Episode 8 — Aptitude and Reasoning / 8.11 — Speed Distance and Time
8.11.c Solved Examples -- Speed, Distance, and Time
Problem 1: Basic Calculation (Easy)
A car travels 240 km in 4 hours. Find its speed.
Given: D = 240 km, T = 4 hours
Speed = D / T = 240 / 4 = 60 km/h
Answer: 60 km/h
Problem 2: Finding Distance (Easy)
A cyclist rides at 15 km/h for 2 hours 30 minutes. How far does he travel?
Given: S = 15 km/h, T = 2 h 30 min = 2.5 hours
D = S x T = 15 x 2.5 = 37.5 km
Answer: 37.5 km
Problem 3: Finding Time (Easy)
How long will it take to cover 450 km at 75 km/h?
Given: D = 450 km, S = 75 km/h
T = D / S = 450 / 75 = 6 hours
Answer: 6 hours
Problem 4: Unit Conversion (Easy)
Convert 90 km/h to m/s.
90 km/h = 90 x (5/18) m/s
= 5 x 5
= 25 m/s
Answer: 25 m/s
Problem 5: Unit Conversion Reverse (Easy)
A sprinter runs at 10 m/s. What is his speed in km/h?
10 m/s = 10 x (18/5) km/h
= 2 x 18
= 36 km/h
Answer: 36 km/h
Problem 6: Average Speed -- Round Trip (Moderate)
A person goes from city A to city B at 40 km/h and returns at 60 km/h. Find the average speed for the entire journey.
Since the distance is the SAME for both legs, use harmonic mean:
Average Speed = 2 x S1 x S2 / (S1 + S2)
= 2 x 40 x 60 / (40 + 60)
= 4800 / 100
= 48 km/h
Answer: 48 km/h
NOTE: The answer is NOT (40+60)/2 = 50 km/h.
Problem 7: Average Speed -- Equal Time (Moderate)
A car travels at 50 km/h for 2 hours and then at 70 km/h for 2 hours. Find the average speed.
Since the TIME is the same for both parts:
Average Speed = (S1 + S2) / 2 = (50 + 70) / 2 = 60 km/h
Verification:
Distance at 50 km/h = 50 x 2 = 100 km
Distance at 70 km/h = 70 x 2 = 140 km
Total distance = 240 km, Total time = 4 hours
Average speed = 240 / 4 = 60 km/h (confirmed)
Answer: 60 km/h
Problem 8: Relative Speed -- Opposite Direction (Moderate)
Two cars start from the same point and travel in opposite directions at 50 km/h and 70 km/h. After how many hours will they be 360 km apart?
They move APART, so relative speed = 50 + 70 = 120 km/h
Time = Distance / Relative speed
= 360 / 120
= 3 hours
Answer: 3 hours
Problem 9: Meeting Problem (Moderate)
A and B are 600 km apart. A starts from city A at 40 km/h and B starts from city B at 60 km/h, both towards each other. When and where do they meet?
A ----------600 km----------> B
40 km/h ----> <---- 60 km/h
Relative speed = 40 + 60 = 100 km/h
Time to meet = 600 / 100 = 6 hours
Meeting point from A = 40 x 6 = 240 km from A
Meeting point from B = 60 x 6 = 360 km from B
Verification: 240 + 360 = 600 km (correct)
Answer: They meet after 6 hours, 240 km from A (360 km from B).
Problem 10: Overtaking Problem (Moderate)
A bus starts from town X at 40 km/h. After 2 hours, a car starts from the same town at 60 km/h. How long will the car take to catch the bus?
Head start of bus = 40 x 2 = 80 km
Relative speed (same direction) = 60 - 40 = 20 km/h
Time for car to catch bus = 80 / 20 = 4 hours
Distance from X where they meet = 60 x 4 = 240 km
Verification: Bus travels for 2+4 = 6 hours at 40 km/h = 240 km (correct)
Answer: The car catches the bus in 4 hours, 240 km from X.
Problem 11: Late and Early Formula (Moderate)
Travelling at 30 km/h, a man reaches his office 10 minutes late. Travelling at 40 km/h, he reaches 5 minutes early. Find the distance.
S1 = 30 km/h, t1 = 10 min late
S2 = 40 km/h, t2 = 5 min early
Total time gap = t1 + t2 = 10 + 5 = 15 min = 15/60 = 1/4 hour
Distance = S1 x S2 x (t1 + t2) / (S2 - S1)
= 30 x 40 x (1/4) / (40 - 30)
= 1200/4 / 10
= 300 / 10
= 30 km
Answer: 30 km
Problem 12: Proportionality (Moderate)
A train covers a distance in 40 minutes at 60 km/h. How long will it take to cover the same distance at 80 km/h?
