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 |
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 = | ❨ | 45 x | 5 | ❩m/sec | = | ❨ | 25 | ❩m/sec. |
18 | 2 |
Time = 30 sec.
Let the length of bridge be x metres.
Then, | 130 + x | = | 25 |
30 | 2 |
⟹ 2(130 + x) = 750
⟹ x = 245 m.
Relative speed = (60 + 40) km/hr = | ❨ | 100 x | 5 | ❩m/sec | = | ❨ | 250 | ❩m/sec. |
18 | 9 |
Distance covered in crossing each other = (140 + 160) m = 300 m.
Required time = | ❨ | 300 x | 9 | ❩sec | = | 54 | sec = 10.8 sec. |
250 | 5 |
Then, distance covered = 2x metres.
Relative speed = (46 - 36) km/hr
= | ❨ | 10 x | 5 | ❩m/sec |
18 |
= | ❨ | 25 | ❩m/sec |
9 |
∴ | 2x | = | 25 |
36 | 9 |
⟹ 2x = 100
⟹ x = 50.
Total distance covered |
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∴ Time taken |
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= 3 min. |
Formula for converting from km/hr to m/s: X km/hr = | ❨ | X x | 5 | ❩ | m/s. |
18 |
Therefore, Speed = | ❨ | 45 x | 5 | ❩m/sec | = | 25 | m/sec. |
18 | 2 |
Total distance to be covered = (360 + 140) m = 500 m.
Formula for finding Time = | ❨ | Distance | ❩ |
Speed |
∴ Required time = | ❨ | 500 x 2 | ❩sec | = 40 sec. |
25 |
Speed = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
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.
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, 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.
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