Question

While boarding an airplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalize the injured and so the plane started late by $30\mathrm{minutes}$. To reach the destination, $1500\mathrm{km}$ away in time, the pilot increased the speed by $100\mathrm{km}/\mathrm{hr}$. Calculate the original speed/hour of the plane in $\mathrm{km}/\mathrm{hr}$.

Answer 1

Step 1: Given parameters:

Consider the original speed of the plane be $x\mathrm{km}/\mathrm{hr}$.

The increased speed of the plane will be $(x+100)\mathrm{km}/\mathrm{hr}$.

The total distance covered is $1500\mathrm{km}$.

Step 2: Calculate the time taken to reach the final destination

Calculate the time using the below expression given as:

$\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

Therefore, the time taken $\left(t_1\right)$ to reach the destination at the original speed is given as:

$$\begin{gathered} t_1=\frac{\mathrm{Distance}\mathrm{covered}}{\mathrm{Original}\mathrm{speed}} \\ {t}_1=\frac{1500}{x}......\left(1\right) \end{gathered}$$

Similarly, the time taken $\left(t_2\right)$ to reach the destination at an increasing speed is given as:

$t_2=\frac{1500}{\left(x+100\right)}......\left(2\right)$

Step 3: Calculate the original speed of the train

It is given that due to some circumstances, the plane is late by $30\mathrm{minutes}$.

Convert minutes to hours

$$\begin{gathered} 1\mathrm{hr}=60\mathrm{minutes} \\ \Rightarrow 1\mathrm{minute}=\frac{1}{60}\mathrm{hr} \\ \Rightarrow 30\mathrm{minutes}=\frac{30}{60}\mathrm{hr} \\ =\frac{1}{2}\mathrm{hr} \end{gathered}$$

Now, according to the question,

Substitute the values from equations (1) and (2).

$$\begin{gathered} t_1-t_2=\frac{1}{2}\mathrm{hr} \\ \frac{1500}{x}-\frac{1500}{\left(x+100\right)}=\frac{1}{2} \\ \frac{1500\left(x+100\right)-1500x}{x\left(x+100\right)}=\frac{1}{2} \\ \frac{1500x+150000-1500x}{x^2+100x}=\frac{1}{2} \\ \frac{150000}{x^2+100x}=\frac{1}{2} \\ \Rightarrow 300000=x^2+100x \\ \Rightarrow x^2+100x-300000=0 \\ \Rightarrow x^2+600x-500x-300000=0 \\ \Rightarrow x\left(x+600\right)-500\left(x+600\right)=0 \\ \Rightarrow \left(x-500\right)\left(x+600\right)=0 \end{gathered}$$

Therefore,

$x–500=0\mathrm{or}x=500$

$x+600=0\mathrm{or}x=–600$

As speed cannot be negative. Thus, the original speed/hour of the plane is $500\mathrm{Km}/\mathrm{hr}$.