Speed, Time and Distance
 Two bullets were fired at a place at an interval of 12 minutes. A person approaching the firing point in his car hears the two sounds at an interval of 11 minutes 40 seconds. The speed of sound is 330 metres per second. What is the approximate speed of the car?

 34 kmph
 32 kmph
 36 kmph
 38 kmph

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If the car were not moving, the person would have heard the two sounds at an interval of 12 minutes. Therefore, the distance travelled by car in 11 minutes 40 seconds is equal to the distance that could have been covered by sound in 12 min – 11 min. 40 seconds = 20 seconds.
Distance covered by sound in 20 seconds
= 330 × 20 = 6600 m
In 11 min 40 seconds
or 700 seconds the car travels 6600 m.
In 1 second the car will travel6600 metre = 66 metre 700 7 ∴ 66 metre per second 7 = 66 × 18 km per hr. 7 5 = 1188 km per hr. 35 = 33 33 km per hr. ≈ 34 kmph 35 Correct Option: A
If the car were not moving, the person would have heard the two sounds at an interval of 12 minutes. Therefore, the distance travelled by car in 11 minutes 40 seconds is equal to the distance that could have been covered by sound in 12 min – 11 min. 40 seconds = 20 seconds.
Distance covered by sound in 20 seconds
= 330 × 20 = 6600 m
In 11 min 40 seconds
or 700 seconds the car travels 6600 m.
In 1 second the car will travel6600 metre = 66 metre 700 7 ∴ 66 metre per second 7 = 66 × 18 km per hr. 7 5 = 1188 km per hr. 35 = 33 33 km per hr. ≈ 34 kmph 35
 A carriage driving in a fog passed a man who was walking at the rate of 6 km per hr. in the same direction. He could see the carriage for 4 minutes and it was visible to him up to a distance of 200 metres. Find the speed of the carriage.

 8.75 kmph
 8.5 kmph
 8 kmph
 9 kmph

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The distance covered by man in 4 minutes
= 6 × 1000 × 4 = 400 metres 60
The distance covered by carriage in 4 minutes
= 200 + 400 = 600 metres∴ 600 × 60 km per hr. 4 1000
= 9 km per hr.Correct Option: D
The distance covered by man in 4 minutes
= 6 × 1000 × 4 = 400 metres 60
The distance covered by carriage in 4 minutes
= 200 + 400 = 600 metres∴ 600 × 60 km per hr. 4 1000
= 9 km per hr.
 Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km per hr. faster than the second train. If after two hours they are 50 km apart, find the average speed of faster train.

 18 kmph
 15 kmph
 20 kmph
 25 kmph

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Let the speed of the second train be x km per hr. Then the speed of the first train is x + 5 km per hr.
Let O be the position of the railway station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = 2 (x + 5) km.
Distance travelled by the 2nd train in 2 hours= OB = 2x km.
By Pythagoras theorem,
AB^{2} + OA^{2} + OB^{2}
⇒ 50^{2} = [2 (x + 5)]^{2} + [2x]^{2}
⇒ 2500 = 4 (x + 5)^{2} + 4x^{2}
⇒ 2500 = 4 (x^{2} + 10x + 25) + 4x^{2}
⇒ 8x^{2} + 40x – 2400 = 0
⇒ x^{2} + 5x – 300 = 0
⇒ x^{2} + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x – 15) (x + 20) = 0
⇒ x = 15, – 20
But x cannot be negative
∴ x = 15
∴ The speed of the second train is 15 km per hr. and the speed of the first train is 20 km per hr.Correct Option: C
Let the speed of the second train be x km per hr. Then the speed of the first train is x + 5 km per hr.
Let O be the position of the railway station from which the two trains leave. Distance travelled by the first train in 2 hours = OA = 2 (x + 5) km.
Distance travelled by the 2nd train in 2 hours= OB = 2x km.
By Pythagoras theorem,
AB^{2} + OA^{2} + OB^{2}
⇒ 50^{2} = [2 (x + 5)]^{2} + [2x]^{2}
⇒ 2500 = 4 (x + 5)^{2} + 4x^{2}
⇒ 2500 = 4 (x^{2} + 10x + 25) + 4x^{2}
⇒ 8x^{2} + 40x – 2400 = 0
⇒ x^{2} + 5x – 300 = 0
⇒ x^{2} + 20x – 15x – 300 = 0
⇒ x (x + 20) – 15 (x + 20) = 0
⇒ (x – 15) (x + 20) = 0
⇒ x = 15, – 20
But x cannot be negative
∴ x = 15
∴ The speed of the second train is 15 km per hr. and the speed of the first train is 20 km per hr.
 In a flight of 600 kms, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km per hr. and the time of flight increased by 30 minutes. Find the duration of flight.

