hours T=6\text{ hours}
ANSWER:
6 hours 6\text{ hours}
🟢 EXAMPLE 4: CONVERT KM/H TO M/S QUESTION: Convert 90 km/h into m/s. SOLUTION:
90×518 90\times\frac{5}{18}
=25 m/s =25\text{ m/s}
ANSWER:
25 m/s 25\text{ m/s}
🟢 EXAMPLE 5: CONVERT M/S TO KM/H QUESTION: Convert 20 m/s into km/h. SOLUTION:
20×185 20\times\frac{18}{5}
=72 km/h =72\text{ km/h}
ANSWER:
72 km/h 72\text{ km/h}
🟢 EXAMPLE 6: AVERAGE SPEED QUESTION: A car travels 120 km in 2 hours and another 180 km in 3 hours. Find the average speed. SOLUTION: Total distance:
120+180=300 km 120+180=300\text{ km}
Total time:
2+3=5 hours 2+3=5\text{ hours}
Average speed:
3005 \frac{300}{5}
=60 km/h =60\text{ km/h}
ANSWER:
60 km/h 60\text{ km/h}
🟢 EXAMPLE 7: AVERAGE SPEED FOR EQUAL DISTANCES QUESTION: A person travels equal distances at 40 km/h and 60 km/h. Find the average speed. SOLUTION: Using:
Average Speed=2aba+b \text{Average Speed} = \frac{2ab}{a+b}
=2×40×6040+60 =\frac{2\times40\times60}{40+60}
=4800100 =\frac{4800}{100}
=48 km/h =48\text{ km/h}
ANSWER:
48 km/h 48\text{ km/h}
🟢 EXAMPLE 8: RELATIVE SPEED QUESTION: Two cars are moving in opposite directions at 50 km/h and 70 km/h. Find their relative speed. SOLUTION: For opposite directions:
Relative Speed=50+70 \text{Relative Speed}=50+70
=120 km/h =120\text{ km/h}
ANSWER:
120 km/h 120\text{ km/h}
🟢 EXAMPLE 9: SAME DIRECTION QUESTION: Two cars are moving in the same direction at 80 km/h and 50 km/h. Find their relative speed. SOLUTION:
Relative Speed=80−50 \text{Relative Speed}=80-50
=30 km/h =30\text{ km/h}
ANSWER:
30 km/h 30\text{ km/h}
🟢 EXAMPLE 10: MEETING PROBLEM QUESTION: Two towns are 300 km apart. Two cars start from the towns towards each other at 60 km/h and 40 km/h. After how much time will they meet? SOLUTION: Relative speed:
60+40=100 km/h 60+40=100\text{ km/h}
Time:
T=300100 T=\frac{300}{100}
=3 hours =3\text{ hours}
ANSWER:
3 hours 3\text{ hours}
🟢 EXAMPLE 11: SPEED INCREASE AND TIME DECREASE QUESTION: A person increases his speed by 25%. Find the percentage decrease in the time taken for the same distance. SOLUTION: Using:
Decrease in Time=x100+x×100 \text{Decrease in Time} = \frac{x}{100+x}\times100
=25125×100 =\frac{25}{125}\times100
=20% =20\%
ANSWER:
20% decrease 20\%\text{ decrease}
🟢 EXAMPLE 12: SPEED DECREASE AND TIME INCREASE QUESTION: A person's speed decreases by 20%. Find the percentage increase in time taken for the same distance. SOLUTION:
Increase in Time=x100−x×100 \text{Increase in Time} = \frac{x}{100-x}\times100
=2080×100 =\frac{20}{80}\times100
=25% =25\%
ANSWER:
25% increase 25\%\text{ increase}
🟢 EXAMPLE 13: TRAIN CROSSING A POLE QUESTION: A train 180 m long is moving at 54 km/h. How long will it take to cross a pole? SOLUTION: Convert speed:
54×518=15 m/s 54\times\frac{5}{18}=15\text{ m/s}
Distance:
180 m 180\text{ m}
Time:
T=18015 T=\frac{180}{15}
=12 seconds =12\text{ seconds}
ANSWER:
12 seconds 12\text{ seconds}
🟢 EXAMPLE 14: TRAIN CROSSING A PLATFORM QUESTION: A train 200 m long crosses a platform 300 m long at a speed of 90 km/h. Find the time taken. SOLUTION: Total distance:
200+300=500 m 200+300=500\text{ m}
Convert speed:
90×518=25 m/s 90\times\frac{5}{18}=25\text{ m/s}
Time:
T=50025 T=\frac{500}{25}
=20 seconds =20\text{ seconds}
ANSWER:
20 seconds 20\text{ seconds}
🟢 EXAMPLE 15: TWO TRAINS OPPOSITE DIRECTIONS QUESTION: Two trains of lengths 150 m and 250 m move in opposite directions at 54 km/h and 36 km/h. Find the time taken to cross each other. SOLUTION: Total length:
150+250=400 m 150+250=400\text{ m}
Relative speed:
54+36=90 km/h 54+36=90\text{ km/h}
Convert:
90×518=25 m/s 90\times\frac{5}{18}=25\text{ m/s}
Time:
T=40025 T=\frac{400}{25}
=16 seconds =16\text{ seconds}
