Let the distance between home and school is 'x'.
Let actual time to reach be 't'.
Thus, x/4 = t + 4 ---- (1)
and x/6 = t - 30 -----(2)
Solving equation (1) and (2)
=> x = 98 mts.
In the same time, they cover 110 km and 90 km respectively.
Therefore, Ratio of their speeds = 110 : 90 = 11 : 9
Let the distance travelled by Tilak in first case or second case = d kms
Now, from the given data,
d/20 = d/22 + 36 min
=> d/20 = d/22 + 3/5 hrs
=> d = 132 km.
Hence, the total distance travelled by him = d + d = 132 + 132 = 264 kms.
Time taken to meet the first time = length of track/relative speed
Given the length of the track is 600 m
The relative speed = 25 + 35 = 60 kmph = 60 x 5/18 m
Therefore Time = = 600/60 x (18/5) = 36 sec.
let 't' be the time after which they met since L starts.
Given K is 50% faster than L
50 t + 1.5*50(t-1) = 300
50 t +75 t = 300 + 75
t = 375 / 125 = 3 hrs past the time that L starts
So they meet at (9 + 3)hrs = 12:00 noon.
By car 240 km at 60 kmph
Time taken = 240/60 = 4 hr.
By train 240 km at 60 kmph
Time taken = 400/100 = 4 hr.
By bus 240 km at 60 kmph
Time taken = 200/50 = 4 hr.
So total time = 4 + 4 + 4 = 12 hr.
and total speed = 240+400+200 = 840 km
Average speed of the whole journey = 840/12 = 70 kmph.
Time taken to cover 600 km = 600/100 = 6 hrs.
Number of stoppages = 600/75 - 1 = 7
Total time of stoppages = 4 x 7 = 28 min
Hence, total time taken = 6 hrs 28 min.
Let the distance traveled be x km.
Then, x/10 - x/15 = 2
3x - 2x = 60 => x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 = 6 hrs.
So, Robert started 6 hours before 2. p.m. i.e., at 8 a.m.
Required speed = 60/5 = 12 kmph.
Let the actual distance travelled be x km.
Then x/8=(x+20)/12
=> 12x = 8x + 160
=> 4x = 160
=> x = 40 km.
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