## Speed, Time and Distance

#### Speed, Time and Distance

1. A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/hr, then the same distance is covered in 48 minutes. The distance between station A and B is
1. 60 km
2. 64 km
3. 80 km
4. 55 km

1. Let the distance between stations be x km, then speed of train

 = 2x = (4x/3) kmph 45 60

 ∴ x = (48/60) 4x − 5 3

 ⇒ 3x = 4 4x − 15 5

⇒  16x – 60 = 15x
⇒  x = 60 km

##### Correct Option: A

Let the distance between stations be x km, then speed of train

 = 2x = (4x/3) kmph 45 60

 ∴ x = (48/60) 4x − 5 3

 ⇒ 3x = 4 4x − 15 5

⇒  16x – 60 = 15x
⇒  x = 60 km

1. When a person cycled at 10 km per hour he arrived at his office 6 minutes late. He arrived 6 minutes early, when he increased his speed by 2 km per hour. The distance of his office from the starting place is
1. 6 km
2. 7 km
3. 12 km
4. 16 km

1. Let the distance be x km.

 ∴ x − x = 12 10 12 60

 ⇒ 6x − 5x = 1 60 5

 ⇒  x = 1 × 60 = 12 km 5

Second Methdod :
Here, S1 = 10, t1 = 6
S2 = 12, t2 = 6
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (10 × 12)(6 + 6) 12 − 10

 = 120 × 12 2

 = 60 × 12 km = 12 km 60

##### Correct Option: C

Let the distance be x km.

 ∴ x − x = 12 10 12 60

 ⇒ 6x − 5x = 1 60 5

 ⇒  x = 1 × 60 = 12 km 5

Second Methdod :
Here, S1 = 10, t1 = 6
S2 = 12, t2 = 6
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (10 × 12)(6 + 6) 12 − 10

 = 120 × 12 2

 = 60 × 12 km = 12 km 60

1. A student goes to school at the rate of
 2 1 km/h and reaches 6 minutes late. 2
If he travels at the speed of 3 km/h. he is 10 minutes early. The distance (in km) between the school and his house is
1. 5
2. 4
3. 3
4. 1

1. Let the required distance be x km.

 x − x = 16 5/2 3 60

 ⇒ 2x − x = 4 5 3 15

 ⇒ 6x − 5x = 4 15 15

⇒  x = 4 km.

Second Methdod :
 Here, S1 = 2 1 , t1 = 6 2

S2 = 3, t2 = 10
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 5 × 3(6 + 10) 2 3 − 5 2

 = 15 × 16 km = 4 km 60

##### Correct Option: B

Let the required distance be x km.

 x − x = 16 5/2 3 60

 ⇒ 2x − x = 4 5 3 15

 ⇒ 6x − 5x = 4 15 15

⇒  x = 4 km.

Second Methdod :
 Here, S1 = 2 1 , t1 = 6 2

S2 = 3, t2 = 10
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 5 × 3(6 + 10) 2 3 − 5 2

 = 15 × 16 km = 4 km 60

1. Walking at 5 km/hr a student reaches his school from his house 15 minutes early and walking at 3 km/hr he is late by 9 minutes. What is the distance between his school and his house ?
1. 5 km
2. 8 km
3. 3 km
4. 2 km

1. Let the required distance be x km.

 Then, x − x = 24 3 5 60

 ⇒ 5x − 3x = 2 15 5

 ⇒ 2x = 2 3

⇒  2x = 2 × 3 ⇒ x = 3 km

Second Methdod :
Here, S1 = 3, t1 = 9
S2 = 5, t2 = 15
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (3 × 5)(9 + 15) 5 − 3

 = 15 × 24 2 60

= 3 km

##### Correct Option: C

Let the required distance be x km.

 Then, x − x = 24 3 5 60

 ⇒ 5x − 3x = 2 15 5

 ⇒ 2x = 2 3

⇒  2x = 2 × 3 ⇒ x = 3 km

Second Methdod :
Here, S1 = 3, t1 = 9
S2 = 5, t2 = 15
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (3 × 5)(9 + 15) 5 − 3

 = 15 × 24 2 60

= 3 km

1. Shri X goes to his office by scooter at a speed of 30km/h and reaches 6 minutes earlier. If he goes at a speed of 24 km/h, he reaches 5 minutes late. The distance of his office is
1. 20 km
2. 21 km
3. 22 km
4. 24 km

1. Let the distance of office be x km.

 ∴ x − x = 11 24 30 60

 ⇒ 5x − 4x = 11 120 60

 ⇒ x = 11 120 60

 ⇒  x = 11 × 120 = 22 km. 60

Second Methdod :
Here, S1 = 24, t1 = 5
S2 = 30, t2 = 6
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 24 × 30(5 + 6) 30 − 24

 = 720 × 11 = 22 km 6 × 60

##### Correct Option: C

Let the distance of office be x km.

 ∴ x − x = 11 24 30 60

 ⇒ 5x − 4x = 11 120 60

 ⇒ x = 11 120 60

 ⇒  x = 11 × 120 = 22 km. 60

Second Methdod :
Here, S1 = 24, t1 = 5
S2 = 30, t2 = 6
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 24 × 30(5 + 6) 30 − 24

 = 720 × 11 = 22 km 6 × 60