Method 1: Direct calculation
Distance = 60 x (40/60) = 40 km
Time at 80 km/h = 40/80 = 0.5 hours = 30 minutes
Method 2: Inverse proportionality
S1/S2 = T2/T1
60/80 = T2/40
T2 = 40 x 60/80 = 30 minutes
Answer: 30 minutes
Problem 13: Speed Increase and Time Saved (Moderate)
By increasing speed by 25%, a person reaches his destination 30 minutes earlier. Find the original time of the journey.
Speed increases by 25% = 1/4
Using the shortcut: Time decreases by 1/(4+1) = 1/5
1/5 of original time = 30 minutes
Original time = 30 x 5 = 150 minutes = 2.5 hours
Answer: 2.5 hours (150 minutes)
Problem 14: Stoppage Problem (Moderate)
A train travels at 72 km/h without stoppages. With stoppages, it covers 60 km in an hour. How many minutes per hour does the train stop?
Without stoppages: 72 km/h
With stoppages: 60 km/h
Distance lost per hour = 72 - 60 = 12 km
Time spent stopped = 12/72 hours = 1/6 hour = 10 minutes
Answer: 10 minutes per hour
Problem 15: Three Speeds for Equal Distances (Advanced)
A man covers 1/3 of his journey at 20 km/h, 1/3 at 30 km/h, and 1/3 at 60 km/h. Find the average speed for the entire journey.
Let each part = D km. Total distance = 3D.
Time for part 1 = D/20
Time for part 2 = D/30
Time for part 3 = D/60
Total time = D/20 + D/30 + D/60
= D(3 + 2 + 1)/60
= 6D/60
= D/10
Average speed = 3D / (D/10) = 30 km/h
Using the formula:
Avg = 3 x S1 x S2 x S3 / (S1.S2 + S2.S3 + S1.S3)
= 3 x 20 x 30 x 60 / (20x30 + 30x60 + 20x60)
= 108000 / (600 + 1800 + 1200)
= 108000 / 3600
= 30 km/h (confirmed)
Answer: 30 km/h
Problem 16: Circular Track -- Same Direction (Advanced)
A and B run on a circular track of length 600 m. A runs at 5 m/s and B at 3 m/s in the same direction. After how many seconds will they meet for the first time?
.----.
/ \
/ 600 m \
| track |
\ /
\ /
'----'
Start (both)
Same direction: Relative speed = 5 - 3 = 2 m/s
Time for first meeting = Track length / Relative speed
= 600 / 2
= 300 seconds
= 5 minutes
Answer: 300 seconds (5 minutes)
Problem 17: Circular Track -- Opposite Direction (Advanced)
Using the same track (600 m), if A and B run in opposite directions at 5 m/s and 3 m/s, when do they first meet?
Opposite direction: Relative speed = 5 + 3 = 8 m/s
Time for first meeting = 600 / 8 = 75 seconds
Answer: 75 seconds
Problem 18: Multiple Meetings Between Two Points (Advanced)
A starts from P and B starts from Q simultaneously. They walk towards each other, meet, and continue to the opposite end, then turn back. Distance PQ = 100 km. A walks at 20 km/h and B at 30 km/h. Where do they meet for the second time?
For the 2nd meeting, total distance covered together = 3 x PQ = 300 km
Time = 300 / (20 + 30) = 300/50 = 6 hours
Distance by A in 6 hours = 20 x 6 = 120 km
A starts from P. In 120 km:
- P to Q = 100 km (reaches Q)
- Turns back: 120 - 100 = 20 km from Q (towards P)
- So 80 km from P
Answer: 80 km from P (or 20 km from Q)
Problem 19: Delayed Start Meeting (Advanced)
Two towns P and Q are 200 km apart. A starts from P at 9:00 AM at 40 km/h. B starts from Q at 10:00 AM at 60 km/h towards P. When and where do they meet?
A starts 1 hour before B.
Distance covered by A in 1 hour = 40 x 1 = 40 km
Remaining distance when B starts = 200 - 40 = 160 km
Relative speed (towards each other) = 40 + 60 = 100 km/h
Time after B starts = 160 / 100 = 1.6 hours = 1 hour 36 minutes
They meet at: 10:00 AM + 1 hr 36 min = 11:36 AM
Distance from P = 40 x 1 + 40 x 1.6 = 40 + 64 = 104 km
Verification:
Distance from Q = 60 x 1.6 = 96 km
Total: 104 + 96 = 200 km (correct)
Answer: 11:36 AM, at 104 km from P.
