Let speed of the 1st car = 7k
and speed of the 2nd car = 8k
According to the question,
8k = 200/5 = 40 [ using, speed = Distance/ Time ]
? k = 40/8 = 5
? Speed of the 1st car = 7k
= 7 x 5 = 35 km/h
Let speed of 1st bus = 5k
and speed of 2nd bus = 3k
According to the question,
5k = 400/8 = 50
? k = 50/5 = 10
? Speed of 2nd bus = 3k = 3 x 10 = 30 km/h
We know that speed and time are inversely proportional.
? Ratio of time taken = 1/2 : 1/3 : 1/5
= 30/2 : 30/3 : 30/5 [ LCM of 2, 3, 5 = 30 ]
= 15 : 10 : 6
Let the speed in return journey = x
According to the question,
6( x + 2 ) = 9x
? 6x + 12 = 9x
? 9x - 6x = 12
? 3x = 12
? x = 12/3 = 4km/h
Let B takes x h to walk D km.
Then, A takes ( x + 4 ) h to walk D km.
With doube of the speed,
A will take ( x+ 4)/2 h.
According to the question,
x - (x + 4)/2 = 2
? 2x - (x + 4) = 4
? 2x - x - 4 = 4
? x = 4 + 4 = 8 h
Let required distance = L
According to the question,
L/16 - L/20 = 15/20
? (5L - 4L)/80 = 1/4
? L = (1/4) x 80 = 20 km
Let th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h
Speed of car = Distance/Time = 200/240/60
= 600/8 = 75 km/h
Speed of jeep = 200/2 = 100 km/h
? Required ratio = 75/100 = 3/4 3 : 4
Let Brijesh Catches Amit after x h.
Then, distance travelled by Amit in (x + 4) h = Distance travelled by Brijesh in x h.
According to the question,
4(x + 4) = 20 x
? 4x + 16 = 20x
? 16x = 16
? x = 16/16 = 1
Distance travelled by Brijesh in1h = 1 x 20 = 20 km
Due to stoppages, it covers 10 km less per hour.
Time taken to cover 10 km = (10/50) x 60 = 12 min
Hence, the train stops on an average 12 min per hour.
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