 1.2 hours
 1 hour
 1.5 hours
 2 hours

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Let the original speed be x kmph
then, new speed = (x – 200) kmph
According to question,
Time taken with new speed – time taken with original speed= 30 min. i.e 1 hr. 2 ∴ 600 − 600 = 1 x − 200 x 2 ⇒ 600 1 − 1 = 1 x − 200 x 2 ⇒ x − x + 200 = 1 x(x − 200) 1200
⇒ 24000 = x (x – 200)
⇒ x^{2} – 200x – 24000 = 0
⇒ x^{2} – 600x + 400x – 24000 = 0
⇒ x (x – 600) + 400 (x – 600) = 0
⇒ (x – 600) (x + 400) = 0
⇒ x = 600, – 400
Speed cannot be negative
Hence, original speed = 600 kmph and duration of flight= 600 hr. = 1 hr. 600 Correct Option: B
Let the original speed be x kmph
then, new speed = (x – 200) kmph
According to question,
Time taken with new speed – time taken with original speed= 30 min. i.e 1 hr. 2 ∴ 600 − 600 = 1 x − 200 x 2 ⇒ 600 1 − 1 = 1 x − 200 x 2 ⇒ x − x + 200 = 1 x(x − 200) 1200
⇒ 24000 = x (x – 200)
⇒ x^{2} – 200x – 24000 = 0
⇒ x^{2} – 600x + 400x – 24000 = 0
⇒ x (x – 600) + 400 (x – 600) = 0
⇒ (x – 600) (x + 400) = 0
⇒ x = 600, – 400
Speed cannot be negative
Hence, original speed = 600 kmph and duration of flight= 600 hr. = 1 hr. 600
 A hare, pursued by a grey hound is 50 of her own leaps before him. While the hare takes 4 leaps, the grey hound takes 3 leaps. In one leap, the hare goes 1.75 metres and the grey hound 2.75 metres. In how many leaps, will the grey hound overtake the hare?

 210 leaps
 220 leaps
 230 leaps
 250 leaps

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Grey hound and hare make 3 leaps and 4 leaps respectively. This happens at the same time.
The hare goes 1.75 metres in 1 leap.
∴ Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres
The grey hound goes 2.75 metres in one leap.
∴ Distance covered by it in 3
leaps
= 3 × 2.75 = 8.25 metres
Distance gained by grey hound in 3 leaps = (8.25 − 7)
= 1.25 metres
Distance covered by hare in 50 leaps = 50 × 1.75 metres
= 87.5 metres
Now, 1.25 metres is gained by grey hound in 3 leaps
∴ 87.5 metres is gained in3 × 87.5 = 210 leaps. 1.25 Correct Option: A
Grey hound and hare make 3 leaps and 4 leaps respectively. This happens at the same time.
The hare goes 1.75 metres in 1 leap.
∴ Distance covered by hare in 4 leaps = 4 × 1.75 = 7 metres
The grey hound goes 2.75 metres in one leap.
∴ Distance covered by it in 3
leaps
= 3 × 2.75 = 8.25 metres
Distance gained by grey hound in 3 leaps = (8.25 − 7)
= 1.25 metres
Distance covered by hare in 50 leaps = 50 × 1.75 metres
= 87.5 metres
Now, 1.25 metres is gained by grey hound in 3 leaps
∴ 87.5 metres is gained in3 × 87.5 = 210 leaps. 1.25