Speed, Time and Distance
 Two rifles are fired from the same place at a difference of 11 min. 45 seconds. But a man who is coming towards the same place in a train hears the second sound after 11 minutes. Find the speed of the train (Assuming speed of sound = 330 m/s).

 72 km/h
 36 km/h
 81 km/h
 108 km/h

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Difference of time = 11 minutes 45 seconds – 11 minutes = 45 seconds
Distance covered by sound in 45 seconds = Distance covered by train in 11 minutes
⇒ 330 × 45 = 11 × 60 × Speed of train⇒ Speed of train = 330 × 45 m/sec. 11 × 60 = 45 × 18 kmph. 2 5
= 81 kmph.Correct Option: C
Difference of time = 11 minutes 45 seconds – 11 minutes = 45 seconds
Distance covered by sound in 45 seconds = Distance covered by train in 11 minutes
⇒ 330 × 45 = 11 × 60 × Speed of train⇒ Speed of train = 330 × 45 m/sec. 11 × 60 = 45 × 18 kmph. 2 5
= 81 kmph.
 A man travels 450 km to his home partly by train and partly by car. He takes 8 hours 40 minutes if he travels 240 km by train and rest by car. He takes 20 minutes more if he travels 180 km by train and the rest by car. The speed of the car in km/hr is

 45
 50
 60
 48

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Let speed of train be x kmph.
Speed of car = y kmph.
Case I,∵ Time = Distance Speed ∴ 240 + 210 = 8 40 = 8 2 x y 60 3 ⇒ 240 + 210 = 26 ...(i) x y 3
Case II,180 + 270 = 9 ..... (ii) x y
By equation (i) × 3 – (ii) × 4,720 + 630 − 720 − 1080 x y x y
= 26 – 36⇒ −450 = –10 y
⇒ y = 45 kmph.Correct Option: A
Let speed of train be x kmph.
Speed of car = y kmph.
Case I,∵ Time = Distance Speed ∴ 240 + 210 = 8 40 = 8 2 x y 60 3 ⇒ 240 + 210 = 26 ...(i) x y 3
Case II,180 + 270 = 9 ..... (ii) x y
By equation (i) × 3 – (ii) × 4,720 + 630 − 720 − 1080 x y x y
= 26 – 36⇒ −450 = –10 y
⇒ y = 45 kmph.
 A plane can cover 6000 km in 8 hours. If the speed is increased by 250 kmph, then the time taken by the plane to cover 9000 km is

 8 hours
 6 hours
 5 hours
 9 hours

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Speed of plane = Distance Time = 6000 = 750 kmph 8
New speed = (750 + 250) kmph = 1000 kmph∴ Required time = 9000 = 9 hours 1000 Correct Option: D
Speed of plane = Distance Time = 6000 = 750 kmph 8
New speed = (750 + 250) kmph = 1000 kmph∴ Required time = 9000 = 9 hours 1000
 A train leaves station A at 5 AM and reaches station B at 9 AM on the same day. Another train leaves station B at 7 AM and reaches station A at 10:30 AM on the same day. The time at which the two trains cross each other is :

 8 : 26 AM
 7: 36 AM
 7: 56 AM
 8 AM

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Here distance is constant.
∴ Speed ∝ 1 Time
∴ Ratio of the speeds of A and B= 7 = 7 : 8 2/4
∴ A’s speed = 7x kmph (let)
B’s speed = 8x kmph
B = 7x × 4 = 28x km.
Let both trains cross each other after t hours from 7 a.m.
According to the question,
7x (t + 2) + 8x × t = 28x
⇒ 7t + 14 + 8t = 28
⇒ 15t = 28 – 14 = 14⇒ t = 14 hours 15 = 14 × 60 minutes 15
= 56 minutes
∴ Required time = 7 : 56 A.M.Correct Option: C
Here distance is constant.
∴ Speed ∝ 1 Time
∴ Ratio of the speeds of A and B= 7 = 7 : 8 2/4
∴ A’s speed = 7x kmph (let)
B’s speed = 8x kmph
B = 7x × 4 = 28x km.
Let both trains cross each other after t hours from 7 a.m.
According to the question,
7x (t + 2) + 8x × t = 28x
⇒ 7t + 14 + 8t = 28
⇒ 15t = 28 – 14 = 14⇒ t = 14 hours 15 = 14 × 60 minutes 15
= 56 minutes
∴ Required time = 7 : 56 A.M.
 If a distance of 50 m is covered in 1 minute, that 90 m in 2 minutes and 130 m in 3 minutes find the distance covered in 15 minutes.

 610 m
 750 m
 1000 m
 650 m

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Distance covered in 2nd minute = 90 – 50 = 40 metre
Distance covered in 3rd minute = 130 – 90 = 40 metre
∴ Required distance
= 50 + 40 × 14
= 50 + 560 = 610 metreCorrect Option: A
Distance covered in 2nd minute = 90 – 50 = 40 metre
Distance covered in 3rd minute = 130 – 90 = 40 metre
∴ Required distance
= 50 + 40 × 14
= 50 + 560 = 610 metre