QUESTION

# In a flight of 6000km an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 400km/hour and time increased by 30 minutes. Find the original duration of the flight.

Hint: We will first write average speed equations of flight before and after it slowed down. We will then equate both the equation by substituting one of the unknown terms and solve it to find the answer.

Complete Step-by-Step solution:
Average speed of an object is the total distance covered by it to the total time taken to cover that distance. Therefore,
$\text{Average speed = }\dfrac{\text{Total distance covered}}{\text{Total time taken}}$.
Now, let us consider the average speed of the flight before it slowed down to be $x$ km/hour, and original time taken by flight to cover 6000km to be $t$ hours.
Therefore, from above equation,
$s=\dfrac{6000}{t}\cdots \cdots \left( i \right)$
Now, after the flight slowed down due to bad weather, time taken by the flight to cover the given distance got increased by 30 minutes, that is, $\dfrac{30}{60}=\dfrac{1}{2}=0.5$ hour. So, the time taken after slowing down will become $t+0.5$ hours. Also, speed is reduced by 400km/hour. So, the new speed will be $s-400$ km/hour. Here, distance travelled by the flight remains constant.
Therefore, speed equation after it slowed down will be,
$s-400=\dfrac{6000}{t+0.5}$
Cross multiplying above equation, we get,
$s=\dfrac{6000}{t+0.5}+400\cdots \cdots \left( ii \right)$
From equation $\left( i \right)$ and $\left( ii \right)$, we found the left hand side of both the equations to be equal. Hence, we can equate the right hand side of both the equations too.
Equating right hand side of equation $\left( i \right)$ and $\left( ii \right)$, we get,
$\dfrac{6000}{t}=\dfrac{6000}{t+0.5}+400$
Taking LCM on the right hand side above equation, we get,
$\Rightarrow \dfrac{6000}{t}=\dfrac{6000+400(t+0.5)}{t+0.5}$
Applying distributive law, we write,
$\Rightarrow \dfrac{6000}{t}=\dfrac{6000+400t+400\times 0.5}{t+0.5}$
$\Rightarrow \dfrac{6000}{t}=\dfrac{6000+400t+200}{t+0.5}$
$\Rightarrow \dfrac{6000}{t}=\dfrac{6200+400t}{t+0.5}$
Cross multiplying above equation, we get,
$\Rightarrow 6000(t+0.5)=t(6200+400t)$
Applying distributive law on both sides of the equation, we write,
$\Rightarrow 6000t+6000\times 0.5=6200t+400{{t}^{2}}$
$\Rightarrow 6000t+3000=6200t+400{{t}^{2}}$
$\Rightarrow 400{{t}^{2}}+6200t-6000t-3000=0$
$\Rightarrow 400{{t}^{2}}+200t-3000=0$
Taking 200 common, we get,
$\Rightarrow 200(2{{t}^{2}}+t-15)=0$
Diving 200 on sides of the equation, we get,
$\Rightarrow 2{{t}^{2}}+t-15=\dfrac{0}{200}=0\cdots \cdots \left( iii \right)$
Formula to solve quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Using this formula in equation $\left( iii \right)$, we get,
$t=\dfrac{-1\pm \sqrt{{{1}^{2}}-4\times 2\times \left( -15 \right)}}{2\times 2}$
$\Rightarrow t=\dfrac{-1\pm \sqrt{1+120}}{4}$
$\Rightarrow t=\dfrac{-1\pm \sqrt{121}}{4}$
Putting value $\sqrt{121}=11$, we get,
$\Rightarrow t=\dfrac{-1\pm 11}{4}$
Separating both + and – terms, we get,
$\Rightarrow t=\dfrac{-1+11}{4}$ or $t=\dfrac{-1-11}{4}$
$\Rightarrow t=\dfrac{10}{4}$ or $t=\dfrac{-12}{4}$
$\Rightarrow t=2.5$ or $t=-3$
Since, $t$ here is time, which cannot be a negative term.
Therefore, $t=2.5$ hours = 2 hours + 0.5 hour = 2 hours and 30 minutes.
Hence, the original duration of the flight is 2 hours and 30 minutes.

Note: In this question, we can also substitute $t$ in speed equations that we got before and after flight slowed down, and solve the equation for $s$. After finding $s$, we can put that value in equation $\left( i \right)$ to find $t$. Also, we can use any method for solving quadratic equations.