Relative speed | = (x + 50) km/hr | |||||||
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Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
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⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Then, the length of the second train is | ❨ | x | ❩ | meters. |
2 |
Relative speed = (48 + 42) kmph = | ❨ | 90 x | 5 | ❩ | m/sec = 25 m/sec. |
18 |
∴ | [x + (x/2)] | = 12 or | 3x | = 300 or x = 200. |
25 | 2 |
∴ Length of first train = 200 m.
Let the length of platform be y meters.
Speed of the first train = | ❨ | 48 x | 5 | ❩ | m/sec = | 40 | m/sec. |
18 | 3 |
∴ (200 + y) x | 3 | = 45 |
40 |
⟹ 600 + 3y = 1800
⟹ y = 400 m.
Relative speed = | = (45 + 30) km/hr | |||||||
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We have to find the time taken by the slower train to pass the DRIVER of the faster train and not the complete train.
So, distance covered = Length of the slower train.
Therefore, Distance covered = 500 m.
∴ Required time = | ❨ | 500 x | 6 | ❩ | = 24 sec. |
125 |
Then, length of the first train = 27x metres,
and length of the second train = 17y metres.
∴ | 27x + 17y | = 23 |
x+ y |
⟹ 27x + 17y = 23x + 23y
⟹ 4x = 6y
⟹ | x | = | 3 | . |
y | 2 |
2 kmph = | ❨ | 2 x | 5 | ❩ | m/sec = | 5 | m/sec. |
18 | 9 |
4 kmph = | ❨ | 4 x | 5 | ❩ | m/sec = | 10 | m/sec. |
18 | 9 |
Let the length of the train be x metres and its speed by y m/sec.
Then, | ❨ | x | ❩ | = 9 and | ❨ | x | ❩ | = 10. |
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∴ 9y - 5 = x and 10(9y - 10) = 9x
⟹ 9y - x = 5 and 90y - 9x = 100.
On solving, we get: x = 50.
∴ Length of the train is 50 m.
Then, speed of the faster train = 2x m/sec.
Relative speed = (x + 2x) m/sec = 3x m/sec.
∴ | (100 + 100) | = 3x |
8 |
⟹ 24x = 200
⟹ x = | 25 | . |
3 |
So, speed of the faster train = | 50 | m/sec |
3 |
= | ❨ | 50 | x | 18 | ❩km/hr |
3 | 5 |
= 60 km/hr.
Speed = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
= | ❨ | 36 x | 5 | ❩m/sec |
18 |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
Speed of the first train = | ❨ | 120 | ❩ | m/sec = 12 m/sec. |
10 |
Speed of the second train = | ❨ | 120 | ❩ | m/sec = 8 m/sec. |
15 |
Relative speed = (12 + 8) = 20 m/sec.
∴ Required time = | [ | (120 + 120) | ] | sec = 12 sec. |
20 |
= | ❨ | 66 x | 5 | ❩m/sec |
18 |
= | ❨ | 55 | ❩m/sec. |
3 |
∴ Time taken to pass the man = | ❨ | 110 x | 3 | ❩sec = 6 sec. |
55 |
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