Question

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

Answer 1

Hint: We have to know the formulae of time is \[\text{time}\left( \text{t} \right)\text{=}\dfrac{\text{distance}\left( \text{d} \right)}{\text{speed}\left( \text{s} \right)}\] when distance and speed is given. Firstly we have to calculate the initial speed of the car. Then we have to calculate the speed when 12km/hr. is increased. After getting both speeds if we subtract them and equal to the less time we will get the equation. Then we will get a quadratic equation from that equation if we solve this we will get the required answer.

Complete step by step answer:
Let us assume the initial speed of the car is \[s=x\] km/hr.
In the question it is stated that the car covers a distance of 400 km at a certain speed.
So the distance is \[d=400\].
We know the formulae of time when distance and speed is given is
\[\text{time}\left( \text{t} \right)\text{=}\dfrac{\text{distance}\left( \text{d} \right)}{\text{speed}\left( \text{s} \right)}\]hr.
Putting the value of “s” and “d” we get,
\[t=\dfrac{400}{x}\] hr.
But the speed was increased by 12km/hr.
Now the speed of the car is
\[s=\left( x+12 \right)\]km/hr.
Now to cover the same distance time will take,
\[t=\dfrac{400}{\left( x+12 \right)}\] hr.
According to the question, if the speed of the car has been 12 km/hr more, the time taken for the journey would have been 1 hour 40 minutes less.
We know 1 hour 40 min = \[1\dfrac{40}{60}\] hr.
Now,
\[\begin{align}
  & \dfrac{400}{x}-\dfrac{400}{\left( x+12 \right)}=1\dfrac{40}{60} \\
 & \Rightarrow \dfrac{400}{x}-\dfrac{400}{\left( x+12 \right)}=\dfrac{5}{3} \\
\end{align}\]
Taking 400 common from numerator we get
\[\Rightarrow 400\left( \dfrac{1}{x}-\dfrac{1}{x+12} \right)=\dfrac{5}{3}\]
Now taking LCM and brake the brackets we get,
\[\begin{align}
  & \Rightarrow \dfrac{x+12-x}{x\left( x+12 \right)}=\dfrac{5}{3\times 400} \\
 & \Rightarrow \dfrac{12}{{{x}^{2}}+12x}=\dfrac{1}{240} \\
\end{align}\]
Now cross multiplying,
\[\begin{align}
  & \Rightarrow {{x}^{2}}+12x=2880 \\
 & \Rightarrow {{x}^{2}}+12x-2880=0 \\
\end{align}\]
We will use the quadratic formula method to solve it. So, we have a formula given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ .
Here, we have $a=1,b=12,c=-2880$ .
Now applying the formula and solving the quadratic equation we get,
$\begin{align}
  & x=\dfrac{-12\pm \sqrt{{{\left( 12 \right)}^{2}}-\left( 4\times 1\times -2880 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm \sqrt{144+11520}}{2} \\
\end{align}$
Taking 144 common outside inside the root, we have
$\begin{align}
  & \Rightarrow x=\dfrac{-12\pm \sqrt{144\left( 1+80 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm 12\sqrt{81}}{2} \\
 & \Rightarrow x=\dfrac{-12\pm \left( 12\times 9 \right)}{2} \\
 & \Rightarrow x=\dfrac{-12\pm 108}{2} \\
\end{align}$
Dividing each term by 2 and simplifying, we have $x=-6\pm 54$.
Considering each value of x, we have
\[\begin{align}
  & \Rightarrow x=-6+54=48 \\
 & and \\
 & \Rightarrow x=-6-54=-60 \\
\end{align}\]

Speed of the car can never be negative so the speed of the car will be 48km/hr.

Note: Students have to remember the formulae of time when speed and distance is given. Students have to be able to solve quadratic equations. After getting both values from the quadratic equation students get confused to choose the value and make silly mistakes. They have to understand that speed can never be negative. By knowing this they can easily eliminate the negative value. And they can solve the question.