Speed, Time and Distance
- 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.?
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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
- 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.
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Distance= Speed x Time = 96 x (100/60) = 160 km
New time = 80/60 h = 4/3 hCorrect 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
- 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 ?
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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
- A train running at the speed of 72 km/hr passes a tunnel completely in 3 minutes. While inside the tunnel, it meets another
direction at the speed of 90km/hr and passes it completely intrain of 3 of its length coming from opposite 4 3 1 seconds. Find the length of the tunnel. 2
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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 metreCorrect 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
- 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.
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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.