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.
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
Relative speed = 60 + 40 = 100 km/h
Time = Distance/speed = 150/ 00 = 3/2 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
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
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
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