Question : A boy increases his speed to $\frac{9}{5}$th 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 usually take?
Option 1: 120
Option 2: 30
Option 3: 90
Option 4: 45
Correct Answer: 90
Solution :
Let the speed of the boy be $x$ km/hr.
Initial time = $t$
So, distance = $tx$
New time = $t-40$
New speed = $x × (\frac{9}{5})=\frac{9x}{5}$
According to the question,
$tx=(t-40)\frac{9x}{5}$
$⇒5t=9t-360$
$⇒ 4t=360$
$\therefore t =90$ minutes
Hence, the correct answer is 90.
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