(A's speed) : (B's speed) = √b : √a = √16 : √9 = 4 : 3.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
∴ 20x + 25(x - 1) = 110
⟹ 45x = 135
⟹ x = 3.
So, they meet at 10 a.m.
Speed = | ❨ | 54 x | 5 | ❩m/sec = 15 m/sec. |
18 |
Length of the train = (15 x 20)m = 300 m.
Let the length of the platform be x metres.
Then, | x + 300 | = 15 |
36 |
⟹ x + 300 = 540
⟹ x = 240 m.
Speed = | ❨ | 72 x | 5 | ❩m/sec | = 20 m/sec. |
18 |
Time = 26 sec.
Let the length of the train be x metres.
Then, | x + 250 | = 20 |
26 |
⟹ x + 250 = 520
⟹ x = 270.
Total distance covered |
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∴ Time taken |
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= 3 min. |
Speed = | ❨ | 60 x | 5 | ❩m/sec | = | ❨ | 50 | ❩m/sec. |
18 | 3 |
Length of the train = (Speed x Time).
∴ Length of the train = | ❨ | 50 | x 9 | ❩m = 150 m. |
3 |
Relative speed = (40 - 20) km/hr = | ❨ | 20 x | 5 | ❩ | m/sec = | ❨ | 50 | ❩ | m/sec. |
18 | 9 |
∴ Length of faster train = | ❨ | 50 | x 5 | ❩ | m = | 250 | m = 27 | 7 | m. |
9 | 9 | 9 |
Then, relative speed of the two trains = 2x m/sec.
So, 2x = | (120 + 120) |
12 |
⟹ 2x = 20
⟹ x = 10.
∴ Speed of each train = 10 m/sec = | ❨ | 10 x | 18 | ❩ | km/hr = 36 km/hr. |
5 |
Speed of the train relative to man | = (63 - 3) km/hr | |||||||
= 60 km/hr | ||||||||
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∴ Time taken to pass the man |
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= 30 sec. |
Speed = | ❨ | 78 x | 5 | ❩ | m/sec | = | ❨ | 65 | ❩ | m/sec. |
18 | 3 |
Time = 1 minute = 60 seconds.
Let the length of the tunnel be x metres.
Then, | ❨ | 800 + x | ❩ | = | 65 |
60 | 3 |
⟹ 3(800 + x) = 3900
⟹ x = 500.
Speed = | ❨ | 300 | ❩ | m/sec = | 50 | m/sec. |
18 | 3 |
Let the length of the platform be x metres.
Then, | ❨ | x + 300 | ❩ | = | 50 |
39 | 3 |
⟹ 3(x + 300) = 1950
⟹ x = 350 m.
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