Question

If the speed of an aeroplane is reduced by 40 km per hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.

Answer 1

Hint: Let us assume that the original speed of the aeroplane is x km/hr. Since we know that, $\text{Speed = }\dfrac{\text{distance}}{\text{time}}$, so find the time taken to cover 1200 km at a speed of x km/hr. Let the time is ${{t}_{1}}$
Now, it is said that speed is reduced by 40 km/hr. Therefore, the new speed is (x – 40) km/hr. So, find the time taken for the reduced speed as ${{t}_{2}}$. Since ${{t}_{2}}$is 20 minutes more than \[{{t}_{1}}\], so get an equation in terms of x and solve for x.

Complete step-by-step answer:
Let us start by assuming that the original speed of the aeroplane was x km/hr.
So, by using the formula $\text{Speed = }\dfrac{\text{distance}}{\text{time}}$, for 1200 km, time taken is:
\[{{t}_{1}}=\dfrac{1200}{x}hr......(1)\]
Now, we have reduced speed as (x – 40) km/hr
So, by using the formula $\text{Speed = }\dfrac{\text{distance}}{\text{time}}$, for 1200 km, time taken is:
\[{{t}_{2}}=\dfrac{1200}{\left( x-40 \right)}hr......(2)\]
As it is mentioned that:
${{t}_{2}}$is 20 minutes more than \[{{t}_{1}}\], so get an equation in terms of x and solve for x.
So, we can write:
\[{{t}_{2}}={{t}_{1}}+\dfrac{20}{60}......(3)\]
Substituting the values, we have
\[\begin{align}
  & \dfrac{1200}{\left( x-40 \right)}=\dfrac{1200}{x}+\dfrac{20}{60} \\
 & \dfrac{1200}{\left( x-40 \right)}-\dfrac{1200}{x}=\dfrac{20}{60} \\
 & \dfrac{1200\left( x-x+40 \right)}{x\left( x-40 \right)}=\dfrac{1}{3} \\
 & 144000={{x}^{2}}-40x \\
 & {{x}^{2}}-40x-144000=0 \\
\end{align}\]
Now, we will solve the above equation using the factorisation method.
$\begin{align}
  & x\left( x-400 \right)+360\left( x-400 \right)=0 \\
 & (x-400)(x+360)=0 \\
 & x=400,-360 \\
\end{align}$
Since speed can not be negative, therefore we neglect the negative value of x.
Hence, the original speed of the aeroplane is 400 km/hr.


Note: Speed is a scalar quantity. So, it only has magnitude. Therefore, it can never be a negative value. Velocity is a vector quantity which has magnitude as well as direction. Therefore, it can be negative too.