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Here is a curated set of questions on Speed, Distance, and Time extracted from the
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key, detailed solutions, and links to relevant video concepts.

Speed, Distance, and Time - Practice Questions

Easy
1.​ A train crosses a pole in 8 seconds. If the length of the train is 200 metres, find the
speed of the train.
○​ a) 75.6 km/h
○​ b) 75.4 km/h
○​ c) 90 km/h
○​ d) 21 km/h
2.​ Walking at 3/4 of his normal speed, a man takes 2.5 hours more than the normal
time. Find the normal time.
○​ a) 7.5 h
○​ b) 6 h
○​ c) 8 h
○​ d) 12 h
3.​ A motor car does a journey in 17.5 hours, covering the first half at 30 km/h and the
second half at 40 km/h. Find the distance of the journey.
○​ a) 684 km
○​ b) 600 km
○​ c) 120 km
○​ d) 540 km
4.​ Without stoppage, a train travels a certain distance with an average speed of 60
km/h, and with stoppage, it covers the same distance with an average speed of 40
km/h. On an average, how many minutes per hour does the train stop during the
journey?
○​ a) 20 min/h
○​ b) 15 min/h
○​ c) 10 min/h
○​ d) 10 min/h

Medium
1.​ Two trains for Mumbai leave Delhi at 6:00 a.m. and 6:45 a.m. and travel at 100 kmph
and 136 kmph respectively. How many kilometres from Delhi will the two trains be
together?
○​ a) 262.4 km
○​ b) 260 km
○​ c) 283.33 km
○​ d) 275 km
2.​ Two trains, Calcutta Mail and Bombay Mail, start at the same time from stations
Kolkata and Mumbai respectively towards each other. After passing each other,
they take 12 hours and 3 hours to reach Mumbai and Kolkata respectively. If the
 Calcutta Mail is moving at the speed of 48 km/h, the speed of the Bombay Mail is
○​ a) 24 km/h
○​ b) 22 km/h
○​ c) 21 km/h
○​ d) 96 km/h
3.​ A passenger train takes 2 hours less for a journey of 300 kilometres if its speed is
increased by 5 kmph over its usual speed. Find the usual speed.
○​ a) 10 kmph
○​ b) 12 kmph
○​ c) 20 kmph
○​ d) 25 kmph
4.​ A plane left half an hour later than the scheduled time and in order to reach its
destination 1500 kilometre away in time, it had to increase its speed by 33.33 per
cent over its usual speed. Find its increased speed.
○​ a) 250 kmph
○​ b) 500 kmph
○​ c) 750 kmph
○​ d) 1000 kmph

Hard
1.​ A man travels from his house to his office at 5 km/h and reaches his office 20
minutes late. If his speed had been 7.5 km/h, he would have reached his office 12
minutes early. Find the distance from his house to his office.
○​ a) 8 km
○​ b) 10 km
○​ c) 7.5 km
○​ d) 12 km
2.​ A train meets with an accident and moves at 3/4 its original speed. Due to this, it is
20 minutes late. Find the original time for the journey beyond the point of accident.
○​ a) 45 minutes
○​ b) 60 minutes
○​ c) 75 minutes
○​ d) 50 minutes
3.​ Two bodies A and B start from opposite ends P and Q of a straight road. They meet
at a point 0.6D from P. Find the point of their fourth meeting. (Where D is the distance
between P and Q)
○​ a) 0.2D from P
○​ b) 0.4D from P
○​ c) 0.6D from Q
○​ d) At P
4.​ A starts walking from a place at a uniform speed of 2 km/h in a particular direction.
After half an hour, B starts from the same place and walks in the same direction as
A at a uniform speed and overtakes A after 1 hour 48 minutes. Calculate the speed
of B.
○​ a) 3 km/h
○​ b) 23/9 km/h
○​ c) 4 km/h
 ○​ d) 2.5 km/h

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Answer Key
Question Answer
1 c
2 a
3 b
4 a
5 c
6 d
7 d
8 d
9 a
10 b
11 a
12 b
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Detailed Solutions
Easy
1.​ Solution: The train has to cover a distance equal to its own length to cross the pole.
Speed of the train = Distance / Time = 200 meters / 8 seconds = 25 m/s. To convert m/s to
km/h, we multiply by 18/5. [cite_start]Speed in km/h = 25 * (18/5) = 90 km/h.
2.​ Solution: When the speed becomes 3/4 of the normal speed, the time taken will be 4/3 of
the normal time. The extra time taken is (4/3)T - T = T/3, where T is the normal time.
Given, T/3 = 2.5 hours. [cite_start]Therefore, the normal time T = 2.5 * 3 = 7.5 hours.
3.​ Solution: Let the total distance be 2D. Time taken for the first half = D/30. Time taken for
the second half = D/40. Total time = (D/30) + (D/40) = (4D + 3D)/120 = 7D/120. Given,
7D/120 = 17.5 hours. D = (17.5 * 120) / 7 = 300 km. [cite_start]Total distance = 2D = 600
km.
4.​ Solution: Let the distance be D. Time without stoppage = D/60. Time with stoppage =
D/40. The extra time taken is due to stoppages: (D/40) - (D/60) = D/120. For a journey of
D/40 hours, the stoppage time is D/120 hours. This means for every 1/40 of an hour of
travel, there is a stoppage of 1/120 of an hour. [cite_start]So, for every hour of travel, the
stoppage is (1/120) / (1/40) = 40/120 = 1/3 of an hour, which is 20 minutes.

