Question

An aeroplane traveled a distance of $400km$ at an average speed of $x$kmph. On the return journey, the speed was increased by $40$kmph. Write down an expression for the time taken for:
(i). The onward journey.
(ii). The return journeys.
If the return journey took $30$ minutes less than the onward journey, write down an equation in $x$ and find its value.

Answer 1

Hint: In this problem we have given the values of the velocity and distance. So from the definition of velocity $v=\dfrac{d}{t}$ where $v$ is the velocity and $d$ is the distance travelled $t$is time taken to travel, we have relation between velocity and distance as $t=\dfrac{d}{v}$. Here we will calculate the time taken by the aeroplane to cover a distance of $400km$ at an average speed of $x$kmph and then we will calculate the time taken by the aeroplane in return journey with the given conditions. Then we will find the difference between two times and equate it to half an hour (from given data) to get an expression. From that expression, we will find the value of $x$.

Complete step-by-step solution:
Given that, the aero plane travels a distance of $400km$ at an average speed of $x$kmph, then the time taken by the aeroplane is
${{t}_{1}}=\dfrac{400}{x}$
In return journey the speed of aero plane is increased by $40kmph$, then the speed of the aero plane in return is $x+40$kmph , then the time taken by aero plane in return journey is
${{t}_{2}}=\dfrac{400}{x-40}$
Now given that the difference between the time taken is $\dfrac{1}{2}hour$, so
$\begin{align}
  & {{t}_{1}}-{{t}_{2}}=\dfrac{1}{2} \\
 &\Rightarrow \left( \dfrac{400}{x} \right)-\left( \dfrac{400}{x+40} \right)=\dfrac{1}{2} \\
 &\Rightarrow 400\left( \dfrac{1}{x}-\dfrac{1}{x+40} \right)=\dfrac{1}{2} \\
 &\Rightarrow 400\left( \dfrac{x+40-x}{x\left( x+40 \right)} \right)=\dfrac{1}{2} \\
 &\Rightarrow 800\left( 40 \right)=x\left( x+40 \right) \\
 &\Rightarrow {{x}^{2}}+40x-32000=0
\end{align}$
Now the required equation is ${{x}^{2}}+40x-32000=0$
For the value of $x$, we are going to solve the above equation, then
$\begin{align}
  & {{x}^{2}}+200x-160x-32000=0 \\
 &\Rightarrow x\left( x+200 \right)-160\left( x+200 \right)=0 \\
 &\Rightarrow \left( x+200 \right)\left( x-160 \right)=0 \\
 &\Rightarrow x=-200\text{ or }160
\end{align}$
$x$ is the average velocity so it can’t be negative hence the value of $x$ is $160$kmph.

Note: Understand the problem carefully and find the required data from given data to form an expression. From that expression find the required value. For problems related to velocity remember the formula $v=\dfrac{d}{t}$. Here we got $\dfrac{1}{2}$ by converting 30mins in hours as we have the speed in km/hr and distance in km, so we need to take care of units, units should be the same during solving the problem otherwise we will get the wrong answer.