Question

An aeroplane flying with a wind of $30\text{ kmph}$ takes $120\text{ mins}$ less to fly $2400\text{ km}$ than what it have taken to fly against the same wind. Find the speed of the plane flying in still air.

Answer 1

Hint: We need to find the speed of the plane flying in still air. Given speed of wind $=30\text{ kmph}$ . Using \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\] , we will get the time taken with the wind \[=\dfrac{\text{2400}}{x\text{+30}}\] and Time taken against the wind \[=\dfrac{\text{2400}}{x-\text{30}}\] . According to the condition, we will get \[\dfrac{\text{2400}}{x\text{+30}}=\dfrac{\text{2400}}{x-\text{30}}-120\min \]
Change minute to hours and solve to get the required speed.

Complete step by step answer:
We need to find the speed of the plane flying in still air.
Let the speed of the plane flying in still air be $x\text{ kmph}$ .
Given, speed of wind $=30\text{ kmph}$
We have, \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\]
From this, \[\text{time}=\dfrac{\text{distance}}{\text{speed}}\]
Hence, time taken with the wind \[=\dfrac{\text{2400}}{x\text{+30}}\] .
Time taken against the wind \[=\dfrac{\text{2400}}{x-\text{30}}\]
Hence, according to the question,
\[\dfrac{\text{2400}}{x\text{+30}}=\dfrac{\text{2400}}{x-\text{30}}-120\min \]
Let us convert $120\text{ mins}$ to hours
$1\min =\dfrac{1}{60}\text{ h}$
$\Rightarrow 120\min =\dfrac{120}{60}\text{ h =2h}$
Hence,
\[\dfrac{\text{2400}}{x\text{+30}}=\dfrac{\text{2400}}{x-\text{30}}-2\]
Rearranging, we get
\[\dfrac{\text{2400}}{x-\text{30}}-\dfrac{\text{2400}}{x\text{+30}}=2\]
Taking LCM, we get
\[\dfrac{\text{2400(}x\text{+30})-\text{2400(}x-\text{30})}{(x-\text{30})(x\text{+30)}}=2\]
This can be written as \[\dfrac{\text{(}x\text{+30})-\text{(}x-\text{30})}{(x-\text{30})(x\text{+30)}}=\dfrac{2}{\text{2400}}\]
Simplifying, we get
\[\dfrac{x\text{+30}-x+\text{30}}{(x-\text{30})(x\text{+30)}}=\dfrac{1}{\text{1200}}\]
Solving gives
\[60\times 1200=(x-\text{30})(x\text{+30)}\]
\[\Rightarrow (x-\text{30})(x\text{+30)=}72000\]
We know that ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$
Hence, the above equation can be written as
\[{{x}^{2}}-900=72000\]
\[\Rightarrow {{x}^{2}}=72000+900=72900\]
Taking the square root, we get
\[x=\pm 270\] .
Speed cannot be negative.
Hence the speed of the plane flying in still air is $270\text{ kmph}$ .

Note:
Be careful with the units given. You will have to convert it first before solving it. There can be a chance of error when writing equations to the given condition, ie. \[\dfrac{\text{2400}}{x\text{+30}}=\dfrac{\text{2400}}{x-\text{30}}-120\min \] , one might add 120 min instead of subtracting it. Also, there can be a chance of error when writing the equation of speed that leads to the entire answer to be wrong.