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Arithmetic II

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Time & Speed

Time And Speed

MODULES

Basics of Ratios
Average Speed
Product Constancy
Relative Speed
Trains
Walking To & Fro
Circular Track
Boats, Races & Clock
Escalators
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Time & Speed 1
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Time & Speed 2
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Time & Speed 3
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Time & Speed 4
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PRACTICE

Time & Speed : Level 1
Time & Speed : Level 2
Time & Speed : Level 3
ALL MODULES

CAT 2025 Lesson : Time & Speed - Trains

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6. Train Problems

6.1 Train crossing a stationary object

If a train of length lll takes time ttt to pass an object of negligible length at speed sss, then l=s×tl = s \times tl=s×t

If a train of length l1l_1l1​ takes time ttt to cross an object of length l2l_2l2​ at speed sss, then l1+l2=s×tl_1 + l_2 = s \times tl1​+l2​=s×t

It must be noted that the train is said to have crossed the stationary object only when the engine of the train meets the objects and crosses until the last wagon crosses the object.

Example 18

If a train travelling at a speed of 144 km/hr crosses a pole in 3 seconds, how long does it take the train to cross a platform which is 200m in length?

Solution

We first convert the speed to m/s as the time is given in seconds.

144144144 km/hr =144×518= 144 \times \dfrac{5}{18}=144×185​ m/s = 40m/s

Let lll metres be the length of the train. Speed s=40s = 40s=40 m/s and time t=3t = 3t=3 seconds.

l=s×tl = s \times t l=s×t ⇒ 40×3=120 40 \times 3 = 12040×3=120 metres

Distance === Train length +++ Platform length =120+200= 120 + 200=120+200 = 320 metres

Time taken to cross the platform =32040= \dfrac{320}{40}=40320​ = 8 seconds

Answer: 888 seconds

6.2 Train crossing another moving object of negligible length or vice-versa

If a train of length lll and a car (or any moving object) of negligible length, moving in the opposite direction at speeds of s1s_1s1​ and s2s_2s2​ respectively, take time ttt to cross each other from the point they meet, then their relative speed is s1s_1s1​ + s2s_2s2​ and the distance is lll.

l=(s1+s2)×tl = (s_1 + s_2) \times tl=(s1​+s2​)×t

Example 19

How many seconds would a 480 metre long train travelling at a speed of 60 km/hr take to completely go past a horse travelling at a speed of 48 km/hr in the opposite direction?

Solution

Relative speed =60+48=108= 60 + 48 = 108=60+48=108 km/hr =108×518 = 108 \times \dfrac{5}{18}=108×185​ m/s = 303030 m/s

Time taken for the 480480480m long train to go past horse =48030=16= \dfrac{480}{30} = 16=30480​=16 seconds

Answer: 161616 seconds

If a train of length lll and a car (or any moving object) of negligible length, moving in the same direction and at speeds of s1s_1s1​ and s2s_2s2​ respectively (where s1>s2s_1 > s_2s1​>s2​), and the train takes time ttt to cross the car from the point they meet, then their relative speed is s1−s2s_1 - s_2s1​−s2​ and the distance is lll.
l=(s1−s2)×tl = (s_1 - s_2) \times tl=(s1​−s2​)×t

Note that, if the car is faster than the train, the relative speed is s2−s1s_2 - s_1s2​−s1​ and the equation is
l=(s1−s2)×tl = (s_1 - s_2) \times tl=(s1​−s2​)×t

Example 20

A train travelling at a speed of 727272 km/hr takes 101010 seconds to cross a car travelling in the opposite direction at a speed of 545454 km/hr. How many seconds would it take the train to overtake another car travelling, in the same direction as the train, at the speed of 21.621.621.6 km/hr?

Solution

As time is given in seconds, we convert the speeds to m/s.

