## Speed, Time and Distance

#### Speed, Time and Distance

1. A person goes from one point to another point with a speed of 5 km/h and comes back to starting point with a speed of 3 km/h. Find the average speed for the whole journey.?
1. 4.5 km/h
2. 4 km/h
3. 4. 25 km/h
4. 3. 75 km/h

1. Average speed = 2AB/(A + B)
= 2 x 5 x 3/(5 + 3)

##### Correct Option: D

Average speed = 2AB/(A + B)
= 2 x 5 x 3/(5 + 3)
= 30/8
= 3.75 km/h

1. A train covers a distance in 1 h 40 min, if it runs at a speed of 96 km/h on an average. Find the speed at which the train must run reduce the time of journey to 1 h 20 min.
1. 120 km/h
2. 90 km/h
3. 144 km/h
4. 150 km/h

1. Distance= Speed x Time = 96 x (100/60) = 160 km
New time = 80/60 h = 4/3 h

##### Correct Option: A

Distance= Speed x Time = 96 x (100/60) = 160 km
New time = 80/60 h = 4/3 h
New speed = 160 x 3/4 = 40 x 3 =120 km/h

1. The average speed of a car is 75 km/h. The driver first decrease its average speed by 40% and then increase it by 50%. What is the new average speed now ?
1. 67.5 km
2. 60 km
3. 90 km
4. 60.5 km

1. Required average speed = 75 x {(100 - 40)/100} x {(100 + 50)/100}

##### Correct Option: A

Required average speed = 75 x {(100 - 40)/100} x {(100 + 50)/100}
= 75 x 3/5 x 3/2
= 67.5 km/h

1. A train running at the speed of 72 km/hr passes a tunnel completely in 3 minutes. While inside the tunnel, it meets another
 train of 3 of its length coming from opposite 4
direction at the speed of 90km/hr and passes it completely in
 3 1 seconds. Find the length of the tunnel. 2
1. 3510 metre
2. 3500 metre
3. 3400 metre
4. 3600 metre

1. Trains are running in opposite directions.
∴  Relative speed = 72 + 90
= 162 kmph

 = 162 × 5 = 45 m/sec 18

Let the length of the first train be = x metre.
∴  Length of the second train
 = 3 x meter. 4

Now,
 distance travelled in 3 1 seconds at 45 m/sec. 2

 = 45 × 7 = 315 metre 2 2

This distance is equal to sum of the lengths of trains.
 ∴  x + 3x = 315 4 2

 ⇒ 4x + 3x = 315 4 2

 ⇒ 7x = 315 4 2

 ⇒  x = 315 × 4 = 90 2 7

Hence, the length of the first train = 90 metre.
Speed of first train = 72 kmph
 = 72 × 5 = 20 m/sec 18

Time taken by the first train to cross the tunnel
= 3 minutes = 180 seconds
∴  Distance covered by it in 180 seconds
= 180 × 20 = 3600 metre
∴  Length of (first train + tunnel) = 3600 metre
∴  Length of tunnel
= 3600 – 90 = 3510 metre

##### Correct Option: A

Trains are running in opposite directions.
∴  Relative speed = 72 + 90
= 162 kmph

 = 162 × 5 = 45 m/sec 18

Let the length of the first train be = x metre.
∴  Length of the second train
 = 3 x meter. 4

Now,
 distance travelled in 3 1 seconds at 45 m/sec. 2

 = 45 × 7 = 315 metre 2 2

This distance is equal to sum of the lengths of trains.
 ∴  x + 3x = 315 4 2

 ⇒ 4x + 3x = 315 4 2

 ⇒ 7x = 315 4 2

 ⇒  x = 315 × 4 = 90 2 7

Hence, the length of the first train = 90 metre.
Speed of first train = 72 kmph
 = 72 × 5 = 20 m/sec 18

Time taken by the first train to cross the tunnel
= 3 minutes = 180 seconds
∴  Distance covered by it in 180 seconds
= 180 × 20 = 3600 metre
∴  Length of (first train + tunnel) = 3600 metre
∴  Length of tunnel
= 3600 – 90 = 3510 metre

1. In a one-kilometre race, A beats B by 15 seconds and B beats C by 15 seconds. If C is 100 metres away from the finishing mark, when B has reached it, find the speed of A.
1. 9.5 m/sec.
2. 9 m/sec.
3. 8 m/sec.
4. 8.3 m/sec

1. Let B take x seconds to run 1000 m.
∴  Time taken by C
= (x + 15) seconds

 ∴ x = 900 = 9 x + 15 1000 10

⇒  10x = 9x + 135
⇒  x = 135 seconds
Now in a one kilometre race, A beats B by 15 seconds.
It means A covers 1000 m in
135 – 15 = 120 seconds
 ∴  Speed of A = 1000 = 25 m/sec 120 3

= 8.3 m/sec.

##### Correct Option: D

Let B take x seconds to run 1000 m.
∴  Time taken by C
= (x + 15) seconds

 ∴ x = 900 = 9 x + 15 1000 10

⇒  10x = 9x + 135
⇒  x = 135 seconds
Now in a one kilometre race, A beats B by 15 seconds.
It means A covers 1000 m in
135 – 15 = 120 seconds
 ∴  Speed of A = 1000 = 25 m/sec 120 3

= 8.3 m/sec.