Question

A car cover a distance of 300 km with some speed. If the speed is increased by 10 km/hr, it will cover the same distance in one hour less. Find the speed of car.

Answer 1

Step 1:

Given data:

Distance covered by the car$=300km$

Increase in speed$=10\mathrm{km}/\mathrm{hr}$

After increasing speed, the same distance will be covered in one hour less.

Step 2:

Form the required equation to calculate the speed of the car further:

Let the speed of car be $x\mathrm{km}/\mathrm{h}$ and the total time of journey be $t\mathrm{hours}$

Then,

$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{time}}$

$$\begin{gathered} \Rightarrow x=\frac{300}{t}...\left(1\right) \\ \Rightarrow x\times t=300 \\ \Rightarrow xt=300...\left(2\right) \end{gathered}$$

Now,

$\mathrm{speed}=x+10\mathrm{km}/\mathrm{hr}$ then $\mathrm{time}=t-1\mathrm{hours}$

According to the given condition,

$$\begin{gathered} x+10=\frac{300}{\left(t-1\right)} \\ \Rightarrow (x+10)(t-1)=300...\left(3\right) \\ \Rightarrow xt-x+10t-10=300 \\ \Rightarrow 300-x+10t=310\left[\mathrm{From}\left(2\right),xt=300\right] \\ \Rightarrow -x+10t=310-300 \\ \Rightarrow -x+10t=10 \\ \Rightarrow 10t-x=10 \\ \Rightarrow 10t-\frac{300}{t}=10\left[From\left(1\right),x=\frac{300}{t}\right] \\ \Rightarrow 10t^2-300=10t \\ \Rightarrow 10t^2-10t-300=0 \\ \Rightarrow t^2-t-30=0 \end{gathered}$$

Step 3:

solve the quadratic equation obtained:

$$\begin{gathered} t^2-t-30=0 \\ \Rightarrow t^2-6t+5t-30=0 \\ \Rightarrow t\left(t-6\right)+5\left(t-6\right)=0 \\ \Rightarrow \left(t-6\right)\left(t+5\right)=0 \\ \Rightarrow t=6\mathrm{or},t=-5 \end{gathered}$$

As$t$cannot be negative,

So, $t=6$

Step 4:

the speed of the car :

$$\begin{gathered} x=\frac{300}{6} \\ =50\mathrm{km}/\mathrm{hr} \end{gathered}$$

Therefore, the speed of the car is$50\mathrm{km}/\mathrm{h}.$