Speed, Time and Distance
- A goods train 158 metres long and travelling at the speed of 32 km per hr. leaves Delhi at 6 am. Another mail train 130 metres long and travelling at the average speed of 80 km per hr. leaves Delhi at 12 noon and follows the goods train. At what time will the mail train completely cross the goods train?
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The goods train leaves Delhi at 6 am and mail train at 12 noon, hence after 6 hours
The distance covered by the goods train in 6 hours at 32 km per hr. = 32 ´ 6 = 192 kms
The relative velocity of mail train with respect to goods train = 80 – 32 = 48 km per hr.
To completely cross the goods train, the mail train will have to cover a distance
= 192 km + 158m + 130m
= 192km + 0.158 km + 0.130 km
= 192.288 km more
Since, the mail train goes 48 kms more in 1 hour.
∴ The mail train goes 192.288 kms more in= 192288 × 1 = 2003 1000 48 500
= 4 hours 21.6 sec.Correct Option: B
The goods train leaves Delhi at 6 am and mail train at 12 noon, hence after 6 hours
The distance covered by the goods train in 6 hours at 32 km per hr. = 32 ´ 6 = 192 kms
The relative velocity of mail train with respect to goods train = 80 – 32 = 48 km per hr.
To completely cross the goods train, the mail train will have to cover a distance
= 192 km + 158m + 130m
= 192km + 0.158 km + 0.130 km
= 192.288 km more
Since, the mail train goes 48 kms more in 1 hour.
∴ The mail train goes 192.288 kms more in= 192288 × 1 = 2003 1000 48 500
= 4 hours 21.6 sec.
- A motor-boat goes 2 km upstream in a stream flowing at 3 km per hr. and then returns
downstream to the starting point in 30 minutes. Find the speed of the motor-boat in still water.
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Let the speed of the motorboat in still water be Z km per hr.
Downstream speed= (Z + 3) km per hr.
Upstream speed = (Z – 3) km per hr.
Total journey time= 30 minutes = 30 hr. = 1 hour 60 2
We can write,2 + 2 = 1 Z − 3 Z + 3 2 or, 2 
(Z + 3) + (Z − 3) 
= 1 (Z − 3) + (Z − 3) 2 or, 2Z = 1 Z2 − 9 4
or, Z2 – 9 = 8Z
or, Z2 – 8Z – 9 = 0
or, Z2 + Z – 9Z – 9 = 0
or, Z(Z + 1) – 9 (Z + 1) = 0
or, (Z + 1) (Z – 9) = 0
∴ Z = – 1 or 9.
Since speed can’t be negative
Therefore, the speed of the motor-boat in still water = 9 km per hr.Correct Option: C
Let the speed of the motorboat in still water be Z km per hr.
Downstream speed= (Z + 3) km per hr.
Upstream speed = (Z – 3) km per hr.
Total journey time= 30 minutes = 30 hr. = 1 hour 60 2
We can write,2 + 2 = 1 Z − 3 Z + 3 2 or, 2 (Z + 3) + (Z − 3) = 1 (Z − 3) + (Z − 3) 2 or, 2Z = 1 Z2 − 9 4
or, Z2 – 9 = 8Z
or, Z2 – 8Z – 9 = 0
or, Z2 + Z – 9Z – 9 = 0
or, Z(Z + 1) – 9 (Z + 1) = 0
or, (Z + 1) (Z – 9) = 0
∴ Z = – 1 or 9.
Since speed can’t be negative
Therefore, the speed of the motor-boat in still water = 9 km per hr.
- A person can row a boat 32 km upstream and 60 km downstream in 9 hours. Also, he can row 40 km upstream and 84 km downstream in 12 hours. Find the rate of the current.
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Let the upstream speed be x km per hr. and downstream speed be y km per hr.
Then, we can write,32 + 60 = 9 x y and, 40 + 84 = 12 x y Let 1 = m and 1 = n x y
The above two equations can now be written as
32 m + 60 n = 9 ...(i)
and, 40 m + 84 n = 12 ...(ii)
7 × (i) – 5 × (ii) gives 24 m = 3or m = 1 or x = 8 km per hr. 8
4 × (ii) – 5 × (i) gives 36 n = 3or n = 1 or y = 12 km per hr. 12
Rate of current= y − x = 12 − 8 = 2 km. per hr. 2 2 Correct Option: D
Let the upstream speed be x km per hr. and downstream speed be y km per hr.
Then, we can write,32 + 60 = 9 x y and, 40 + 84 = 12 x y Let 1 = m and 1 = n x y
The above two equations can now be written as
32 m + 60 n = 9 ...(i)
and, 40 m + 84 n = 12 ...(ii)
7 × (i) – 5 × (ii) gives 24 m = 3or m = 1 or x = 8 km per hr. 8
4 × (ii) – 5 × (i) gives 36 n = 3or n = 1 or y = 12 km per hr. 12
Rate of current= y − x = 12 − 8 = 2 km. per hr. 2 2
- A boatman takes his boat in a river against the stream from a place A to a place B where AB is 21 km and again returns to A. Thus he takes 10 hours in all. The time taken by him downstream in going 7 km is equal to the time taken by him against stream in going 3 km. Find the speed of river.
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Let the speed of boat and river be x km per hr. and y km per hr. respectively.
Then, The speed of boatman downstream = (x + y) km per hr.
and the speed of boatman upstream = (x – y) km per hr.
