Question

A car cover a distance of $400km$ at a certain speed. Had the speed $12km/h$ more, the time taken for the journey would have been $1$ hour $40$ minutes less. Find the original speed of the car.

Answer 1

Let the speed of the car be $xkm/hr$ and the time taken at that speed to cover $400km$ be $t$ minutes. Therefore, according to the first condition:

Distance $=$ Speed $\times$ Time

$\Rightarrow 400=xt\Rightarrow t=\frac{400}{x}....\left(1\right)$

As per the second case, the distance remains the same, speed is increased to $(x+12)km/hr$ and the time reduces to $(t-100)$ minutes.

$\therefore 400=(x+12)(t-100)\Rightarrow 400=(x+12)(\frac{400}{x}-100)\Rightarrow 4=(x+12)(\frac{4}{x}-1)$

$\Rightarrow 4=4-x+\frac{48}{x}-12\Rightarrow \frac{48}{x}-x=12$

$\Rightarrow 48-x^2=12x$

$\Rightarrow x^2+12x-48=0$

$\therefore x=\frac{-12\pm \sqrt{(-12{)}^2-4(-48)}}{2}=\frac{-12\pm \sqrt{336}}{2}=\frac{-12\pm 4\sqrt{21}}{2}$

$\therefore x=-6+2\sqrt{21}$ or $x=-6-2\sqrt{21}$

Since, speed cannot be negative, we neglect the second answer.

Hence, the original speed of the car is $2\sqrt{21}-6\approx 3.165km/hr$