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
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
Ratio of speeds of A and B = 3 : 4
? Ratio of time taken = 4 : 3
Let time taken by A and B be 4k and 3k, respectively.
Then, according to the question,
4k - 3k = 20
? k = 20
Hence, time taken by A = 4 x 20 = 80 min
And time taken by B = 3 x 20 = 60 min
Let time taken by A = Y
Let speed of C = x
Then, speed of B = 3x
? speed of A = 6x
Now, ratio of speeds of A and C = Ratio of time taken by C and A
6x : x = 56 : y
? 6x/x = 56/y
? y = 56/6 = 92/6
= 91/3 min
Given, A = 6 km/h, B = 3 km/ h
According to the formula, Average speed = 2AB/(A + B)
? Required average speed = 2 x 6 x 3/(6 + 3)
= 36/9 = 4 km/h
Let L km is covered in T h.
Then, first speed = L/T km/h
Again, L/2 km is covered in 4T h.
? New speed = (L/2 x 1/4T ) = (L/8T) km/ h
Ratio of speed = L/T : L/8T = 1 : 1/8 = 8 : 1
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
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