Speed, Time and Distance
 A train met with an accident 3 hours after starting, which detains it for one hour, after which it proceeds at 75% of its original speed. It arrives at the destination 4 hours late. Had the accident taken place 150 km further along the railway line,
Find the length of the trip and the original speed of the train.the train would have arrived only 3 1 hours late. 2

 1100 km ; 100 kmph
 1200 km ; 100 kmph
 1200 km ; 90 kmph
 1600 km ; 90 kmph

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Let A be the starting point, B the terminus. C and D are points where accidents take place.
∴ 0.75 = 3 4
By travelling a 3/4 of its original speed, the train would take 4/3 of its usual time i.e., 1/3 more of the usual time.
∴ 1/3 of the usual time taken to travel the distance CB.
= 4 – 1 = 3 hrs. ...(i)
and 1/3 of the usual time taken to travel the distanceDB = 3 1 − 1 = 2 1 hrs. ...(ii) 2 2
Subtracting equation (ii) from (i)
we can write,
1/3 of the usual time taken to travel the distanceCD = 3 − 2 1 = 1 hr. 2 2
∴ Usual time taken to travelCD (150 km) = 1 2 1 3 = 3 hr. 2
Usual speed of the train= 3 = 9 hrs. 1 3
Total time = 3 + 9 = 12 hrs.
∴ Length of the trip = 12 × 100 = 1200 km.Correct Option: B
Let A be the starting point, B the terminus. C and D are points where accidents take place.
∴ 0.75 = 3 4
By travelling a 3/4 of its original speed, the train would take 4/3 of its usual time i.e., 1/3 more of the usual time.
∴ 1/3 of the usual time taken to travel the distance CB.
= 4 – 1 = 3 hrs. ...(i)
and 1/3 of the usual time taken to travel the distanceDB = 3 1 − 1 = 2 1 hrs. ...(ii) 2 2
Subtracting equation (ii) from (i)
we can write,
1/3 of the usual time taken to travel the distanceCD = 3 − 2 1 = 1 hr. 2 2
∴ Usual time taken to travelCD (150 km) = 1 2 1 3 = 3 hr. 2
Usual speed of the train= 3 = 9 hrs. 1 3
Total time = 3 + 9 = 12 hrs.
∴ Length of the trip = 12 × 100 = 1200 km.
 A train meets with an accident after travelling 30 kms, after which it moves
at the destination 45 minutes late. Had the accident happened 18 kms further on, it would have been 9 minutes before. Find the distance of journey and original speed of the train.with 4 th of its original speed and arrives 5

 120 km ; 25 kmph
 125 km ; 25 kmph
 130 km ; 30 kmph
 120 km ; 30 kmph

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Let the original speed be x and distance be y
Case I.
Time taken by train to travel30 km = 30 x Time taken by train after accident = y − 30 4/5x Total time taken = 30 + y − 30 x 4/5x
Case II :
Time taken by train to travel48 km = 48 x Time taken by train after accident = y − 48 4/5x Total time taken = 48 + y − 48 x 4/5x 30 + y − 30 − 48 + y − 48 x 4/5x x 4/5x = 9 [∵ Difference between time is 9 minutes] 60 y − 30 − y − 48 + 30 − 48 4/5x 4/5x x x = 9 60 y − y − 30 + 48 + (− 18) = 9 4/5x x 60 5(18) − 18 = 9 4x 4x 60 ⇒ 90 − 72 = 9 4x 60 x = 18 × 60 = 30 4 × 9
Hence, original speed = 30 kmph
Also,30 + y − 30 = y + 45 x 4/5x x 60
[Original time + 45 minute = New time]
⇒ 3x – y = –30
⇒ 3(30) – y = –30
⇒ i.e. Distance = 120 kmCorrect Option: D
Let the original speed be x and distance be y
Case I.
Time taken by train to travel30 km = 30 x Time taken by train after accident = y − 30 4/5x Total time taken = 30 + y − 30 x 4/5x
Case II :
Time taken by train to travel48 km = 48 x Time taken by train after accident = y − 48 4/5x Total time taken = 48 + y − 48 x 4/5x 30 + y − 30 − 48 + y − 48 x 4/5x x 4/5x = 9 [∵ Difference between time is 9 minutes] 60 y − 30 − y − 48 + 30 − 48 4/5x 4/5x x x = 9 60 y − y − 30 + 48 + (− 18) = 9 4/5x x 60 5(18) − 18 = 9 4x 4x 60 ⇒ 90 − 72 = 9 4x 60 x = 18 × 60 = 30 4 × 9
Hence, original speed = 30 kmph
Also,30 + y − 30 = y + 45 x 4/5x x 60
[Original time + 45 minute = New time]
⇒ 3x – y = –30
⇒ 3(30) – y = –30
⇒ i.e. Distance = 120 km
 Ram starts his journey from Bombay to Pune and simultaneously Mohan starts from Pune to Bombay. After crossing each other they finish their remaining
What is Mohan’s speed if Ram’s speed is 20 km per hr. ?journey in 6 1 and 4 hours respectively. 4

