Since, ratio of speeds of Meena and teena is 2 : 3.
? Ratio of time taken = 3 : 2
If, Teena takes x min to walk from A to B, then Meena takes (x + 20) min.
? (x + 20)/x = 3/2 ? 2x + 40 = 3x
? x = 40 min
Hence, Meena takes 60 min walking at her usual speed and at double the speed, she would take 30 min.
Let the speed of the train during returning journey be x km/h
Speed during onward journey = x + 25x/100 = 5x/4 km/h.
Distance coverd in onward journey = 800 / 2 = 400 km
Total time taken = Covered distance / Speed
Time taken by train in onward journey = 400/(5x/4)
and time taken in returning journey = 400/x
Thus, 400/(5x/4) + 400/x = 16
? 320/x + 400/x = 16
? 16x = 720
? x = 45 km/h
Speed of the train in the onward journey = 5 x 45/4 = 56.25 km/h
Suppose length of the train = L m
Speed of the train = 60 km/h = 60 x 1000 = 60000 m/h
Length of tunnel = 1.5 km = 1500 m
Time taken by train = 2 min = 1/30 h
Time = Distance/Speed
? 1/30 = L + 1500/60000
? l = 500 m
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
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 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.
Relative speed of express train to local train = 65 - 29 = 36 km/h
= 36 x (5/18) m/s = 10 m/s
? Length of faster train = 10 x 16 = 160 m
Let total distance covered in the whole journey = L km
? 2L/25 + 21L/50 + 2 = L
? L = 4
? Total distance covered = 4 km
Number of gaps between 25 telephone posts = 24
Distance travelled in 1 min = 40 x 24 = 960 m
Speed = (960 x 60)/1000 = 57.6 km/h
Let speed of walking be V km/h.
Total time taken = (7.5/3) + 2 = 4.5 h
Total distance covered = (7.5 + 2V) km
? (7.5 + 2V)/4.5 = 4
? 7.5 + 2V = 18
? 2V = 10.5
? V = 5.25
? Speed of walking = 5.25 km/h
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
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