Suppose they meet after 'h' hours
Then
3h + 4h = 17.5
7h = 17.5
h = 2.5 hours
So they meet at => 4 + 2.5 = 6:30 pm
Speed = 44 kmph x 5/18 = 110/9 m/s
We know that, Time = distance/speed
Time = (360 + 140) / (110/9)
= 500 x 9/110 = 41 sec.
Initial speed = 80km/hr
Total distance = 80 x 10 = 800km
New speed = 800/4 =200km/hr
Increase in speed = 200 - 80 = 120km/hr
Speed of Man = 4.5 kmph
Speed of stream = 1.5 kmph
Speed in DownStream = 6 kmph
Speed in UpStream = 3 kmph
Average Speed = (2 x 6 x 3)/9 = 4 kmph.
Let the distance and original speed be 'd' km and 'k' kmph respectively.
d/0.8k - d/k = 20/60 => 5d/4k - d/k = 1/3
=> (5d - 4d)/4k = 1/3 => d = 4/3 k
Time taken to cover the distance at original speed
= d/k = 4/3 hours = 1 hour 20 minutes.
For this we have to find the LCM of 24, 36 and 30
LCM of 24, 36 and 30 = 360 sec
360/60 min = 6 minutes.
Let the total distance be 'x' km.
Time taken to cover remaining 40% of x distance is
But given time taken to cover first 60% of x distance is
x=80 km.
Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:
x = 40 km.
Let p be the speed of man in kmph
According to the given data in the question,
Distance travelled by bus in 10 min with 20 kmph == Distance travelled by man in 8 min with (20 + p) kmph in opposite direction
=> 20 x 10/60 = 8/60 (20 + p)
=> 200 = 160 + 8p
=> p = 40/8 = 5 kmph.
Speed of tractor = 360/12 = 30 kmph
Speed of jeep = 250 x 30/100 = 75 kmph
But, given ratio of speed of car, jeep and tractor is 3 : 5 : 2
Speed of car = 3 x 75/5 = 45 kmph
Required, average speed of car and jeep = 75 + 45/2 = 60 kmph.
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