Let the distance be D km
Difference in timings = 8 min = 8/60 h = 2/15 h
? D/3 - D/4 = 2/15
? D/12 = 2/15
? D = 2 x 12/15 = 1.6 km
New speed = 8/11 of usual speed
New time = 11/8 of usual time
? 11/8 of usual time = 11/2 h
? Usual time = (11 x 8)/(2 x 11) = 4 h
? Time saved = 51/2- 4
= 11/2
Average speed = [(30/60) x 40 + (45/60) x 60 + (2 x 70)] / [(30/60) + (45/60) + 2]
= (20 + 45 + 140)/[(30 + 45 + 120)/60]
= (205/195) x 60 km/h
= 63.07 km/h
= 63 km/h
Since, the boys now walks at 5/7 of usual speed, so he will take 7/5 of his usual time.
? Extra time = (7/5 - 1) x Usual time = 6 min (known)
? 2/5 x Usual time = 6
? Usual time = 15 min
Relative speed = 60 + 40 = 100 km/h
Time = Distance/speed = 150/ 00 = 3/2 h
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
Relative speed of policeman with thief = (10 - 8) = 2 km/h
Now, time taken to cover 100 m with speed of 2 km/h or 2000 m/h
Now, Speed = Distance / Time
? Time = D/S = 100/2000
= 1/20 h or 60/20 min = 3 min
Speed upstream + Speed downstream = 2 x 71/2 = 15 km/h
Since, the time taken is in the ratio 2 : 1
hence the speed will be in the ratio 1 : 2
? Speed upstream = (1/3) x 15 = 5 km/h
and speed downstream = (2/3) x 15 = 10 km/h
? Speed of stream = (10 - 5)/2 = 2.5 km/h
Let the distance between the two ports = L km
Then, speed downstream = L/4
and speed upstream = L/5
? Speed of the stream = 1/2(L/4 - L/5)
? 1/2[(5L - 4L)/20] = 2
? L = 80 km
Let the speed of boat in still water be V km/h. Then,
35/(V - 1) + 35/(V + 1) = 12
? 35(V + 1 + V - 1) = 12(V2 - 1)
? 6V2 - 35V - 6 = 0
? (V - 6)(6V + 1) = 0
? V = 6 (V = - 1/6 is not possible)
So, speed of boat in still water is 6 km/h.
Clearly, motorboat takes 2 h go from Mumbai to Goa and 4 h to go from Goa to Mumbai. i.e.,
Time taken to row upstream = 2 x Time taken to row downstream
? n = 2
Let the speed of stream be V km/h.
According to the formula,
Speed of boat = V(n + 1)/(n - 1)
? 30 = V(2 + 1)/(2 - 1)
? V = 30/3 = 10 km/h
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