A man increases his speed $\frac{12}{7}$ times of its original speed.  By doing this he reaches his office 20 minutes before the usual time.  Find the time he usually takes to reach his destination at normal speed.

48 mins.

S ∝ $\frac{1}{t}$,  when distance is same.

Speed    →   7      :     12

Time      →  12     :      7

                    ↓             ↓

                 Difference = 5

                                    ↓×4

               Given time = 20  mins.   

               Usual time = 12 × 4 = 48 mins.