Question

An aeroplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Its usual speed is ___ km/hr.

Answer 1

Hint: Here we will first assume the usual speed to be any variable. Then we will find the time taken by the plane with usual speed. We will find the increase in the speed and then we will find the value of the time taken by the plane with the increased speed. Then we will find the difference between these two time values and equate it with the given value. From there, we will get the equation and after solving the equation, we will get the value of the usual speed of the plane.

Formula used:
 \[{\rm{Distance}} = {\rm{speed}} \times {\rm{time}}\]
Complete step-by-step answer:
Let \[x\] be the usual speed of the plane.
Distance covered by the plane in the journey \[ = 1200{\rm{km/hr}}\]
Now, we will find the time taken by the plane with the usual speed.
We know the formula to calculate the distance is given by \[{\rm{Distance}} = {\rm{speed}} \times {\rm{time}}\].
The time taken by the plane with the usual speed \[ = \dfrac{{1200}}{x}{\rm{hr}}\] ………… \[\left( 1 \right)\]
Now, speed is increased by \[100{\rm{km/hr}}\]. Therefore,
New speed of the plane \[ = \left( {x + 100} \right){\rm{km/hr}}\]
Time taken by the plane to cover the same distance but with increased speed \[ = \dfrac{{1200}}{{x + 100}}{\rm{hr}}\]
It is also given that after increasing the speed, the plane takes 1 hour less to complete the journey.
Therefore, mathematically we can write it as,
\[\dfrac{{1200}}{x} - \dfrac{{1200}}{{x + 100}} = 1\]
On taking 1200 common, we get
\[ \Rightarrow 1200\left( {\dfrac{1}{x} - \dfrac{1}{{x + 100}}} \right) = 1\]
Taking LCM inside the bracket, we get
\[ \Rightarrow 1200\left( {\dfrac{{x + 100 - x}}{{x \cdot \left( {x + 100} \right)}}} \right) = 1\]
On further simplification, we get
\[ \Rightarrow 1200\left( {\dfrac{{100}}{{x \cdot \left( {x + 100} \right)}}} \right) = 1\]
On cross multiplying the terms, we get
\[ \Rightarrow 120000 = {x^2} + 100x\]
On subtracting 120000 on both sides, we get
\[\begin{array}{l} \Rightarrow 120000 - 120000 = {x^2} + 100x - 120000\\ \Rightarrow 0 = {x^2} + 100x - 120000\end{array}\]
On rearranging the terms, we get
\[ \Rightarrow {x^2} + 100x - 120000 = 0\]
Now, we will factorize this quadratic equation.
\[ \Rightarrow {x^2} + 400x - 300x - 120000 = 0\]
On further simplification, we get
\[ \Rightarrow x\left( {x + 400} \right) - 300\left( {x + 400} \right) = 0\]
We can see that the new terms have a common factor, we get
\[ \Rightarrow \left( {x - 300} \right)\left( {x + 400} \right) = 0\]
On further simplification, we get
\[x = - 400\] or \[x = 300\].
But we know that the value of speed can’t be negative.
Therefore, \[x = 300\].
Hence, the usual speed of the plane is equal to \[300{\rm{km/hr}}\]

Note: To solve this problem, we need to know the basic formulas to calculate the distance. We need to keep in mind that the distance covered by any object is always equal to the product of the speed of that object and the time taken by the object to cover that distance. So, we can say that the speed and time are inversely proportional to each other. That means when the speed of an object increases the time taken to reach the distance decrease or vice versa.