Let Pele x km in 1 hour. so Maradona takes (2h-40min)=1 h 20 min to cover x km. Let speed of Maradona and Pele be M and P respectively than
x =M x (4/3) and x= P x 1
=> M/P = 3/4
Again
=> k= 25
M= 3k = 75 km/h and P= 4k = 100 km/h
( Through options it is very easy to solve)
u = k , v= 3k
=> k = 8km/h
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.
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.
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.
To find the minimum distance, we have to get the LCM of 75, 80, 85
Now, LCM of 75, 80, 85 = 5 x 15 x 16 x 17 = 20400
Hence, the minimum distance each should walk so that thay can cover the distance in complete steps = 20400 cms = 20400/100 = 204 mts.
Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 40 km at a speed of 8kmph
time taken indivisually:
=> 6/1.5 = 4 m
=> 8/2 = 4 m
=> 40/8 = 5 m
Average speed of man= total distance/ total time
=> 54/13 =
kmph
Given speed of the bike after servicing = 55 kmph
Time taken for travelling some distance at 55 kmph = 5 hrs
Then,
Distance = Speed x Time = 55 x 5 = 275 kms.
Now,
Speed of the bike before servicing = 35 kmph
Distance = 275 kms
Now, Time = Distance/Speed = 275/35 = 7.85 hrs =~ 8 hrs.
Speed of lorry = = 30 kmph
Speed of van = = 75 kmph
Speed of bike = = 45 kmph
Therefore, now required average speed of bike and van = = 60 kmph.
Given that,
1st train speed = 62 km/hr
2nd train speed = 44 km/hr
Hence the relative speed of two trains is : 62-44 = 18 km/hr = m/s = 5 m/s.
There fore, the first train crosses the second train i.e (85+75)mts of distance with a speed of 5m/s in,
time = = 32 sec.
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