A person can walk a certain distance and drive back in 6 h. He can also walk both ways in 10 h. How much time will he take 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

Length of the bridge = Distance travelled by the person in 4 min = Speed x Time
Speed = 54 km/h = 54 x (5/18) = 3 x 5 = 15 m/s
Time = 4 min = 4 x 60 = 240 s
? Required length = 15 x 240 = 3600 m

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

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