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?
Answer & explanation
Correct answer: option 3
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 .