Question

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

Answer 1

$\underline{\underline{\Huge{\textbf{Solution:}}}}$

Given,
A truck covers a distance of 150km at a certain average speed.

Then it covers another 200km at an average speed which is 20km/hr more than the first speed.

Time taken by the truck to cover the total distance = 5 Hrs. ————[1]



$\underline{\large{\textsf{Step-1:}}}$
Assume a Variable for first speed of the truck

Let the first speed of the truck be 'x' kmph

$\underline{\large{\textsf{Step-2:}}}$
Find the Second Speed of the truck

Second Speed of the truck = First speed of the truck + 20

$\implies$ Second Speed of the truck = x+20



$\underline{\large{\textsf{Step-3:}}}$
Express the time taken by the train to cover 150 km and 200km in terms of Speed.

We have,
$\bigstar \mathbf{Speed = \dfrac{Distance\: covered}{time\: taken}}$

$\implies time\: taken = \dfrac{Distance\: covered}{Speed}$

$\therefore$

time taken by truck to cover 150 km = $\dfrac{150}{x}$ ————[2]

time taken by truck to cover 200 km = $\dfrac{200}{x+20}$ ————[3]



$\underline{\large{\textsf{Step-4:}}}$
Express the Time taken to cover the total distance in terms of first speed.

$\implies$ Total time taken = $\dfrac{150}{x} + \dfrac{200}{x+20}$

From[1]

$\implies \dfrac{150}{x} + \dfrac{200}{x+20} = 5$ ————[4]



$\underline{\large{\textsf{Step-5:}}}$
Simplify [4]

$\implies \dfrac{150(x+20)+200(x)}{x(x+20)} = 5$

$\implies \dfrac{150(x+20)+200(x)}{x(x+20)} = 5$

$\implies \dfrac{150x+3000+200x}{x^{2}+20x} = 5$

$\implies 150x+3000+200x = 5(x^{2}+20x)$

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

$\implies 5x^{2}+100x - 350x -3000= 0$

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

$\implies 5(x^{2} - 50x -600)= 0$

$\implies x^{2} - 50x -600= 0$ ————[5]



$\underline{\large{\textsf{Step-6:}}}$
Factorise [4]

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

$x^{2} \times (-600) = -600x^{2}$
$-600x^{2} = (-60x) \times 10x$
$-60x+10x = -50x$

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

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

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

$\implies x-60 = 0$

$\implies x = 60$

While equating the term (x+10) to zero, a negative value is obtained. And speed can't be negative.

$\therefore$
$\blacksquare \textsf{The First speed of the truck = \underline{\large{\textbf{60\: kmph}}}}$