A boy increases his speed to $\frac{9}{5}$ times of his original speed. By doing this, he reaches his school 40 minutes before the usual time. How much time (in minutes) does he take usually?

90

As , Speed is inversely proportional to Time .

According to question ,

Ratio               Initial     :      Final

Speed                 5        :       9

Time                   9        :       5

Now ,

4R = 40 minutes

1R = 10 minutes

9R = 9 × 10 = 90 minutes

So , time taken by the boy usually is 90 minutes .