Time taken by boatman in going 21 km downstream= 21 hours x + y
Time taken by boatman in going 21 km upstream= 21 hrs. x − y
According to the question,= 21 = 21 = 10 ...(i) x + y x − y Now, time taken for 7 kms downstream = 7 hrs. x + y and time taken for 3 kms upstream = 3 hrs. x − y
According to the question7 − 3 = 0 ...(ii) x + y x − y
By (ii) × 7 + (i)49 − 21 = 21 = 21 = 10 x + y x − y x + y x − y ⇒ 70 = 10 x + y
⇒ x + y = 7 ...(iii)
Putting x + y = 7 in equation (ii)
we have7 − 3 = 0 7 x − y ⇒ 1 − 3 = 0 x − y
⇒ x – y = 3 ...(iv)
On adding (iii) and (iv), we have
2x = 10
⇒ x = 5
∴ y = 7 – x = 7 – 5 = 2
∴ Speed of river = 2 km per hr.Correct Option: A
Let the speed of boat and river be x km per hr. and y km per hr. respectively.
Then, The speed of boatman downstream = (x + y) km per hr.
and the speed of boatman upstream = (x – y) km per hr.
Time taken by boatman in going 21 km downstream= 21 hours x + y
Time taken by boatman in going 21 km upstream= 21 hrs. x − y
According to the question,= 21 = 21 = 10 ...(i) x + y x − y Now, time taken for 7 kms downstream = 7 hrs. x + y and time taken for 3 kms upstream = 3 hrs. x − y
According to the question7 − 3 = 0 ...(ii) x + y x − y
By (ii) × 7 + (i)49 − 21 = 21 = 21 = 10 x + y x − y x + y x − y ⇒ 70 = 10 x + y
⇒ x + y = 7 ...(iii)
Putting x + y = 7 in equation (ii)
we have7 − 3 = 0 7 x − y ⇒ 1 − 3 = 0 x − y
⇒ x – y = 3 ...(iv)
On adding (iii) and (iv), we have
2x = 10
⇒ x = 5
∴ y = 7 – x = 7 – 5 = 2
∴ Speed of river = 2 km per hr.
- A motorist and a cyclist start from A to B at the same time. AB is 18 km. The speed of motorist is 15 m per hr. more than the cyclist. After covering half the distance, the motorist rests for 30 minutes and thereafter his speed is reduced by 20%. If the motorist reaches the destination B, 15 minutes earlier than that of the cyclist, then find the speed of the cyclist.
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Let the speed of the cyclist be x km per hr.
Speed of the motorist= (x + 15) km per hr.
Time taken by the motorist to cover half of the distance= 18 = 9 hrs. 2 × (x + 15) x + 15
After covering 9 kms, the speed of motorist gets reduced by 20%∴ New speed = (x + 15) × 80 100 = 4(x + 15) km per hr. 5
Time taken by the motorist to cover the remaining half distance= 9 × 5 = 45 hrs. 4 (x + 15) 4 (x + 15)
Total time taken by the motorist= 9 + 1 + 45 hrs. x + 15 2 4 (x + 15) Total time taken by the cyclist = 18 hrs. x Motorist reaches 15 minutes, i.e., 1 hr. earlier. 4 ∴ 18 − 9 − 1 − 45 = 1 x x + 15 2 4 (x + 15) 4 ⇒ 18 × 4 (x + 15) − 36x − 2x (x + 15) − 45x = 1 4x (x + 15) 4
⇒ 72x + 1080 – 36x – 2x2 – 30x – 45x = x2 + 15
⇒ 3x2 + 54x – 1080 = 0
⇒ x2 + 18x – 360 = 0
⇒ x2 + 30x – 12x – 360 = 0
⇒ x (x + 30) – 12 (x + 30) = 0
⇒ (x + 30) (x – 12) = 0
⇒ x = – 30, 12
The speed cannot be negative.
∴ The speed of the cyclist = 12 km per hr.Correct Option: B
Let the speed of the cyclist be x km per hr.
Speed of the motorist= (x + 15) km per hr.
Time taken by the motorist to cover half of the distance= 18 = 9 hrs. 2 × (x + 15) x + 15
After covering 9 kms, the speed of motorist gets reduced by 20%∴ New speed = (x + 15) × 80 100 = 4(x + 15) km per hr. 5
Time taken by the motorist to cover the remaining half distance= 9 × 5 = 45 hrs. 4 (x + 15) 4 (x + 15)
Total time taken by the motorist= 9 + 1 + 45 hrs. x + 15 2 4 (x + 15) Total time taken by the cyclist = 18 hrs. x Motorist reaches 15 minutes, i.e., 1 hr. earlier. 4 ∴ 18 − 9 − 1 − 45 = 1 x x + 15 2 4 (x + 15) 4 ⇒ 18 × 4 (x + 15) − 36x − 2x (x + 15) − 45x = 1 4x (x + 15) 4
⇒ 72x + 1080 – 36x – 2x2 – 30x – 45x = x2 + 15
⇒ 3x2 + 54x – 1080 = 0
⇒ x2 + 18x – 360 = 0
⇒ x2 + 30x – 12x – 360 = 0
⇒ x (x + 30) – 12 (x + 30) = 0
⇒ (x + 30) (x – 12) = 0
⇒ x = – 30, 12
The speed cannot be negative.
∴ The speed of the cyclist = 12 km per hr.
