A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed

Question:

A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.

Solution:

Let the first speed of the truck be x km/h.

$\therefore$ Time taken to cover $150 \mathrm{~km}=\frac{150}{x} \mathrm{~h} \quad$ (Time $\left.=\frac{\text { Distance }}{\text { Speed }}\right)$

New speed of the truck = (x + 20) km/h

$\therefore$ Time taken to cover $200 \mathrm{~km}=\frac{200}{x+20} \mathrm{~h}$

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h

$\therefore \frac{150}{x}+\frac{200}{x+20}=5$

$\Rightarrow \frac{150 x+3000+200 x}{x(x+20)}=5$

$\Rightarrow 350 x+3000=5\left(x^{2}+20 x\right)$

$\Rightarrow 350 x+3000=5 x^{2}+100 x$

$\Rightarrow 5 x^{2}-250 x-3000=0$

$\Rightarrow x^{2}-50 x-600=0$

$\Rightarrow x^{2}-60 x+10 x-600=0$

$\Rightarrow x(x-60)+10(x-60)=0$

$\Rightarrow(x-60)(x+10)=0$

$\Rightarrow x-60=0$ or $x+10=0$

$\Rightarrow x=60$ or $x=-10$

∴ x = 60                  (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.