Two men A and B travel from point P to Q, a distance of 84 km at 12 km/h and 16 km/h, respectively. B reaches Q and returns immediately and meet A at R. Find the distance form P to R.

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

Let total distance = x km
According to the question,
x/3 - x/4 = 1/2;
? x/12 = 1/2
? x = 6 km

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, 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