Option 2 : 0.9 hours

**Given :**

Person walks with (11/18) of his usual speed

He is late by 21 minutes

**Concept used :**

Time = Distance/Speed

Time is inversely proportional to speed.

**Calculations :**

Let the usual time be 't' and usual speed be 's'

For speed s time will be 't' and for speed (11/18)s time will be (18/11)t (time is inversely proportional to speed)

According to the question

Time with slower speed - time with faster speed = 21 minutes

(18/11)t - t = 21

⇒ (7/11)t = 21

⇒ 33 minutes

Time when he is late = usual time + 21

⇒ 33 + 21

⇒ 54 minutes

Time in hours = 54/60 = 0.9 hours

**∴ Option 2 will be the correct answer.**

__Alternate Method__

Let the initial speed be 18x

His new speed is (11/18) × 18x = 11x

Ratio of initial to changed speed = 18 : 11

As time is inversely proportional to speed

⇒ Ratio of time taken = 11 : 18

Let the time taken initially be 11x minutes and time taken after changed speed is 18x minutes

According to question

18x - 11x = 7x = 21

⇒ x = 3

Time taken when he is late = 18x = 18 × 3 = 54 minutes = 0.9 hours

**∴ Time taken when he is late is 0.9 hours**