A girl increases her speed to 9/4 times her original speed. So, she reaches her school 15 minutes before the usual time. What is her usual time?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.45 Hours
For the same distance, time is inversely proportional to speed. If the speed becomes \(\frac{9}{4}\) times, the time becomes \(\frac{4}{9}\) of the usual time.
Let the usual time be \(T\) minutes.
New time \(\displaystyle =\frac{4}{9}T\), so the time saved \(\displaystyle =T-\frac{4}{9}T=\frac{5}{9}T\)
\(\displaystyle \frac{5}{9}T=15\), so \(T=27\) minutes
27 minutes \(\displaystyle =\frac{27}{60}\) hour \(=\) 0.45 hours