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
Given that, W + D = 6 ...(i)
[ w = Time taken while walking and
D = Time taken while driving ]
From Eq. (i)
5 +D = 6
? D = 1
2D = 2 x 1 = 2
? He will take 2 h to drive both ways.
According to the formula,
Average speed = 2AB/(A + B)
Here, A = 5 km/h, and B = 2 km/h
? Required average speed = (2 x 5 x 2)/7 = 20/7
= 26/7 km/h
Total distance to covered in 10 h = 80 km
But it covers 40 km in 3/5th of time, i.e, 40 km in 6 h.
? Required time = 10 - 6 = 4 h
And remaining distance = 40 km
Thus, required speed = 40/4 =10 km/h
Let the total distance = L
Then, according to the question,
(2L/3)/ 4 + (1 - 2/3)L/5 = 84/60
? 2L/12 + L/15 = 84/60
? L/6 + L/15 = 84/60
? 5L + 2L = 42
? 7L = 42
? L = 42/7 = 6 km
Given, x = 16 and Y = 25
According to the formula,
1st man's speed : 2nd man's speed = ? y : ? x = ? 25 : ? 16 = 5 : 4
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, 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 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
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
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