A person goes from home to the office at 80% of his usual speed. After a 2-hour rest, he returns with the usual speed. If he covers 2000 km of distance during the whole journey, it takes 52 hours to complete it. Find the speed at which he goes to the office.

Given:

The new speed is 80% of the usual speed

Total distance cover 2000 km in 52 hours

Take 2 hour's rest then return

Formula used: 

Average speed  = 2ab/(a + b)

a = new speed , b = usual speed

Calculation:

New speed = \(\frac{80}{100}\)(usual speed)

⇒ new speed/usual speed = 4/5

New speed = 4u, usual speed = 5u

Time excluding the rest hour =  52 - 2 = 50h

Average speed  = 2000/50   

Average speed = 40 km/h

⇒ \(2\ ×\ 4u ×\ 5u \over 4u\ +\ 5u\) = 40

⇒ 40u²/9u = 40

⇒ u/9 = 1

⇒ u = 9
Since,  km/h × 5/18 = m/s

New speed = 4u

New speed = 9 × 4 = 36 × 5/18 = 10 m/s  

∴ The correct answer is 10 m/s