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
Given, x = 16 and Y = 25
According to the formula,
1st man's speed : 2nd man's speed = ? y : ? x = ? 25 : ? 16 = 5 : 4
Relative speed of policeman = ( 20 - 16 ) x 5/18 = 10/9 m/s
To catch the thief, the policeman in has to gain 200 m = 200 x 9/10 = 180 s
Actual distance covered by policeman in 180 s = 180 x 50/9 = 100 m
? Distance covered by the thief = 1000 - 200 = 800m
Given, D = 84 km, a = 12 km/h and b = 16 km/ h
According to the formula
Distance traveled by A = PR = 2D x a/(a + b)
= 2 x 84 x 12/(12 + 16) = (2 x 84 x 12)/28
= 2 x 6 x 6 = 72 km
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 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
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