At first glance it might seem that this problem cannot be solved because we do not have enough
information. It can be solved as long as you double whatever amount you start with. If we start with
$100, then P = $100 and FV = $200.
FV=P(1+r/n)^nt
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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