 28 kmph
 24 kmph
 25 kmph
 30 kmph

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Suppose that Ram and Mohan meet at A. Let Ram’s speed be x km per hr. and Mohan’s speed be y km per hr.Then AP = 25 x km x
and AB = 4y km.
Now, time taken by Ram in going fromB to A = 4y x
and the time taken by Mohan in going fromP to A = 25x . 4y
Obviously time taken is equal∴ 4y = 25x x 4y
⇒ 16y^{2} = 25x^{2}⇒ y^{2} = 25 x^{2} 16 ⇒ y = 5 x 4 ⇒ y = 5 x 4
Here, x = 20 km per hr.
∴ y = Mohan’s speed= 5 × 20 = 25 km per hr. 4 Correct Option: C
Suppose that Ram and Mohan meet at A. Let Ram’s speed be x km per hr. and Mohan’s speed be y km per hr.Then AP = 25 x km x
and AB = 4y km.
Now, time taken by Ram in going fromB to A = 4y x
and the time taken by Mohan in going fromP to A = 25x . 4y
Obviously time taken is equal∴ 4y = 25x x 4y
⇒ 16y^{2} = 25x^{2}⇒ y^{2} = 25 x^{2} 16 ⇒ y = 5 x 4 ⇒ y = 5 x 4
Here, x = 20 km per hr.
∴ y = Mohan’s speed= 5 × 20 = 25 km per hr. 4
 A car driving in the morning fog passes a man walking at 4 km per hr. in the same direction. The man can see the car for 3 minutes and visibility is upto a distance of 130 metres. Find the speed of the car.

 7.5 kmph
 6.6 kmph
 6 kmph
 7 kmph

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Distance covered by man in 3 minutes
= 4 × 1000 m × 3 minutes 60 minutes
=200 metres
Total distance covered by the car in 3 min.
= (200 + 130) m = 330 metres∴ Speed of the car = 330 m per min. 3
= 110 m per minutes= 110 1000 1 60 = 33 km per hr. 5
or 6.6 kmphCorrect Option: B
Distance covered by man in 3 minutes
= 4 × 1000 m × 3 minutes 60 minutes
=200 metres
Total distance covered by the car in 3 min.
= (200 + 130) m = 330 metres∴ Speed of the car = 330 m per min. 3
= 110 m per minutes= 110 1000 1 60 = 33 km per hr. 5
or 6.6 kmph
 A and B start simultaneously at 5 km per hr. and 4 km per hr. from P and Q, 180 kms apart, towards Q and P respectively. They cross each other at M and after reaching Q and P turn back immediately and meet again at N. Find the distance MN.

 45 km
 40 km
 35 km
 42 km

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When A and B cross each other at M for the first time, they have together covered the whole distance PQ = 180 km.
When they meet again at N, they have together covered total distance equal to 3 times of PQ = 3 × 180 = 540 km.PM = 5 × 180 = 100 km 5 + 4
[Distance covered by each will be in the ratio of their speeds)QP + PN = 4 × 540 = 240 km 5 + 4
or PN = 240 – QP = 240 – 180 = 60 km.
Then, MN = PM – PN
= 100 – 60 = 40 km.Correct Option: B
When A and B cross each other at M for the first time, they have together covered the whole distance PQ = 180 km.
When they meet again at N, they have together covered total distance equal to 3 times of PQ = 3 × 180 = 540 km.PM = 5 × 180 = 100 km 5 + 4
[Distance covered by each will be in the ratio of their speeds)QP + PN = 4 × 540 = 240 km 5 + 4
or PN = 240 – QP = 240 – 180 = 60 km.
Then, MN = PM – PN
= 100 – 60 = 40 km.