## Speed, Time and Distance

#### Speed, Time and Distance

1. If a train runs at 70 km/hour, it reaches its destination late by 12 minutes. But if it runs at 80 km/ hour, it is late by 3 minutes. The correct time to cover the journey is
1. 58 minutes
2. 2 hours
3. 1 hour
4. 59 minutes

1. Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes

 = 9 hour = 3 hour 60 20

 ∴ x − x = 3 70 80 20

 ⇒ x − x = 3 7 8 2

 ⇒ 8x − 7x = 3 56 2

 ⇒ x = 3 56 2

 ⇒  x = 3 × 56 = 84 km 2

∴   Required correct time
 = 84 hours – 12 minutes 70

 = 84 × 60 − 12 minutes 70

= 72 – 12 = 60 minutes
= 1 hour

##### Correct Option: C

Distance of journey = x km
Difference of time = 12 – 3 = 9 minutes

 = 9 hour = 3 hour 60 20

 ∴ x − x = 3 70 80 20

 ⇒ x − x = 3 7 8 2

 ⇒ 8x − 7x = 3 56 2

 ⇒ x = 3 56 2

 ⇒  x = 3 × 56 = 84 km 2

∴   Required correct time
 = 84 hours – 12 minutes 70

 = 84 × 60 − 12 minutes 70

= 72 – 12 = 60 minutes
= 1 hour

1. A train travelling at a speed of 55 km/hr travels from place X to place Y in 4 hours. If its speed is increased by 5 km/hr., then the time of journey is reduced by
1. 25 minutes
2. 35 minutes
3. 20 minutes
4. 30 minutes

1. Distance between stations X and Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph

 ∴   Required time = 220 60

 = 11 hours 3

= 3 hours 40 minutes.
= 4 hours – 3 hours 40 minutes
= 20 minutes

##### Correct Option: C

Distance between stations X and Y = Speed × Time
= 55 × 4 = 220 km.
New speed = 55 + 5 = 60 kmph

 ∴   Required time = 220 60

 = 11 hours 3

= 3 hours 40 minutes.
= 4 hours – 3 hours 40 minutes
= 20 minutes

1. If a boy walks from his house to school at the rate of 4 km per hour, he reaches the school 10 minutes earlier than the scheduled time. However, if he walks at the rate of 3 km per hour, he reaches 10 minutes late. Find the distance of his school from his house.
1. 5 km
2. 4 km
3. 6 km
4. 4.5 km

1. Let the distance of school be x km, then

 x − x = 20 3 4 60

 ⇒ x = 1 ⇒ x = 12 = 4 km 12 3 3

Second Methdod :
Here, S1 = 3, t1 = 10
S2 = 4, t2 = 10
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (3 × 4)(10 + 10) 4 − 3

 = 12 × 20 = 4 km 60

##### Correct Option: B

Let the distance of school be x km, then

 x − x = 20 3 4 60

 ⇒ x = 1 ⇒ x = 12 = 4 km 12 3 3

Second Methdod :
Here, S1 = 3, t1 = 10
S2 = 4, t2 = 10
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (3 × 4)(10 + 10) 4 − 3

 = 12 × 20 = 4 km 60

1. Walking at a speed of 5 km/hr, a man reaches his office 6 minutes late. Walking at 6 km/hr, he reaches there 2 minutes early. The distance of his office is
1. 3 km
2. 4 km
3. 3.5 km
4. 2 km

1. Let the distance of the office be x km, then

 x − x = 8 5 6 60

 ⇒ 6x − 5x = 2 30 15

⇒  x = 2 × 2 = 4 km

Second Methdod :
Here, S1 = 5, t1 = 6
S2 = 6, t2 = 2
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (6 × 5)(6 + 2) 6 − 5

 = 30 × 8 = 4 km 60

##### Correct Option: B

Let the distance of the office be x km, then

 x − x = 8 5 6 60

 ⇒ 6x − 5x = 2 30 15

⇒  x = 2 × 2 = 4 km

Second Methdod :
Here, S1 = 5, t1 = 6
S2 = 6, t2 = 2
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (6 × 5)(6 + 2) 6 − 5

 = 30 × 8 = 4 km 60

1. A train covers a distance of 10 km in 12 minutes. If its speed is decreased by 5 km/hr, the time taken by it to cover the same distance will be :
1. 10 minutes
2. 13 minutes 20 sec
3. 13 minutes
4. 11 minutes 20 sec

1.  Speed of train = Distance Time

 = 10 kmph 12 60

 = 10 × 60 = 50 kmph 12

New speed = 45 kmph
 ∴   Required time = 10 hour 45

 = 2 × 60 minutes 9

 = 40 minutes 3

or 13 minutes 20 seconds

##### Correct Option: B

 Speed of train = Distance Time

 = 10 kmph 12 60

 = 10 × 60 = 50 kmph 12

New speed = 45 kmph
 ∴   Required time = 10 hour 45

 = 2 × 60 minutes 9

 = 40 minutes 3

or 13 minutes 20 seconds