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 | = (x + 50) km/hr | |||||||
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Distance covered = (108 + 112) = 220 m.
∴ | 220 | = 6 | ||
|
⟹ 250 + 5x = 660
⟹ x = 82 km/hr.
Jar A has 36 litres of mixture of milk and water in the respective ratio of 5 : 4 => Quantity of milk in Jar A = 5/9 x 36 = 20 litres Quantity of water in Jar A = 36 - 20 = 16 litres Let quantity of water in Jar B = x litres => Quantity of milk in Jar B = (20 - x) litres Acc. to ques, =>[20 + (20-x)]/(16+x) = 5/3 => 120?3x = 80+5x => 5x +3x = 120?80 => 8x = 40 => 5 litres.
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
Ratio of milk and water = 2 : 1
Quantity of milk = 60 X 2/3 = 40 litre
Quantity of water = 20 litre
To make ratio, 1: 2, we have to double the water that of milk
So, water should be 80 litre.
That means 80 ? 20 = 60 litre water to be added.
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 |
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