Let x km . be covered in y hrs.
then, 1st speed = (x / y) km/hr.
2nd speed = [(x/2) / 2y)] km/hr.
= (x/4y) km/hr.
? Ratio of speed = x/y : x/4y = 1 :1/4 = 4: 1
Speed of the train = (36 x 5)/18 = 10 m/sec.
Distance = (Time x Speed ) = (8 x 10 ) = 80 meters
? Length of the train = 80 meters
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 |
= | ❨ | 36 x | 5 | ❩m/sec |
18 |
= 10 m/sec.
Distance to be covered = (240 + 120) m = 360 m.
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.
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.
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 |
⟹ x = | ❨ | 18700 x 115 | ❩ | = 25300. |
85 |
Hence, S.P. = Rs. 25,300.
Since 653xy is divisible by 2 and 5 both, so y = 0.
Now, 653x is divisible by 8, so 13x should be divisible by 8.
This happens when x = 6.
∴x + y = (6 + 0) = 6.
(2n + 3)2 - (2n + 1)2 = (2n + 3 + 2n + 1) (2n + 3 - 2n - 1)
= (4n + 4) x 2
= 8(n + 1), which is divisible by 8.
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