727272 km/hr =72×518= 72 \times \dfrac{5}{18}=72×185​ m/s =20= 20=20 m/s

545454 km/hr =54×518= 54 \times \dfrac{5}{18}=54×185​ m/s =15= 15=15 m/s

Time t1=10t_1 = 10t1​=10 seconds and with the train and the car moving in the same direction, relative speed s1=20+15=35s_1 = 20 + 15 = 35s1​=20+15=35 m/s. Let the length of the train be lll metres.

l=s×t1 l = s \times t_1 l=s×t1​ ⇒ l=35×10=350 l = 35 \times 10 = 350l=35×10=350 metres

Speed of the second car =21.6×518=6= 21.6 \times \dfrac{5}{18} = 6=21.6×185​=6 m/s

As the train and the second car are moving in the same direction, relative speed s2=20−6=14s_2 = 20 - 6 = 14s2​=20−6=14 m/s. Let time taken to cross be t2t_2t2​.

∴350=14×t2\therefore 350 = 14 \times t_2 ∴350=14×t2​ ⇒ t2=25 t_2 = 25t2​=25 seconds

Answer: 252525 seconds

6.3 Train crossing another train

If two trains of length l1l_1l1​ and l2l_2l2​ moving in the opposite direction and at speeds of s1s_1s1​ and s2s_2s2​ respectively, take time ttt to cross each other from the point they meet, then their relative speed is s1+s2s_1 + s_2s1​+s2​ and the distance is l1+l2l_1 + l_2l1​+l2​.

l1+l2=(s1+s2)×t l_1 + l_2 = (s_1 + s_2) \times t l1​+l2​=(s1​+s2​)×t

If two trains of length l1l_1l1​ and l2l_2l2​ moving in the same direction and at speeds of s1s_1s1​ and s2s_2s2​ respectively (where s1>s2s_1 > s_2s1​>s2​), and the faster train takes time ttt to cross the slower train from the point they meet, then their relative speed is s1−s2s_1- s_2s1​−s2​ and the distance is l1+l2l_1 + l_2l1​+l2​.

l1+l2=(s1−s2)×t l_1 + l_2 = (s_1 - s_2) \times t l1​+l2​=(s1​−s2​)×t

Example 21

A 150015001500 metre long train running at the speed of 606060 m/s crosses another train running in opposite direction at the speed of 656565 m/s in 282828 seconds. What is the length of the other train?

Solution

Let the length of the other train be xxx. As all the units are in metres and seconds, there is no need for conversion.

Since the two trains are running in the opposite directions, sum of their speeds is the relative speed i.e., s=60+65=125s = 60 + 65 = 125s=60+65=125 m/s and the total distance to be covered by the trains to cross each other equals to sum of the lengths of the two trains.

Time taken for the two trains to cross each other =Total DistanceRelative Speed= \dfrac{\text{Total Distance}}{\text{Relative Speed}}=Relative SpeedTotal Distance​

⇒ 1500+x125=28 \dfrac{1500 + x}{125} = 281251500+x​=28

⇒ 1500+x=3500 1500 + x = 35001500+x=3500

⇒ x=2000 x = 2000x=2000 m

Answer: 200020002000 m

Example 22

Two trains that are 555 km apart and travelling towards each other at speeds of 727272 km/hr and 126126126 km/hr take take 111 minute and 404040 seconds to cross each other. If the length of one of the trains is 200200200 metres, how long is the other train?

Solution

Distance to be covered will be the sum of the distance between the trains and the lengths of each of the two trains. Let the length of the other train be xxx. So, distance to be covered is 5000+200+x=5200+x5000 + 200 + x = 5200 + x5000+200+x=5200+x metres.

Converting the speeds to m/s we get the two trains to be travelling at 202020 m/s and 353535 m/s. As they are moving in the same direction, relative speed =20+35=55= 20 + 35 = 55=20+35=55 m/s. And, the time taken is 111 minute 404040 seconds or 100100100 seconds.

D === S ×\times× T ⇒ 5200+x=55×100 5200 + x = 55 \times 1005200+x=55×100

⇒ x=300 x = 300x=300 metres

Answer: 300300300 metres

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