Circumferences means one resolutions .
Therefore, distance covered in 10 resolutions =300 x 10 = 30 m
i.e., 30 meters in 6 seconds.
? Speed of wheel = 30/6 m/s = 5 m/s
? 5 m/s = 5 x (18/5) = 18 km/h
Let the length of train be x m, then
x/10 = (120 + x)/18
? x = 150m
Relative speed = 50 - 40 = 10 km/h = 50/18 m/s
? Time taken = Sum of length of the trains / Relative speed
= (200/50) x 18 = 72 sec.
Relative speed = 20 + 1 = 21 km/h = 21 x 5/18 = 35/6 m/s
Time = Length of train / Relative speed = (350/35) x 6 = 60 s
Time = Distance advanced / Relative speed
2 = 2x / (30 - x)
? x = 15 km/h
Let the original speed be x km/hr.
Then, [ 840/x - 840/( x + 10) ] = 2
? 840 (x + 10) - 840x = 2x (x+10)
? x2 + 10x - 4200 = 0
? (x + 70) (x - 60 ) = 0
? x = 60 km/hr.
(speed of wind) / (speed of car) = (Time utilised) / (time saved)
? 332/x = 332/28
? x = 28 m/s
Time = Total distance / Relative speed
4.5/60 hr. = (450/1000) / x
? x = 6 km/h
Relative speed = Speed of car - Speed of man
6 = x - 6
? x = 12 km/h
Let they meet after x h.
Then, according to the question,
3x + 4x = 17.5
? 7x = 17.5
? x = 17.5 / 7 = 2.5 h
So, they meet 2.5 h after 8:00 am.
It means they meet at 10:30 am.
Let distance between Manipur and Dispur = D km
Average speed of train from Manipur = D/4 km/h
Average speed of train from Dispur = 2D/7 km/h
Let they meet T h after 6:00 am.
Then, according to the question,
(D/4 x T) + 2D/7 x (T - 2) = D
? T/4 + 2(T - 2)/7 = 1
? 7T + 8(T - 2) = 28
? 15T = 44
? T = 44/15 h = 2h 56 min
Clearly, trains meet 2 h 56 min after 6:00 am.
It means the trains meet at 8:56 am.
Let the distance traveled on foot be x km
Then, distance converted by bicycle = (45 - x) km
? x/3 + (45 -x) /8 = 83/4 = 35/4
? (8x + 135 - 3x)/24 = 35/4
? 5x + 135 = 210
? 5x = 75
? x = 15
? Distance converted by bicycle = (45 - 15) = 300 km
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