Let the original speed be x km/hr.
Then, [ 840/x - 840/( x + 10) ] = 2
? 840 (x + 10) - 840x = 2x (x+10)
? x2 + 10x - 4200 = 0
? (x + 70) (x - 60 ) = 0
? x = 60 km/hr.
Suppose they meet after y hours .
Then, 21y - 16y = 60
? y = 12
?required distance = (16 x 12 + 21 x 12) km.
= 444 km.
Distance covered by thief in (1/2 ) hour = 20 km.
Now , 20 km is compensated by the owner at a relative speed of 10 km/hr in 2 hours so, he overtake the thief at 4 p.m.
Let the distance be x km.
Then, x/3 - x/4 = 30/60
? (4x - 3x ) / 12 = 1/2
? x = 6 km.
Let C' s speed = x km/hr.
Then, B's speed = 3x km/hr.
and A's speed = 6x km/hr.
? Ratio of speed of A, B, C
= 6x : 3x : x = 6 : 3 : 1
Ratio of times taken = 1/6 : 1/3 : 1 or 1 : 2 : 6
? 6 : 1 : : 42 : t
? 6t = 42
? t = 7 min.
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
Time = Distance advanced / Relative speed
2 = 2x / (30 - x)
? x = 15 km/h
Relative speed = 20 + 1 = 21 km/h = 21 x 5/18 = 35/6 m/s
Time = Length of train / Relative speed = (350/35) x 6 = 60 s
Relative speed = 50 - 40 = 10 km/h = 50/18 m/s
? Time taken = Sum of length of the trains / Relative speed
= (200/50) x 18 = 72 sec.
Let the length of train be x m, then
x/10 = (120 + x)/18
? x = 150m
Circumferences means one resolutions .
Therefore, distance covered in 10 resolutions =300 x 10 = 30 m
i.e., 30 meters in 6 seconds.
? Speed of wheel = 30/6 m/s = 5 m/s
? 5 m/s = 5 x (18/5) = 18 km/h
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