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.
60(x + 1) = 64x
X = 15
60 x 16 = 960 km
We know that
Time = Distance/speed
Required time = (10 1/4)/2 + (10 1/4)/6
= 41/8 + 41/6
= 287/24 = 11.9 hours.
Distance travelled by Ramu = 45 x 4 = 180 km
Somu travelled the same distance in 6 hours.
His speed = 180/6 = 30 km/hr
Hence in the conditional case, Ramu's speed = 45 - 9 = 36 km/hr and Somu's speed = 30 + 10 = 40km/hr.
Therefore travel time of Ramu and Somu would be 5 hours and 4.5 hours respectively.
Hence difference in the time taken = 0.5 hours = 30 minutes.
Relative Speed = 5 ¾ - 4 ½ = 1 ¼
Time = 3 ½ h.
Distance = 5/4 x 7/2 = 35/8 = 4 3/8 km.
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.
In the same time, they cover 110 km and 90 km respectively.
Therefore, Ratio of their speeds = 110 : 90 = 11 : 9
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.
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.
Comments
There are no comments.Copyright ©CuriousTab. All rights reserved.