Question : A bus left 60 minutes later than the scheduled time, but to reach its destination 48 km away in time, it had to increase the speed by 4 km/hr from the usual speed. What is the usual speed (in km/hr) of the bus?
Option 1: 9
Option 2: 12
Option 3: 15
Option 4: 8
Correct Answer: 12
Solution :
Let the usual speed be $x$ km/hr.
According to the question,
$\frac{48}{x}-\frac{48}{x+4}=\frac{60}{60}$
$⇒x^2+4x-192=0$
$⇒x^2+(16-12)x-192=0$
$⇒(x+16)(x-12)=0$
Since speed cannot be negative,
$\therefore x= 12$ km/hr
Hence, the correct answer is 12.
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