Problem 20: Ratio-Based Problem (Advanced)
The speeds of A and B are in the ratio 5:3. A takes 24 minutes less than B to cover a certain distance. Find the time taken by each and the distance if A's speed is 50 km/h.
Speed ratio = 5 : 3
Time ratio = 3 : 5 (inverse of speed ratio)
Difference in ratio = 5 - 3 = 2 parts
2 parts = 24 minutes
1 part = 12 minutes
Time by A = 3 x 12 = 36 minutes = 36/60 = 3/5 hour
Time by B = 5 x 12 = 60 minutes = 1 hour
If A's speed = 50 km/h:
Distance = 50 x (3/5) = 30 km
Verification:
B's speed = 50 x (3/5) = 30 km/h (using speed ratio)
Time for B = 30/30 = 1 hour = 60 min (correct, 24 min more than A)
Answer: A takes 36 min, B takes 60 min, Distance = 30 km
Problem 21: Journey in Parts (Advanced)
A person covers the first half of a journey at 40 km/h, one-third of the remaining at 20 km/h, and the rest at 10 km/h. Find the average speed.
Let total distance = D
1st part: D/2 at 40 km/h
Remaining = D/2
2nd part: (1/3)(D/2) = D/6 at 20 km/h
3rd part: (2/3)(D/2) = D/3 at 10 km/h
Times:
T1 = (D/2) / 40 = D/80
T2 = (D/6) / 20 = D/120
T3 = (D/3) / 10 = D/30
Total time = D/80 + D/120 + D/30
Finding LCM of 80, 120, 30:
LCM = 240
Total time = 3D/240 + 2D/240 + 8D/240 = 13D/240
Average speed = D / (13D/240) = 240/13 = 18.46 km/h
Answer: 240/13 km/h (approximately 18.46 km/h)
Problem 22: Speed Increase to Reach On Time (Advanced)
A man walks at 5 km/h and reaches his office 20 minutes late. If he walks at a certain speed, he reaches 10 minutes early. The distance to his office is 15 km. Find the required speed.
Time at 5 km/h = 15/5 = 3 hours = 180 minutes
He is 20 minutes late, so actual required time = 180 - 20 = 160 minutes
To reach 10 minutes early, he must complete in = 160 - 10 = 150 minutes
= 150/60 = 2.5 hours
Required speed = 15 / 2.5 = 6 km/h
Answer: 6 km/h
Problem 23: Two People Walking with a Gap (Advanced)
A starts walking at 4 km/h. After 3 hours, B starts cycling at 10 km/h in the same direction. After how many hours (from B's start) will B be 6 km ahead of A?
When B starts, A has covered = 4 x 3 = 12 km (A is ahead by 12 km)
Relative speed of B over A = 10 - 4 = 6 km/h
B first needs to close the 12 km gap and then get 6 km ahead.
Total relative distance = 12 + 6 = 18 km
Time = 18 / 6 = 3 hours (from B's start)
Verification:
After 3 hours from B's start:
A has walked total 6 hours: 4 x 6 = 24 km from start
B has cycled 3 hours: 10 x 3 = 30 km from start
Difference = 30 - 24 = 6 km (B is ahead by 6 km, correct!)
Answer: 3 hours after B starts
Problem 24: Percentage Speed Problem (Advanced)
Due to bad weather, a plane's speed is reduced by 20%, arriving 36 minutes late. Find the usual time of the journey.
Speed is reduced by 20% = 1/5
Using shortcut: Time increases by 1/(5-1) = 1/4
1/4 of usual time = 36 minutes
Usual time = 36 x 4 = 144 minutes = 2 hours 24 minutes
Verification:
Let usual speed = S, usual time = 144 min
New speed = 0.8S
New time = D / 0.8S = D / (4S/5) = 5D/(4S) = 5/4 x 144 = 180 min
Delay = 180 - 144 = 36 min (correct)
Answer: 144 minutes (2 hours 24 minutes)
Problem 25: Complex Meeting Problem (Advanced)
A and B walk towards each other from two points 72 km apart. A walks at 4 km/h and B at 2 km/h. A has a dog that runs at 10 km/h. The dog starts with A, runs to B, then turns back to A, then to B again, and keeps running back and forth until A and B meet. Find the total distance run by the dog.
This classic problem has an elegant solution:
Time for A and B to meet = 72 / (4 + 2) = 72 / 6 = 12 hours
The dog runs for the SAME 12 hours (it never stops running).
Distance run by dog = Speed of dog x Time
= 10 x 12
= 120 km
Answer: 120 km
NOTE: You do NOT need to calculate each back-and-forth trip.
The total time is the key insight.
Previous: 8.11.b Tips, Tricks, and Shortcuts Next: 8.11 Practice MCQs