Question

A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500km away in time, it has to increase its speed by 250kmph from its usual speed. Find its usual speed.
A plane leaves 30 minutes later than the scheduled time and in order to reach its destination 1500km away in time, it has to increase its speed by 250kmph from its usual speed. Find its usual speed.

Answer 1

Hint: In this question, we are given the distance the plane has to travel and speed it should increase to reach its destination on time as it flew off 30 minutes late. We have to find its usual speed. For this, we will suppose the usual speed to be x km/hr and then use the given information to form an equation. The equation will be formed using the distance time formula that is given by $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$. The equation formed will be quadratic and hence we will solve it using the split middle term method.

Complete step by step answer:
Here, we are given the distance the plane has to travel. The plane took off 30 minutes late so it has to increase its speed by 250 kmph. We need to calculate its usual speed. For this, let us suppose that its usual speed was x kmph. Now, distance is given as 1500km. Therefore, time taken by plane usually would have been given $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$ that is ${{\text{t}}_{1}}=\dfrac{\text{1500}}{\text{x}}$.
But now, the plane took off 30 minutes late and speed has increased, so now speed will be given by (x+250) kmph. Also, distance remains same, so new time taken by plane will be ${{\text{t}}_{2}}=\dfrac{\text{1500}}{\text{x+250}}$.
As we know, the plane was 30 minutes late, therefore, the difference between new time and old time will be 30 minutes. We can see ${{\text{t}}_{1}}-{{\text{t}}_{2}}=30\text{ minutes}$. Since, we are dealing with speed to be in km/hr. So time should be in hours too, so let us convert 30 minutes to hours.
1 hour = 60 minutes.
$\therefore 1\text{ minute}=\dfrac{1}{60}\text{hours}$
And 30 minutes =$\dfrac{30}{60}\text{hours}=\dfrac{1}{2}\text{hours}$.
Hence, the plane was delayed by $\dfrac{1}{2}$ hour.
Now our equation becomes \[\Rightarrow \dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{1}{2}\].
Taking LCM on left side and simplifying we get:
\[\begin{align}
  & \Rightarrow \dfrac{1500\left( x+250 \right)-1500x}{x\left( x+250 \right)}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{1500x+375000-1500x}{{{x}^{2}}+250x}=\dfrac{1}{2} \\
 & \Rightarrow \dfrac{375000}{{{x}^{2}}+250x}=\dfrac{1}{2} \\
\end{align}\]
Cross multiplying we get:
\[\begin{align}
  & \Rightarrow 750000={{x}^{2}}+250x \\
 & \Rightarrow {{x}^{2}}+250x-750000=0 \\
\end{align}\]
Now let us split the middle term in order to form the factor of this equation.
\[\begin{align}
  & \Rightarrow {{x}^{2}}+1000x-750x-750000=0 \\
 & \Rightarrow x\left( x+1000 \right)-750\left( x+1000 \right)=0 \\
 & \Rightarrow \left( x-750 \right)\left( x+1000 \right)=0 \\
\end{align}\]
Hence, x = 750 and x = -1000.
Since x was supposed to be usual speed and speed cannot be negative, therefore x will be equal to 750.

Hence, usual speed of plane is 750 km/hr.

Note: Students should not forget to convert minutes into hours. Unit should remain same for every term otherwise answer would be incorrect. While solving quadratic equation, students can also use quadratic formula given for equation $a{{x}^{2}}+bx+c=0$ as $\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
For this sum, a = 1, b = 250 and c = -750000. So,
\[\begin{align}
  & \Rightarrow x=\dfrac{-250\pm \sqrt{62500-4\left( -750000 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-250\pm \sqrt{3062500}}{2} \\
 & \Rightarrow x=\dfrac{-250\pm 1750}{2} \\
 & \Rightarrow x=\dfrac{-250+1750}{2}\text{ and }x=\dfrac{-250-1750}{2} \\
 & \Rightarrow x=\dfrac{1500}{2}\text{ and }x=\dfrac{-2000}{2} \\
 & \Rightarrow x=750\text{ and }x=-1000 \\
\end{align}\]