speed = 3 x ( 5/18 ) m/sec.
= 5 / 6 m/sec.
? Distance covered in 2 min. = ( 5 / 6) x 2 x 60 m
= 100 m.
? Length of diagonal = 100 m
Area of the field = 1/2 x diagonal2
=1/2 x 100 x 100 m2
= 5000 m2
= 50 acres.
Speed in cm/minute = (Speed in km/hr x 1000 x100 )/60
= 47.52x(50/3)
=79200 cm/min
And Circumference of circle = 2?r
=2 x ( 22 / 7 ) x 21
=132
No. of revolutions = ( Speed in cm/minute ) / circumference of circle in cm
=79200 / 132
=600 rpm
Let the actual speed of train be x and actual time taken be y
Then new speed of train = 5x/6
Therefore, new time taken = 6y/5 (as distance is same in both case)
Given, 6y/5 - y = 1/6 hr , therefore actual time = 50 min
Required average speed = 75 x {(100 - 40)/100} x {(100 + 50)/100}
= 75 x 3/5 x 3/2
= 67.5 km/h
Distance= Speed x Time = 96 x (100/60) = 160 km
New time = 80/60 h = 4/3 h
New speed = 160 x 3/4 = 40 x 3 =120 km/h
Average speed = 2AB/(A + B)
= 2 x 5 x 3/(5 + 3)
= 30/8
= 3.75 km/h
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
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 the distance be x km.
Then, x/3 - x/4 = 30/60
? (4x - 3x ) / 12 = 1/2
? x = 6 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.
Suppose they meet after y hours .
Then, 21y - 16y = 60
? y = 12
?required distance = (16 x 12 + 21 x 12) km.
= 444 km.
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