Medium
1.​ Solution: The first train starts 45 minutes (0.75 hours) before the second train. In this
time, the first train covers a distance of 100 * 0.75 = 75 km. Now, the second train starts
 and its relative speed with respect to the first train is 136 - 100 = 36 kmph. The time taken
for the second train to catch up with the first train is 75 / 36 hours. Distance from Delhi
where they will meet = Speed of second train * time taken to catch up [cite_start]= 136 *
(75 / 36) = 283.33 km.
2.​ Solution: Let the speeds of Calcutta Mail and Bombay Mail be S1 and S2, and the time
taken to reach their destinations after meeting be T2 and T1 respectively. We have the
formula: S1/S2 = sqrt(T2/T1). Here, S1 = 48 km/h, T2 = 3 hours, T1 = 12 hours. 48/S2 =
sqrt(3/12) = sqrt(1/4) = 1/2. [cite_start]S2 = 48 * 2 = 96 km/h.
3.​ Solution: Let the usual speed be 's' kmph. The time taken is 300/s. When the speed is
increased by 5 kmph, the new speed is (s+5) kmph. The new time taken is 300/(s+5).
Given, 300/s - 300/(s+5) = 2. 300(s+5 - s) / (s(s+5)) = 2 300 * 5 = 2s(s+5) 750 = s(s+5)
[cite_start]By inspection, s = 25 kmph.
4.​ Solution: Let the usual speed be 's' and the usual time be 't'. st = 1500. When the speed
is increased by 33.33% (or 1/3), the new speed is (4/3)s. The new time is t - 0.5. (4/3)s * (t
- 0.5) = 1500 (4/3)s * (t - 0.5) = st (4/3)(t-0.5) = t 4t - 2 = 3t => t = 2 hours. Usual speed =
1500/2 = 750 kmph. [cite_start]Increased speed = 750 * (4/3) = 1000 kmph.

Hard
1.​ Solution: Let the distance be D and the normal time be T. When the speed is 5 km/h, the
time taken is T + 20/60 = T + 1/3. So, D = 5(T + 1/3). When the speed is 7.5 km/h, the
time taken is T - 12/60 = T - 1/5. So, D = 7.5(T - 1/5). Equating the two expressions for D:
5T + 5/3 = 7.5T - 1.5 2.5T = 5/3 + 1.5 = 5/3 + 3/2 = 19/6 T = 19/15 hours. [cite_start]D =
5(19/15 + 1/3) = 5(24/15) = 8 km.
2.​ Solution: When the speed becomes 3/4 of the original speed, the time taken becomes
4/3 of the original time. The extra time taken is (4/3)T - T = T/3. Given, T/3 = 20 minutes.
[cite_start]So, the original time T = 60 minutes.
3.​ Solution: For the first meeting, let the speeds of the bodies be S_A and S_B. Since they
meet at 0.6D from P, the ratio of their speeds is S_A/S_B = 0.6D/0.4D = 3/2. For the
fourth meeting, the total distance covered by both bodies together will be D + 3*(2D) =
7D. Distance covered by A = (3/5) * 7D = 4.2D. Distance covered by B = (2/5) * 7D =
2.8D. To find the meeting point, we can track the position of A. After covering 4D, A will be
at P. It then covers another 0.2D. [cite_start]So, the fourth meeting point is 0.2D from P.
4.​ Solution: When B starts, A has already been walking for half an hour and has covered a
distance of 2 * 0.5 = 1 km. B overtakes A in 1 hour 48 minutes, which is 1.8 hours. In this
time, the distance covered by B relative to A is 1 km. So, the relative speed of B with
respect to A is 1 / 1.8 = 10/18 = 5/9 km/h. Let the speed of B be S_B. S_B - S_A = 5/9
S_B - 2 = 5/9 [cite_start]S_B = 2 + 5/9 = 23/9 km/h.

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Video Solution Concepts
Here are some links to YouTube videos that explain the core concepts of Speed, Distance, and
Time, which will be helpful for solving these types of problems:
●​ Speed, Distance and Time - Basics and Formulas
●​ Problems on Trains - Concepts and Tricks
 ●​ Boats and Streams - Upstream and Downstream Concepts
●​ Relative Speed - Same and Opposite Direction

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