ANSWER:
16 seconds 16\text{ seconds}
🟢 EXAMPLE 16: TWO TRAINS SAME DIRECTION QUESTION: Two trains of lengths 120 m and 180 m move in the same direction at 72 km/h and 54 km/h. Find the time taken by the faster train to cross the slower train. SOLUTION: Total length:
120+180=300 m 120+180=300\text{ m}
Relative speed:
72−54=18 km/h 72-54=18\text{ km/h}
Convert:
18×518=5 m/s 18\times\frac{5}{18}=5\text{ m/s}
Time:
T=3005 T=\frac{300}{5}
=60 seconds =60\text{ seconds}
ANSWER:
60 seconds 60\text{ seconds}
🟢 EXAMPLE 17: BOAT DOWNSTREAM QUESTION: The speed of a boat in still water is 15 km/h and the speed of the stream is 3 km/h. Find the downstream speed. SOLUTION:
D=B+S D=B+S
D=15+3 D=15+3
D=18 km/h D=18\text{ km/h}
ANSWER:
18 km/h 18\text{ km/h}
🟢 EXAMPLE 18: BOAT UPSTREAM QUESTION: The speed of a boat in still water is 15 km/h and the speed of the stream is 3 km/h. Find the upstream speed. SOLUTION:
U=B−S U=B-S
U=15−3 U=15-3
U=12 km/h U=12\text{ km/h}
ANSWER:
12 km/h 12\text{ km/h}
🟢 EXAMPLE 19: FIND BOAT AND STREAM SPEEDS QUESTION: A boat travels downstream at 20 km/h and upstream at 12 km/h. Find the speed of the boat in still water and the speed of the stream. SOLUTION: Boat speed:
B=D+U2 B=\frac{D+U}{2}
B=20+122 B=\frac{20+12}{2}
B=16 km/h B=16\text{ km/h}
Stream speed:
S=D−U2 S=\frac{D-U}{2}
S=20−122 S=\frac{20-12}{2}
S=4 km/h S=4\text{ km/h}
ANSWER:
Boat speed=16 km/h \text{Boat speed}=16\text{ km/h}
Stream speed=4 km/h \text{Stream speed}=4\text{ km/h}
🟢 EXAMPLE 20: BOAT AND STREAM TIME QUESTION: A boat moves downstream at 18 km/h. How much time will it take to cover 72 km? SOLUTION:
T=DS T=\frac{D}{S}
T=7218 T=\frac{72}{18}
T=4 hours T=4\text{ hours}
ANSWER:
4 hours 4\text{ hours}
🟢 EXAMPLE 21: RACE PROBLEM QUESTION: A can run 100 m in 10 seconds and B can run 100 m in 12 seconds. Find the ratio of their speeds. SOLUTION: For the same distance:
S=DT S=\frac{D}{T}
Therefore:
SA:SB=10010:10012 S_A:S_B = \frac{100}{10}:\frac{100}{12}
=10:253 =10:\frac{25}{3}
Multiply by 3:
30:25 30:25
=6:5 =6:5
ANSWER:
6:5 6:5
🟢 EXAMPLE 22: RACE MARGIN QUESTION: A can run 100 m in 10 seconds while B takes 12 seconds. When A finishes the race, how far behind is B? SOLUTION: A's speed:
SA=10010=10 m/s S_A=\frac{100}{10}=10\text{ m/s}
In 10 seconds, B covers:
10012×10 \frac{100}{12}\times10
=8313 m =83\frac{1}{3}\text{ m}
Distance by which B is behind:
100−8313 100-83\frac{1}{3}
=1623 m =16\frac{2}{3}\text{ m}
ANSWER:
1623 m 16\frac{2}{3}\text{ m}
🟢 EXAMPLE 23: CIRCULAR TRACK QUESTION: Two runners run on a circular track of length 400 m in opposite directions at 8 m/s and 12 m/s. After how much time will they meet? SOLUTION: Relative speed:
8+12=20 m/s 8+12=20\text{ m/s}
Time:
T=40020 T=\frac{400}{20}
=20 seconds =20\text{ seconds}
ANSWER:
20 seconds 20\text{ seconds}
🟢 EXAMPLE 24: CIRCULAR TRACK SAME DIRECTION QUESTION: Two runners run on a circular track of length 500 m in the same direction at 10 m/s and 6 m/s. After how much time will the faster runner catch the slower runner? SOLUTION: Relative speed:
10−6=4 m/s 10-6=4\text{ m/s}
Time:
T=5004 T=\frac{500}{4}
=125 seconds =125\text{ seconds}
ANSWER:
125 seconds 125\text{ seconds}
🟢 EXAMPLE 25: MULTI-STAGE JOURNEY QUESTION: A person travels 120 km at 40 km/h and then 180 km at 60 km/h. Find the average speed for the entire journey. SOLUTION: Time for first part:
T1=12040=3 hours T_1=\frac{120}{40}=3\text{ hours}
Time for second part:
T2=18060=3 hours T_2=\frac{180}{60}=3\text{ hours}
Total distance:
120+180=300 km 120+180=300\text{ km}
Total time:
3+3=6 hours 3+3=6\text{ hours}
Average speed:
3006 \frac{300}{6}
=50 km/h =50\text{ km/h}
ANSWER:
50 km/h 50\text{ km/h}