## Speed, Time and Distance

#### Speed, Time and Distance

1.  A car can cover a certain distance in 4 1 hours. 2
If the speed is increased by 5 km/hour, it would take
 1 hour less to cover the same distance. 2

Find the slower speed of the car.
1. 50 km/hour
2. 40 km/hour
3. 45 km/hour
4. 60 km/hour

1. Let the initial speed of the car be x kmph and the distance be y km.

 Then, y = 9 x     ...(i) 2

and, y = 4 (x + 5)     ...(ii)
 ∴ 9x = 4 (x + 5) 2

⇒   9x = 8x + 40
⇒   x = 40 kmph

##### Correct Option: B

Let the initial speed of the car be x kmph and the distance be y km.

 Then, y = 9 x     ...(i) 2

and, y = 4 (x + 5)     ...(ii)
 ∴ 9x = 4 (x + 5) 2

⇒   9x = 8x + 40
⇒   x = 40 kmph

1. A boy is late by 9 minutes if he walks to school at a speed of 4 km/hour. If he walks at the rate of 5 km/hour, he arrives 9 minutes early. The distance to his school is
1. 9 km
2. 5 km
3. 4 km
4. 6 km

1. Let the required distance be x km.
According to the question,

 x − x = 18 4 5 60

 ⇒ 5x − 4x − 3 20 10

 ⇒  x = 3 × 20 = 6 km 10

Second Methdod :
Here, S1 = 4, t1 = 9
S2 = 5, t2 = 9
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (4 × 5)(9 + 9) 5 − 4

 = 20 × 18 = 6 km 60

##### Correct Option: D

Let the required distance be x km.
According to the question,

 x − x = 18 4 5 60

 ⇒ 5x − 4x − 3 20 10

 ⇒  x = 3 × 20 = 6 km 10

Second Methdod :
Here, S1 = 4, t1 = 9
S2 = 5, t2 = 9
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (4 × 5)(9 + 9) 5 − 4

 = 20 × 18 = 6 km 60

1. A student walks from his house at
 a speed of 2 1 km per hour and reaches his school 2
6 minutes late. The next day he increases his speed by 1 km per hour and reaches 6 minutes before school time. How far is the school from his house ?
1.  5 km 4
2.  7 km 4
3.  9 km 4
4.  11 km 4

1. Let the required distance be x km.
Difference of time

 = 6 + 6 = 12 minutes = 1 hr. 5

According to the question,
 ⇒ x − x = 1 ⇒ 2x − 2x = 1 5/2 7/2 5 5 7 5

 ⇒ 14x − 10x − 1 35 5

 ⇒ 4x = 1 ⇒ x = 35 = 7 km. 35 5 20 4

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

 Here, S2 = 3 1 , t2 = 6 2

 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 5 × 7 × (6 + 6) 2 2 5 − 7 2 2

 = 35 × 12 = 7 km 4 60 4

##### Correct Option: B

Let the required distance be x km.
Difference of time

 = 6 + 6 = 12 minutes = 1 hr. 5

According to the question,
 ⇒ x − x = 1 ⇒ 2x − 2x = 1 5/2 7/2 5 5 7 5

 ⇒ 14x − 10x − 1 35 5

 ⇒ 4x = 1 ⇒ x = 35 = 7 km. 35 5 20 4

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

 Here, S2 = 3 1 , t2 = 6 2

 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = 5 × 7 × (6 + 6) 2 2 5 − 7 2 2

 = 35 × 12 = 7 km 4 60 4

1. If a train runs at 40 km/hour, it reaches its destination late by 11 minutes. But if it runs at 50 km/hour, it is late by 5 minutes only. The correct time (in minutes) for the train to complete the journey is
1. 13
2. 15
3. 19
4. 21

1. If the distance be x km, then

 x − x = 6 40 50 60

 ⇒ x − x = 1 4 5

⇒  x = 20 km.
 ∴   Required time = 20 hour – 11 minutes 40

 = 1 × 60 − 11 minutes 2

= 19 minutes

##### Correct Option: C

If the distance be x km, then

 x − x = 6 40 50 60

 ⇒ x − x = 1 4 5

⇒  x = 20 km.
 ∴   Required time = 20 hour – 11 minutes 40

 = 1 × 60 − 11 minutes 2

= 19 minutes

1. A student rides on bicycle at 8 km/hour and reaches his school 2.5 minutes late. The next day he increases his speed to 10 km/ hour and reaches school 5 minutes early. How far is the school from his house ?
1.  5 km 8
2. 8 km
3. 5 km
4. 10 km

1. Let x km. be the required distance.
Difference in time = 2.5 + 5 = 7.5 minutes

 = 7.5 hrs. = 1 hrs. 60 8

 Now, x − x = 1 8 10 8

 ⇒ 5x − 4x = 1 40 8

 ⇒  x = 40 = 5 km. 8

Second Methdod :
Here, S1 = 8, t1 = 2.5
S2 = 10, t2 = 5
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (8 × 10)(2.5 + 5) 10 − 8

 = 40 × 7.5 = 5 km 60

##### Correct Option: C

Let x km. be the required distance.
Difference in time = 2.5 + 5 = 7.5 minutes

 = 7.5 hrs. = 1 hrs. 60 8

 Now, x − x = 1 8 10 8

 ⇒ 5x − 4x = 1 40 8

 ⇒  x = 40 = 5 km. 8

Second Methdod :
Here, S1 = 8, t1 = 2.5
S2 = 10, t2 = 5
 Distance = (S1 ×S2)(t1 + t2) S2 −S1

 = (8 × 10)(2.5 + 5) 10 − 8

 = 40 × 7.5 = 5 km 60